Download Selected Chapters from Number Theory and Algebra

Selected Chapters from Number Theory and Algebra
A project under construction
Franz Rothe
Department of Mathematics
University of North Carolina at Charlotte
Charlotte, NC 28223
[email protected]
December 28, 2012
Allnumalg\numalg.tex
Contents
I
Number Theory
3
1 Euclid’s Number Theory
1.1
The Euclidean algorithm . . . . .
1.2
How many owl-primes are there?
1.3
Prime numbers . . . . . . . . . .
1.4
Rational and irrational . . . . . .
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2 Fermat Primes
2.1
A short paragraph about Fermat numbers
2.2
More about Fermat numbers . . . . . . .
2.3
Pseudo random Number Generation . . .
2.4
Pseudo prime and Carmichael numbers . .
1
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4
4
7
9
12
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13
13
14
22
22
3 Arithmetic of the Complex Numbers
3.1 Arctan identities . . . . . . . . . . . . . .
3.2 Gaussian primes . . . . . . . . . . . . . . .
3.3 Sums of two squares . . . . . . . . . . . .
3.4 Roots . . . . . . . . . . . . . . . . . . . .
3.5 The solution of the reduced cubic equation
3.6 The square root of a complex number . . .
3.7 Pythagorean triples . . . . . . . . . . . . .
3.8 Roots of unity . . . . . . . . . . . . . . . .
II
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Algebra
26
26
27
31
36
39
43
44
47
55
1 Polynomials
1.1 Rational zeros of polynomials . . .
1.2 Gauss’ Lemma . . . . . . . . . . .
1.3 Eisenstein’s irreducibility criterium
1.4 Descartes’ Rule of Signs . . . . . .
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2 Tschebychev polynomials
55
55
56
60
63
67
3 Polynomials over the Complex Numbers
3.1 A proof of the Fundamental Theorem of Algebra
3.2 Meditation on Descartes’ rule of signs . . . . . .
3.3 Estimation of zeros with Rouché’s Theorem . .
3.4 Perron’s irreducibility criterium . . . . . . . . .
3.5 The limiting case . . . . . . . . . . . . . . . . .
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71
71
72
77
78
79
4 The Residue Theorem and its Consequences
82
4.1 The residue theorem and the logarithmic residue theorem . . . . . . . . . 82
4.2 Comparing two functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
2
Part I
Number Theory
3
1
Euclid’s Number Theory
1.1
The Euclidean algorithm
10 Problem 1.1. Calculate the greatest common divisor of 765 and 567. Find
integers s and t such that gcd (765, 567) = s · 765 − t · 567.
Answer. The extended Euclidean algorithms is used to calculate the greatest common
divisor. In an additional parallel calculation, one gets the greatest common divisor as
integer combination of the two given numbers.
row 0:
row 1:
row 2:
row 3:
row 4:
row 5:
row 5+1:
765 567 = 1 rem
567 : 198 = 2 rem
198 : 171 = 1 rem
171 : 27 = 6 rem
27 :
9 = 3 rem
198
171
27
9
0
1 0
0 1
1 1
2 3
3 4
20 27
63 85
Indeed, gcd (765, 567) = 9 = (−20) · 765 + 27 · 567. Hence s = −20 and t = −27.
Remark. The optional extra row 5 + 1 does not contain a division, only sM +1 and tM +1
are calculated. This is a convenient check, since 63 · 765 − 85 · 567 = 0.
10 Problem 1.2. Use the last problem to calculate the least common multiple
lcm (765, 567).
Answer.
lcm (765, 567) = 765 ·
567
= 765 · 63 = 48 195
gcd (765, 567)
10 Problem 1.3. Given that
x ≡ 19
mod 765
and x ≡ 1
mod 567
Find the smallest positive solution for x.
Answer. One can determine the unknown integer x modulo the least common multiple of
lcm (765, 567). To get a solution, one needs p and q such that x = 19+p·765 = 1+q·567.
Since 19−1
= 2 is an integer, the congruence is solvable. We need to multiply the result
9
of the problem above by this integer 2 and get
9 = (−20) · 765 + 27 · 567
18 = (−40) · 765 + 54 · 567
18 + 40 · 765 = 54 · 567
19 + 40 · 765 = 1 + 54 · 567
30 619 = 30 619
4
We get a solution x = 30 619, which turns out to be the smallest one, in this example.
1
10 Problem 1.4. Calculate the greatest common divisor of 367 and 47. Find
integers s and t such that gcd (367, 47) = s · 367 + t · 47.
Answer. The extended Euclidean algorithms is used to calculate the greatest common
divisor. In an additional parallel calculation, one gets the greatest common divisor as
integer combination of the two given numbers. Here is the example:
row 0:
row 1:
row 2:
row 3:
row 4:
row 5:
row 5+1:
367 : 47 = 7 rem
47 : 38 = 1 rem
38 : 9 = 4 rem
9:
2 = 4 rem
2:
1 = 2 rem
1
0
0
1
1
7
1
8
5 39
21 164
47 367
32
9
2
1
0
Indeed, gcd (367, 74) = 1 = (−21) · 367 + 164 · 47. The optional extra row 5 + 1 does
not contain a division, only sM +1 and tM +1 are calculated. This is a convenient check,
since 47 · 367 − 367 · 47 = 0.
10 Problem 1.5. Find integers s and t such that 22s + 8t = 2. Find the least
common multiple of 8 and 22.
Answer. One can get immediately gcd (22, 8) = 2. Hence the least common multiple of
22 and 8 is 88. The extended Euclidean algorithm gives the greatest common divisor as
linear combination:
row
row
row
row
0:
1:
2:
3:
22 : 8 = 2 rem
8 : 6 = 1 rem
6 : 2 = 3 rem
6
2
0
1
0
1
1
0
1
2
3
Indeed, gcd (22, 8) = 2 = (−1) · 22 + 3 · 8.
10 Problem 1.6. Given that
x≡3
mod 8
and x ≡ 5
mod 22
determine the unknown integer x modulo the least common multiple of 8 and 22.
Answer. One needs s and t such that 3 + 8s = 5 + 22t. This works for 3 + 8 · 3 =
5 + 22 · 1 = 27. Hence both
x ≡ 27 mod 8 and x ≡ 27 mod 22
what implies x ≡ 27 mod 88
1
One does not always get immediately the smallest solution. The values p and q are not needed any
more.
5
10 Problem 1.7. Calculate and prime factor the number 215 −1−(23 −1)(25 −1).
Answer. (2p − 1) is always a divisor of (2p·q − 1 for any p and q. The prime factoring is
32 767
15
3
5
− 1 = 7 · 31 · 150 = 2 · 3 · 52 · 7 · 31 = 32 550
2 − 1 − (2 − 1)(2 − 1) = 7 · 31 ·
7 · 31
Proposition 1.1 (Euclid’s Lemma). If a prime number divides the product of two
integers, the prime number divides at least one of the two integers.
Proof. Let p be the prime number, and the integers a and b. We assume that p divides
the product ab, but p does not divide a. We need to show that p divides b.
Because p does not divide a, we get gcd (a, p) = 1. By the extended Euclidean
algorithm, there exist integers s, t such that
1 = sa + tp
Hence
b
ab
= s + tb
p
p
Because p divides ab, the right hand side is an integer. Hence p divides b, as to be
shown.
10 Problem 1.8. Prove that a nonconstant integer polynomial assumes infinitely
many composite values P (n), for appropriately chosen natural numbers n.
Proposition 1.2. A nonconstant integer polynomial assumes infinitely many composite
values P (n), for appropriately chosen natural numbers n.
Proof. Let d ≥ 1 be the degree of the given polynomial. By the fundamental theorem
of algebra, the polynomial P can assume any value at most d times. 2 Hence there exist
at most 3d natural numbers n for which P (n) is either 0, −1 or 1. Let a be natural
number such that P (a) 6= 0, ±1. Let p be any prime factor of P (a) and put P (a) = pq.
It is easy to see that for all integer k, the difference P (a + kp) − P (a) is divisible by
p. Indeed, this follows by the binomial theorem. For any given polynomial with integer
coefficients cl
P (x) =
P (a + kp) − P (a) =
d
X
l=0
d
X
cl x l
cl (a + kp)l − al
l=0
=
d
X
l=0
" d
#
l l X
X X
l l−j
l
cl
a (kp)j = p ·
cl
al−j k j pj−1
j
j
j=1
j=1
l=0
P (a + kp) − P (a) = p · m
2
Indeed, we only need to know for this proof that any value can be assumed at most as many times
as the degree of the polynomial.
6
where m is an integer abbreviating the last double sum. Since P (a) = pq, we conclude
P (a + kp) = p(m + q) is divisible by p for all integer k.
There exist at most 3d integers k for which P (a + kp) is either 0, −p or p. Hence
there exist infinitely many integers k such that a + kp ≥ 0 and P (a + kp) 6= 0, ±p is
divisible by p and hence composite.
1.2
How many owl-primes are there?
The values of the polynomial n2 − n + 41 for n = 1, 2, 3, . . . 40 all turn out to be primes.
This curious observation goes back to Euler.
10 Problem 1.9. As a warm-up, find the three smallest primes larger than 41
which are not in the list of 40 primes just mentioned.
Answer. These are the first three primes not in the range: 59, 67, 73 6= n2 − n + 41 for
any natural number n.
We now address the question which other primes besides 41 have the curious property
noted by Euler.
Definition 1.1 (Owl-prime). I call a prime p an owl-prime 3 if the values of the
polynomial n2 − n + p for n = 1, 2, 3, . . . p − 1 all turn out to be primes.
10 Problem 1.10. As a warm-up, find the three smallest owl-primes.
Now we have the obvious question: are there further owl-primes between 5 and 41?
Proposition 1.3. For any prime p > 5 equivalent are
(a) For any natural number n, the values n2 − n + p are divisible neither by 2, 3 nor 5.
(b) p is congruent to either 11 or 17 modulo 30.
Remark. Note this proposition gives only a necessary condition for a prime being an
owl-prime.
Reason. Clearly n2 − n + p is odd since n2 − n is always even, and every prime p > 2 is
odd.
If we go through the natural numbers n = 1, 2, 3, . . . , the value of n2 − n is either
divisible by three, or it is congruent to 2 modulo 3. The first case occurs for n ≡ 0 or
n ≡ 1 modulo 3. The second case occurs for n ≡ 2 mod 3. To assure that n2 − n + p
is never divisible by 3 for any n, we need to have p ≡ 2 mod 3.
From the following small table, we get the possible values of n − n2 mod 5 and n − n2
mod 7:
3
I have chosen this name in honest of Euler
7
n
1
2
3
4
5
6
7
n − n2
0
−2
−6
−12
−20
−30
−42
n − n2 mod 5
0
3
4
3
0
0
3
n − n2 mod 7
0
5
1
2
1
5
0
We see 0, 3, 4 are the possible values of n − n2 mod 5. To avoid that n2 − n + p is
divisible by 5, we need that n − n2 and p are different modulo 5. Hence p has to be
congruent to either 1 or 2 modulo 5.
10 Problem 1.11. If a number p is odd, p ≡ 2 mod 3, and p ≡ 1 mod 5,
determine the number p modulo 30.
Similarly, assume p is odd, p ≡ 2 mod 3, and p ≡ 2 mod 5, and determine p
modulo 30.
Answer. In the first case, we get p ≡ 11 mod 30. In the second case, we get p ≡ 17
mod 30.
10 Problem 1.12. Find the next two owl-primes larger than 5.
10 Problem 1.13. What are the values possible for p mod 7 of an owl-prime
p.
Answer. We have already obtained the possible values of n − n2 mod 7 are 0, 1, 2, 5.
Let p > 7 be any prime. To avoid that n2 − n + p is divisible by 7, we need that n − n2
and p are different modulo 7. Hence p has to be congruent to either 3, 4 or 6 modulo 7.
From the last remark, we already see that 47 is not an owl-prime. Indeed, 7 divides
n − n + 47 for n = 2— since both 47 ≡ 5 mod 7 and 2 − 22 ≡ 5 mod 7.
2
Open Problem. Show that 2, 3, 5, 11, 17 and 41 are the only owl-primes, or find a
larger owl-prime.
8
1.3
Prime numbers
Proposition 1.4 (Euclidean Property). If a number c divides the product ab and
gcd (c, a) = 1, then c divides b.
Standard proof. By the extended Euclidean algorithm, there exist integers s, t such that
1 = sa + tc
b
ab
= s + tb
c
c
The second line results by multiplication of both sides with cb . Because c divides ab, the
right hand side is an integer. Hence c divides b, as to be shown.
Second proof. Both numbers a and c are divisors of both products ab and ac. Hence
both a and c are divisors of the greatest common divisor
G := gcd (ab, ac)
Hence the integer
(1.1)
q :=
ac
G
is a divisor of both a and c. Hence q is a divisor of gcd (a, c) = 1, which was assumed
to be one. Hence q = 1, and ac = G is a divisor of ab. This implies that c is a divisor
of b, as to be shown.
Definition 1.2 (prime number). A prime number is an integer p ≥ 2, which is
divisible only by 1 and itself.
Euclid—and many other mathematicians—have shown that there exist infinitely
many prime numbers. We put them into the increasing sequence
p1 = 2 , p 2 = 3 , p 3 = 5 , p 4 = 7 , . . .
It is rather easy to see that for every positive integer, there exists a decomposition into
prime factors. Let a and b be any positive integers. There exist sequence αi ≥ 0 and
βi ≥ 0, with index i = 1, 2, . . . and only finitely many terms nonzero such that
Y
Y β
(1.2)
a=
piαi , b =
pi i
i≥1
i≥1
The uniqueness of the prime decomposition turns out harder to prove. Astonishingly,
the proof depends on Euclid’s lemma, the proof of which in turn relies on the extended
Euclidean algorithm.
9
Proposition 1.5 (Euclid’s Lemma). If a prime number divides the product of two
integers, the prime number divides at least one of the two integers.
Reason. Let p be the prime number, and the integers be a and b. We assume that p
divides the product ab, but p does not divide a. We need to show that p divides b.
Because p does not divide a, the definition of a prime number implies gcd (a, p) = 1.
By the extended Euclidean algorithm, there exist integers s, t such that
1 = sa + tp
Hence
ab
b
= s + tb
p
p
Because p divides ab, the right hand side is an integer. Hence p divides b, as to be
shown.
Proposition 1.6 (Monotonicity). Let a and b have the prime decompositions
Y β
Y
pi i
(1.3)
a=
piαi , b =
i≥1
i≥1
The number b is a divisor of a if and only if βi ≤ αi for all i ≥ 1.
Reason. If βi ≤ αi for all i ≥ 1, then a = qb with
Y α −β
q=
pi i i
i≥1
and hence b is a divisor of a.
Conversely, assume that b is a divisor of a. We need to show that βi ≤ αi for all
i ≥ 1. Proceed by induction on b. If b = 1, then βi = 0 for all i ≥ 1, and the assertion
is true.
Here is the induction step ”b < n 7→ b = n”: Let pi be any prime factor of b ≥ 2,
which means that βi ≥ 1. Because pi divides b and b divides a, the prime pi divides a.
By Euclid’s Lemma Proposition 1.5, pi is a divisor of one of the primes pj occurring in
the prime decomposition of a. Hence αj ≥ 1. Because different primes cannot divide
each other, this implies i = j and pi = pj . Hence pbj < n is a divisor of paj . By the
induction assumption, this implies βi ≤ αi for all i 6= j, as well as βj − 1 ≤ αj − 1 and
hence βi ≤ αi for all i ≥ 1.
Proposition 1.7 (Uniqueness of prime decomposition). The prime decomposition
of any positive integer is unique.
Reason. Assume
a=
Y
piαi and a =
i≥1
Y
piβi
i≥1
Because a divides a, the fact given above both tells that βi ≤ αi and αi ≤ βi for all
i ≥ 1. Hence βi = αi for all i ≥ 1.
10
Proposition 1.8. Let a and b have the prime decompositions (1.3). The prime decompositions of the greatest common divisor and least common multiple are
Y min[α ,β ]
i i
(1.4)
gcd (a, b) =
pi
i≥1
lcm (a, b) =
(1.5)
Y
max[αi ,βi ]
pi
i≥1
10 Problem 1.14. Check these formulas for a = 1001, b = 4221.
Proof of Proposition 1.8. Let g be the righthand side of equation (1.5). We need to
check properties (i) and (ii) defining the greatest common divisor.
(i) g divides both a and b.
Check. This is clear, because both min[αi , βi ] ≤ αi and min[αi , βi ] ≤ βi for all
i ≥ 1.
(ii) If any positive integer h divides both a and b, then h divides the greatest common
divisor g.
Check. Let
(1.6)
h=
Y
piγi
i≥1
be the prime decomposition of h. Because h divides both a and b, monotonicity
implies that both γi ≤ αi and γi ≤ βi for all i ≥ 1. Hence γi ≤ min[αi , βi ] for all
i ≥ 1, which easily implies that h is a divisor of g.
Let l be the righthand side of equation (1.5). We need to check properties (i) and (ii)
defining the least common multiple.
(i) The number l is a multiple of both a and b.
Check. This is clear, because both max[αi , βi ] ≥ αi and max[αi , βi ] ≥ βi for all
i ≥ 1.
(ii) If any positive integer k is a multiple of both a and b, the integer k is a multiple of
the least common multiple l.
11
Check. Let
(1.7)
k=
Y
piγi
i≥1
be the prime decomposition of k. Because k is a multiple of both a and b, monotonicity implies that both γi ≥ αi and γi ≥ βi for all i ≥ 1. Hence γi ≥ max[αi , βi ]
for all i ≥ 1, which easily implies that k is a multiple of l.
1.4
Rational and irrational
Proposition 1.9. For any natural numbers r ≥ 1 and a ≥ 1, the r-th root
a rational number if it is even an integer.
√
r
a is only
Proof. The assertion is clear for r = 1 or a = 1, hence we may assume r ≥ 2 and a ≥ 2.
Now assume that
√
m
r
a=
n
r
n a = mr
with natural m, n. We show that any prime number p dividing n has to divide m, too.
Hence after cancelling common factors, we get n = 1, and hence the root is an integer.
Now assume that the prime p divides n. Hence pr divides nr , which in turn divides
r
an = mr . Hence pr divides mr . Hence, by Euclid’s Lemma p divides m, as claimed.
As already explained, the assertion follows.
Proposition 1.10. For any natural
numbers r ≥ 2 and any a ≥ 2, which is not the
√
r
r-th power of an integer, the root a is irrational. Especially, the roots of the primes
are all irrational.
√
Proof. If the root r a is rational, it is even an integer m, and hence a = mr is the r-th
power of m.
√
Take the contrapositive: If a is not the r-th power of an integer, the root r a is
irrational.
12
2
Fermat Primes
2.1
A short paragraph about Fermat numbers
Definition 2.1 (Fermat number, Fermat prime). The numbers
n
Fn = 22 + 1
for any integer n ≥ 0 are called Fermat numbers. A Fermat number which is prime is
called a Fermat prime.
Lemma 2.1 (Fermat). If p is any odd prime, and p − 1 is a power of two, then p = Fn
is a Fermat prime.
Proof. Assume p = 1 + 2a and a ≥ 3 odd. Because of the fatorization a = 2n · (2b + 1)
one can factor
n ·(2b+1)
p = 1 + 22
n
n
n ·2
= (1 + 22 )(1 − 22 + 22
n ·2b
− · · · + 22
)
If p is an odd prime, the only possibility is b = 0 and p = Fn .
The only known Fermat primes are Fn for n = 0, 1, 2, 3, 4. They are 3, 5, 17, 257, and
65537, see4 .
Fermat numbers and Fermat primes were first studied by Pierre de Fermat, who
conjectured 1654 in a letter to Blaise Pascal that all Fermat numbers are prime, but
told he had not been able to find a proof. Indeed, the first five Fermat numbers are
easily shown to be prime. However, Fermat’s conjecture was refuted by Leonhard Euler
in 1732 when he showed, as one of his first number theoretic discoveries:
5
F5 = 22 + 1 = 232 + 1 = 4294967297 = 641 · 6700417.
Later, Euler proved that every prime factor p of Fn must have the form p = i · 2n+1 + 1.
Almost hundred years later, Lucas showed that even p = j · 2n+2 + 1. These results
give a vague hope that one could factor Fermat numbers and perhaps settle Fermat’s
conjecture to the negative.
There are no other known Fermat primes Fn with n > 4. However, little is known
about Fermat numbers with large n. In fact, each of the following is an open problem:
1. Is Fn composite for all n > 4? By this anti-Fermat hypothesis 3, 5, 17, 257, and
65537 would be the only Fermat primes.
2. Are there infinitely many Fermat primes? (Eisenstein 1844)
3. Are there infinitely many composite Fermat numbers?
4
sequence A019434 in OEIS
13
4. Are all Fermat numbers square free?
As of 2012, the next twenty-eight Fermat numbers, F5 through F32 , are known to be
composite. As of February 2012, only F0 to F11 have been completely factored. For
complete information, see Fermat factoring status by Wilfrid Keller, on the internet at
http://ww.prothsearch.net/fermat.html
Main Theorem (Gauss-Wantzel Theorem). A regular polygon with n sides is constructible if and only if
n = 2h p1 · p2 · · · ps
where p2 · · · ps is a product of different Fermat primes.
If the anti-Fermat hypothesis is true, there are exactly five Fermat primes, and hence
exactly 31 regular constructible polygons with an odd number of sides.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
2.2
n
f actored
3
F0
5
F1
15
3·5
17
F2
51
3 · 17
85
5 · 17
255
3 · 5 · 17
257
F3
771
3 · 257
1 285
5 · 257
3 855
3 · 5 · 257
4 369
17 · 257
13 107
3 · 17 · 257
21 845
5 · 17 · 257
65 535 3 · 5 · 17 · 257
65 537
F4
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
n
f actored
196 611
3 · 65 537
327 685
5 · 65 537
983 055
3 · 5 · 65 537
1 114 129
17 · 65 537
3 342 387
3 · 17 · 65 537
5 570 645
5 · 17 · 65 537
16 711 935
3 · 5 · 17 · 65 537
16 843 009
257 · 65 537
50 529 027
3 · 257 · 65 537
84 215 045
5 · 257 · 65 537
252 645 135
3 · 5 · 257 · 65 537
286 331 153
17 · 257 · 65 537
858 993 459
3 · 17 · 257 · 65 537
1 431 655 765
5 · 17 · 257 · 65 537
4 294 967 295 3 · 5 · 17 · 257 · 65 537
More about Fermat numbers
10 Problem 2.1. Prove by induction that F0 · F1 · · · Fn−1 = Fn − 2 for all natural
numbers n ≥ 1.
Answer. Basic step: For n = 1, both sides of the formula are equal to 3, since
3 = F0 = F1 − 2 = 5 − 2.
14
Induction step ”n → n + 1”: The formula is assumed to hold for n as written. It is
shown for n replaced by n + 1, as follows:
F0 · F1 · · · Fn = [F0 · F1 · · · Fn−1 ] · Fn
by recursive definition of the product
= (Fn − 2)Fn
by the induction assumption
n
n
n
n+1
2
= 22 − 1 22 + 1 = 22
− 12 = 22 − 1 = Fn+1
We have checked the asserted formula step for n + 1. From the basic step and the
induction step together, we conclude by the principle of induction that the formula
holds for all n ≥ 1.
10 Problem 2.2. Use the last problem to conclude Goldbach’s Theorem, which
states that any two different Fermat numbers are relatively prime. Conclude there exist
infinitely many primes.
Answer. Assume that p is a common prime factor of the Fermat numbers Fk and Fn with
0 ≤ k < n. We could conclude that p is a divisor of F0 · F1 · · · Fn−1 = Fn − 2, and hence
of both Fn − 2 and of Fn . This is only possible for p = 2. But all Fermat numbers are
odd and cannot have divisor 2. This argument excludes that any two Fermat numbers
have a common prime factor.
For any natural number n, let pn be the smallest (or any) prime factor of Fn . The
primes pn are all different since the Fermat numbers are relatively prime. Hence there
exist infinitely many primes.
Corollary. In any arithmetic sequence 1 + i 2k with i = 1, 2, 3, . . . there exist infinitely
many primes.
Reason. Any prime factor of Fn has the form pn = 1+i 2n+2 . Hence all pn with n ≥ k −2
are contained in the arithmetic sequence 1 + i 2k with i = 1, 2, 3, . . . .
Proposition 2.1 (Euler 1770). Any prime factor of the Fermat number Fn has the
form pn = 1 + i 2n+1 .
Reason. Assume that p is a prime factor of the Fermat number Fn . Let h > 1 be the
smallest integer such that 2h ≡ 1 mod p. This power h is also called the order of 2
modulo p. The definition of Fn implies
n
22 ≡ −1
2n
2
≡ −1
n+1
≡1
mod Fn
2n+1
≡1
mod p
mod Fn and 22
mod p
and 2
Since h is a divisor of 2n+1 but not 2n , we get h = 2n+1 .
Fermat’s Little Theorem implies 2p−1 ≡ 1 mod p, and hence h is a divisor of p − 1.
Together we conclude that 2n+1 is a divisor of p − 1, as to be shown.
It took more than hundred years, until Édouard Lucas improved Euler’s result.
15
Proposition 2.2 (Lucas 1878). Any prime factor of the Fermat number Fn with n ≥ 2
has the form pn = 1 + j 2n+2 .
Reason. We continue the reasoning from Euler’s Proposition 2.1. Assume that p is
a prime factor of the Fermat number Fn and n ≥ 2. As a consequence of Euler’s
Proposition, we see that 8 is a divisor of p − 1. Hence one can calculate the Legendre
symbol
(p−1)(p+1)
2
8
= (−1)
=1
p
and one gets from Euler’s criterium
2
p−1
2
2
≡
=1
p
mod p
Hence the order h = 2n+1 of 2 modulo p is actually a divisor of (p − 1)/2. In the end
we conclude that 2n+2 is a divisor of p − 1, as to be shown.
Remark. It turns out occur many times that the Fermat number Fn has a prime factor
1 + j 2n+2 with j odd. This is known to happen for
n = 5, 6, 7, 9, 10, 11, 12, 15, 17, 18, 19, 21, 23, 25, 27, 29, 30, 31, 32
and many more cases with n ≥ 33. In this sense, Lucas’ result turns out to be optimal.
On the other hand, F8 has no prime factor with j odd.
Theorem 2.1 (Pépin’s test (1877)). For n ≥ 1, the Fermat number Fn is prime if
and only if
3(Fn −1)/2 ≡ −1 mod Fn
In 1905 and 1909, J.C. Morehead and A.E.Western used Pépins test to prove that
F7 and F8 are composite. As of 2001, no factor is known for the Fermat numbers Fn
with n = 14, 20, 22, 24. These numbers were proved composite only with Pépin’s test.
Here is a proof and some results related to this test.
Proposition 2.3 (Sufficiency of a Pépin-like test). Assume there exists a natural
number a such that
a(Fn −1)/2 ≡ −1 mod Fn
Then the Fermat number Fn is prime.
Proof. The assumption implies
(2.1)
a(Fn −1)/2 ≡ −1
mod Fn
and
16
aFn −1 ≡ 1
mod Fn
Let p be any prime factor of the Fermat number Fn . Let h > 1 be the smallest integer
such that ah ≡ 1 mod p. Since the congruences (2.1) hold modulo p, too, we see that
h divides Fn − 1 but not (Fn − 1)/2, and hence h = Fn − 1. The Little Fermat Theorem
implies
ap−1 ≡ 1 mod p
and hence h is a divisor of p − 1. We conclude
1 + h = Fn ≤ p ≤ Fn
Hence Fn = p is a prime, as to be shown.
Lemma 2.2. No matter whether Fn is prime or not, the Jacobi symbols are
a
Fn
=
= −1
a
Fn
for a = 3 and all n ≥ 1, as well as a = 5 and 7 for all n ≥ 2.
The case a = 3, n ≥ 1. We calculate the Legendre symbol
n
n
Fn
1 + 22 mod 3
1 + (−1)2
2
=
=
=
≡ 2(3−1)/2 ≡ −1
3
3
3
3
mod 3
From the quadratic reciprocity for the Jacobi symbols we get,—no matter whether Fn
is prime or not:
(Fn −1)(3−1)
Fn
Fn
3
4
=
=
(−1)
= −1
Fn
3
3
since Fn − 1 is divisible by 4.
The case a = 5, n ≥ 2. We calculate the Legendre symbol
!
! 2n−1
2n−1
Fn
1+4
mod 5
1 + (−1)
2
=
=
=
≡ 2(5−1)/2 ≡ −1
5
5
5
5
The case a = 7, n ≥ 2. The Legendre symbols obey the recurrence
!
! n−2
n−2
Fn
1 + 16 2
mod 7
1 + 22
mod 7
Fn−2
=
=
=
7
7
7
7
17
mod 5
for all n ≥ 2, which allows an induction starting with
17
3
=
≡ 3(7−1)/2 ≡ −1 mod 7
7
7
5
≡ 5(7−1)/2 ≡ −1 mod 7
7
From the quadratic reciprocity for the Jacobi symbols we get,—no matter whether Fn
is prime or not:
(Fn −1)(a−1)
Fn
a
Fn
4
=
=
(−1)
= −1
Fn
a
a
since Fn − 1 is divisible by 4 and a = 5, 7 are odd.
Proposition 2.4 (Necessity of a Pépin-like test). Assume the Fermat number Fn
is prime. Then
a(Fn −1)/2 ≡ −1 mod Fn
holds for a = 3 in the case n ≥ 1, and a = 5 and 7 in the case n ≥ 2.
Proof. Because of the assumption that Fn is prime, we can use Euler’s criterium and
get from Euler’s criterium
Fn −1
Fn
2
≡
a
= −1 mod Fn
a
as claimed.
Theorem 2.2 (Lucas-Lehmer). Any number m ≥ 2 is prime if and only if there exists
a primarity witness. A witness is a natural number a with the following two properties:
(i) am−1 ≡ 1 mod m;
(ii) a(m−1)/p 6≡ 1 mod m
for all prime divisors p of m − 1.
If m = Fn is a Fermat number, the only prime divisor of m − 1 is the number 2.
Thus we obtain sufficiency for the following Pépin-like test:
Proposition 2.5 (A more general Pépin-like test). Assume there exists a natural
number a such that
aFn −1 ≡ 1 mod Fn
but
Then the Fermat number Fn is prime.
18
a(Fn −1)/2 6≡ 1 mod Fn
Proof of the Lucas-Lehmer Theorem. The Theorem is true for m = 2 because assumption (i) holds for a = 1, and assumption (ii) is an empty truth. Assume that a is a
witness for m ≥ 3 and let h > 1 be the smallest integer such that ah ≡ 1 mod m.
Assumption (i) yields that h is a divisor of m − 1. By assumption (ii), we conclude
h = m − 1.
Moreover, the witness a is relatively prime to m. The Euler-Fermat Theorem tells
that
aφ(m) ≡ 1 mod m
where φ(m) is the Euler totient function. Hence the order h is a divisor of φ(m). The
inequalities
m − 1 = h ≤ φ(m) ≤ m − 1
imply φ(m) = m − 1. Hence m is a prime number.
Proth’s Theorem is a slight generalization of Pépin’s test.
Theorem 2.3 (Proth’s Theorem 1878). To test whether the number N is prime,
one chooses a base a relatively prime to N for which the Jacobi symbol is
a
= −1
N
Sufficient condition for primarity: Assume additionally that N = k · 2m + 1 with
k ≤ 1 + 2m . If
(2.2)
a(N −1)/2 ≡ −1
mod N
then N is prime.
Necessary condition for primarity: No restriction on k needs to be assumed. If
the above congruence (2.2) does not hold, then the number N is composite.
Reason for the sufficient condition: Assume that p is a prime factor of the given number N . Let h > 1 be the smallest integer such that ah ≡ 1 mod p. The assumed
congruence (2.2) implies that h is a divisor of N − 1 = k · 2m but not (N − 1)/2. The
Little Fermat Theorem implies that h is a divisor of p − 1. Hence we get the inequalities
2m ≤ h ≤ p − 1
The assumption k ≤ 1 + 2m implies N ≤ 22m + 2m + 1 < (1 + 2m )2 . Hence
√
N < 1 + 2m ≤ p ≤ N
holds for each prime divisor of N . Since any composite number has a prime divisor less
or equal its square root, this is only possible if N is a prime number.
19
The necessary condition . —is a direct consequence of Euler’s Theorem about the Legendre symbol.
Here are some further less important remarks about Fermat numbers.
Lemma 2.3. For any Fermat number Fn 6= 3, 5
2(Fn −1)/2 ≡ 1
mod Fn
Proof. By definition
n
22 ≡ −1 mod Fn
n
22·2 ≡ 1 mod Fn
and
For all n ≥ 2 we know that n + 1 ≤ −1 + 2n and hence
n
2n+1 = 2 · 2n ≤ 2−1+2 = (Fn − 1)/2
n
22·2 ≡ 2(Fn −1)/2 ≡ 1
mod Fn
Lemma 2.4. For any Fermat number Fn 6= 3, 5, 17
(F −1)/2
Fn−1n
≡1
mod Fn
Proof. Just calculate:
n−1
2
Fn−1
= (1 + 22
n−1
)2 = 1 + 2 · 22
(F −1)/2
Fn−1n
n
n−1
+ 22 ≡ 21+2
(1+2n−1 )(Fn −1)/4
≡ 2
mod Fn
mod Fn
Since n ≥ 3, we know that n ≤ n − 4 + 2n and hence
n
2 · 2n = 2n+1 ≤ (1 + 2n−1 )2−2+2 = (1 + 2n−1 )(Fn − 1)/4
hence
n
22·2 ≡ 1
(1+2n−1 )(F
2
n −1)/4
(F −1)/2
≡ Fn−1n
≡1
mod Fn
implies
mod Fn
Lemma 2.5. Any Fermat number Fn is a pseudo prime for the base 2 as well as the
base Fn−1 .
In 1964, Rotkiewicz showed that the product of any number of prime or composite
Fermat numbers will be a Fermat pseudo prime to the base 2.
20
Lemma 2.6. Let n > m ≥ 2. No matter whether the Fermat numbers are prime or
not, the Jacobi symbols are
Fm
Fn
=
=1
Fm
Fn
Proof.
n
m
Fn = 1 + 22 = 1 + 22
2n−m
n−m
≡ 1 + (−1)2
≡2
mod Fm
for n > m. We calculate the Jacobi symbol
(Fm −1)(Fm +1)
Fn
2
8
=1
=
= (−1)
Fm
Fm
since Fm − 1 is divisible by 8 for m ≥ 2. From the quadratic reciprocity for the Jacobi
symbols we get,—no matter whether Fn is prime or not:
(Fn −1)(Fm −1)
Fn
Fm
Fn
4
=
(−1)
=1
=
Fn
Fm
Fm
again since Fm − 1 is divisible by 8 for m ≥ 2.
Lemma 2.7. The number 2 is not a primitive root of any Fermat prime except 3. If
Fn 6= 3, 5, 17 is a Fermat prime, none of the Fermat numbers 17 ≤ Fm < Fn with
2 ≤ m < n is a primitive roots modulo Fn .
Lemma 2.8. Any Fermat prime Fn has (Fn − 1)/2 primitive roots. Equivalent are:
1. r is a primitive root of Fn .
2. r is a quadratic non-residue of Fn .
3. r(Fn −1)/2 ≡ −1 mod Fn
The primitive roots of any Fermat prime Fn 6= 3 are are
r ≡ 3 · 9j
with j = 1, 2, . . . , (Fn − 1)/2.
21
mod Fn
2.3
Pseudo random Number Generation
Very large Fermat primes are of particular interest in data encryption. (see Linear
congruential generator, RANDU)
Fermat primes are particularly useful in generating pseudo-random sequences of
numbers in the range 1 . . . N , where N is a power of 2. The most common method used
is to take any seed value between 1 and P −1, where P is a Fermat prime. Now multiply
this by a number A, which is greater than the square root of P and is a primitive root
modulo P (i.e., it is not a quadratic residue). Then take the result modulo P . The
result is the new value for the random number generator.
Vj+1 = (A × Vj ) mod P
This procedure is useful in computer science since most data structures have members
with 2X possible values. For example, a byte has 256 = 28 possible values 0 . . . 255.
Therefore to fill a byte (or any number of bytes, indeed) with random values, a random
number generator which produces values 1 to 256 can be used, subtracting 1 from each
output. This method produces only pseudo random values as, after P − 1 repetitions,
the sequence repeats. A poorly chosen multiplier A which is not a primitive root results
in the sequence repeating sooner than P − 1.
2.4
Pseudo prime and Carmichael numbers
Definition 2.2 (Pseudo prime and Carmichael number). Any number m ≥ 2 for
which
(2.3)
am−1 ≡ 1
mod m
is called a pseudo prime of base a ≥ 2.
Any number m ≥ 2 such that equation (2.3) holds for all a relatively prime to m is
called a Carmichael number.
The Carmichael numbers are those which cannot be proved to be composite with the
help of the Little Fermat Theorem—if one agrees to use only basis a relatively prime to
m.
Lemma 2.9. Let a be any positive or negative integer. If k divides l, then ak − 1 divides
al − 1. Moreover
gcd [ak − 1, al − 1] = agcd (k,l) − 1
for any k and l.
Proof. To check the first part, assume that l = ks. The geometric series with quotient
q := ak yields
al − 1 = q s − 1 = (q − 1)(1 + q + q 2 + · · · + q s−1 )
22
Hence al − 1 is divisible by ak − 1 = q − 1.
We see from this first step that agcd (k,l) − 1 is a divisor of gcd [ak − 1, al − 1]. The
converse divisibility can be checked via the Euclidean algorithm. By the extended
Euclidean algorithm, there exist natural numbers s and t such that sk − tl = gcd (k, l)
(possibly after switching k with l). Assume that d divides both ak − 1 and al − 1. By
the first part of the Lemma, the number d divides both ask − 1 and atl − 1, and hence
their difference
(ask − 1) − (atl − 1) = (ask−tl − 1)atl = (agcd (k,l) − 1)atl
The base a and atl are relatively prime to ak − 1 and hence to d. Hence d divides
agcd (k,l) − 1. Since this reasoning applies for any common divisor of ak − 1 and al − 1,
we conclude that gcd [ak − 1, al − 1] is a divisor of agcd (k,l) − 1.
Proposition 2.6. A number m is a pseudo prime with base a if and only if it has the
following property:
(*) if the prime power ps divides m, then ps divides agcd (m−1,p−1) − 1.
Lemma 2.10. Especially, if a number m is a pseudo prime with base a and the prime
power ps divides m, then ps divides ap−1 − 1.
Sufficiency of (*). Assume that the base a satisfies the assumption (*). Let ps be any
prime power dividing m and put g := gcd (p − 1, m − 1). Since g divides m − 1, we
conclude by the Lemma 2.9 that ag − 1 divides am−1 − 1. The assumption (*) tells that
ps divides ag − 1 and hence am−1 − 1 by the Lemma 2.9.
The simultaneous congruences am−1 ≡ 1 mod ps for all prime power divisors ps of
m together imply am−1 ≡ 1 mod m.
Necessity of (*). We assume that m is a pseudo prime with base a and thus am−1 ≡ 1
mod m. To check property (*), let ps be any prime power dividing m. This assumption
s
implies am−1 − 1 ≡ 0 mod ps , too. The Euler-Fermat Theorem yields aφ(p ) − 1 =
s−1
ap (p−1) − 1 ≡ 0 mod ps . Hence
agcd (m−1,p
s−1 (p−1))
≡1
mod ps
Since p is a divisor of m, it is not a divisor of m − 1. Thus m − 1 and ps−1 are relatively
prime and hence
gcd [m − 1, ps−1 (p − 1)] = gcd (m − 1, p − 1)
Hence
agcd (m−1,p−1) ≡ 1
mod ps
for all prime powers ps dividing m, as to be shown.
Lemma 2.11. A Carmichael number cannot be divisible neither by any odd prime square
nor by 4.
23
Proof. Assume that m is a Carmichael number and divisible by the prime power ps . By
the second part of Proposition 2.11, we know that ps divides ap−1 − 1. We now show
that it is impossible that any prime square divides m.
Excluding the case p ≥ 3 is odd and s ≥ 2: In the case m that is divisible by an
odd prime square, we choose a to be a primitive root modulo p2 , which we know
to exist since p is odd. In that case p2 is not a divisor of ap−1 − 1, contradicting
that ps divides ap−1 − 1.
Excluding the special case p = 2 and s ≥ 2: We need to exclude that m is divisible by 4. We choose a = 3. Now 4 is not a divisor of 2 = ap−1 − 1, contradicting
that ps divides ap−1 − 1.
Proposition 2.7. Assume that the number m has the property
(i) if the prime p divides m, then p − 1 divides m − 1;
Assume furthermore that
(a) the base a is relatively prime to m;
(b) if the prime power ps divides m, then ps divides ap−1 − 1.
Then m is a pseudo-prime of base a.
Proof. Let ps be any prime power divisor of m. By assumption (b), ps divides ap−1 − 1.
By assumption (i), p − 1 divides m − 1 and hence the Lemma 2.9 implies that ap−1 − 1
divides am−1 − 1. Together we see that ps divides am−1 − 1.
The simultaneous congruences am−1 ≡ 1 mod ps for all prime power divisors ps of
m together imply am−1 ≡ 1 mod m.
Proposition 2.8. Assume assumption (i) from Proposition 2.7 holds for the composite
number
Y
m=
psi i with different primes pi .
The number of bases a for which the m is a pseudo prime is equal to
Y
−1 +
(pi − 1)
A power of two is never a pseudo prime.
Proof. For all odd prime factors p, we choose a primitive root r modulo ps . Choose any
integers 0 ≤ t < p − 1, not all zero. By the Chinese Remainder Theorem, the system of
simultaneous congruences
s−1
a ≡ rt·p
mod ps
24
for all prime factors p has a solution, unique modulo m. It is now easy to check by
means of the Euler-Fermat Theorem that a satisfies the assumptions (a) and (b) of
Proposition 2.7. Hence a is a base for the pseudo prime m. The procedure exhausts all
possible choices of the bases a.
Proposition 2.9. A number m is a Carmichael number if and only if it has the following
two properties:
(i) if the prime p divides m, then p − 1 divides m − 1;
(ii) the number m is square free.
Necessity. We assume that m is a Carmichael number and thus am−1 ≡ 1 mod m for
all a relatively prime to m. To check property (i), let p ≥ 3 be any odd prime factor of
m. We choose a to be a primitive root modulo p. By the Lemma 2.9
gcd [am−1 − 1, ap−1 − 1] = agcd (m−1,p−1) − 1
By assumption am−1 − 1 ≡ 0 mod p and the Little Fermat Theorem yields ap−1 − 1 ≡ 0
mod p. Hence
agcd (m−1,p−1) ≡ 1 mod p
Since a is a primitive root, this implies gcd (m − 1, p − 1) = p − 1. Hence p − 1 is a
divisor of m − 1, confirming item (i).
The property (ii) has been check above already.
Sufficiency. Let a be any base relatively prime to m and p be a prime divisor of m. By
the Little Fermat Theorem a prime p divides ap−1 − 1. Since we have assume that p − 1
divides m − 1, the Lemma 2.9 yields that ap−1 − 1 divides am−1 − 1.
Since m is assumed to be square free, the simultaneous congruences am−1 ≡ 1 mod p
for all prime divisors p of m together imply am−1 ≡ 1 mod m, as claimed.
Much more has been found out about Carmichael numbers. There exist only finitely
Carmichael numbers with three prime factors which can be constructed. The remaining
ones have at least four prime factors. Alford et. al. (1994) have proved that the number
C(n) of Carmichael numbers less than n has the asymptotics C(n) ∼ n2/7 for large n.
Proposition 2.10. The Carmichael numbers with three prime factors are
(6k + 1)(12k + 1)(18k + 1)
were k has to be chosen such that all three factors are primes.
25
3
Arithmetic of the Complex Numbers
Definition 3.1 (Gaussian integer ). A complex number with integer real- and imaginary part is called a Gaussian integer.
3.1
Arctan identities
What is remarkable about the following products?
(a) (2 + i) · (3 + i) = 5 + 5i
(b) (2 + i)2 · (7 − i) = 25 + 25i
(c) (3 + i)2 · (7 + i) = 50 + 50i
(d) (5 + i)4 · (239 − i) = 114244 + 114244i
Each time, the real- and imaginary parts turn out to be equal. Do there exists more
examples with this special property?
For any positive x > 0, the principal argument of the complex number x + i is
Arg (x + i) = arctan x1 . Too, Arg (1 + i) = arctan 1 = π4 . Taking the arguments of the
formulas (a) through (d), one gets:
(a)
(b)
(c)
(d)
1
1
+ arctan
2
3
1
1
2 arctan − arctan
2
7
1
1
2 arctan + arctan
3
7
1
1
4 arctan + arctan
5
239
arctan
π
4
π
=
4
π
=
4
π
=
4
=
Of course, left- and right-hand side of these formulas could still differ by an integer
multiple of 2π. But it is clear that this cannot happen in the given four examples. Do
there exist more similar formulas?
Open Problem (May be difficult). How many solutions has the equation
(3.1)
c2 + 1 = 2N 2
with natural numbers N and c?
Open Problem (May be difficult). Are there more than four solutions to the equation
(3.2)
M
Y
(c2k + 1) = 2N 2
k=1
with natural numbers N and M ≥ 2, and ck ≥ 2 for k = 1, . . . , M ?
26
3.2
Gaussian primes
It is straightforward to see that the Gaussian integers are a ring. The units in any ring
are defined as its invertible elements. Clearly there are just four units 1, i, −1, −i in
the ring of Gaussian integers. Furthermore, division with remainder is possible. Hence
the Euclidean algorithm works in the Gaussian integers. Consequently, any Gaussian
integer can be decomposed uniquely into a product of irreducible elements—unique up
multiplication by the four units. Hence the Gaussian integers are a unique factorization domain,
abbreviated UFD. The irreducible elements are called
Theorem 3.1 (Gaussian primes). The Gaussian primes are:
(a) the number 1 + i.
(b) the usual primes p = 3, 7, 11, 19, . . . which are p ≡ 3 mod 4.
(c) all pairs p + iq, p − iq where p2 + q 2 = r is a prime r ≡ 1 mod 4.
Indeed, for every usual prime r = 5, 13, 17, 29, . . . which is r ≡ 1 mod 4 there exists a
unique decomposition into a sum of two integer square.
The decomposition r = p2 + q 2 becomes unique by the additional requirement that the
(negative or positive) integer p ≡ 1 mod 4 and q is even and positive.
Here is a table with the smallest examples. Too, I list the squares a + ib = (p + iq)2 .
r
p
q
a
b
5
1 2 −3
4
13
−3 2
5 −12
17
1 4 −15
8
29
5 2
21
20
37
1 6 −35
12
41
5 4
9
40
53
−7 2
45 −28
61
5 6 −11
60
73
−3 8 −53 −48
89
5 8 −39
80
97
9 4
65
72
101
1 10 −99
20
109 −3 10 −91 −60
113 −7 8 −15 −58
137 −11 4 105 −88
At that point, my heating system had been repaired, and I stopped.
10 Problem 3.1. Prove that there are infinitely many primes p ≡ 3 mod 4.
27
Answer. Let p1 , . . . pN be any primes congruent to 3 modulo 4. The number
P := −1 + 4
N
Y
pν
ν=1
has the following properties:
(a) P is odd.
(b) P ≡ 3 mod 4.
(c) P is not divisible by pν for any ν = 1 . . . N .
By (a) and (b), P cannot be the product of primes which are all congruent to 1 modulo
4. Hence there exists a prime p∗ congruent to 3 modulo 4 dividing P . By item (c), this
prime p∗ is different from p1 , . . . pN . Hence there exist at least N + 1 different primes
equivalent to 3 modulo 4. Since we know that such primes exist, there do exist infinitely
many.
10 Problem 3.2. Prove that there are infinitely many primes r ≡ 1 mod 4.
Answer. Let r1 , . . . rN be any primes congruent to 1 modulo 4. Define the number
P := 1 + 4
N
Y
rν2
ν=1
Indeed P has in the Gaussian integers the factoring
#
# "
"
N
N
Y
Y
rν
P = 1 + 2i
rν · 1 − 2i
ν=1
ν=1
Neither of the two factor is divisible by 1 + i nor any prime p ≡ −1 mod 4. Hence we
get the following properties:
(a) P is odd.
(b) P ≡ 1 mod 4.
(c) P is not divisible by rν for any ν = 1 . . . N .
(d) P is not divisible by any prime p ≡ 3 mod 4.
By (a) and (d), P is the product of primes congruent to 1 modulo 4 only. By item (c),
all these factors are different from r1 , . . . rN . Hence there exist at least N + 1 different
primes equivalent to 1 modulo 4. Since we know that such primes exist, there do exist
infinitely many.
28
10 Problem 3.3 (The little forgotten theorem). Assume that
√
iπk
(3.3)
z = a + ib = a2 + b2 e l
with real integers a, b, k, l such that a and b are relatively prime, and k and l are relatively
prime. How many different nonzero values can z 6= 0 assume? Prove that there are non
besides the obvious eight ones.
Answer. The complex number z can only assume the eight values 1, i, −1, −i and ±1±i,
but no other values.
Because of the assumption gcd (a, b) = 1, the extended Euclidean algorithm yields
integers s, t such that sa − tb = 1, and hence (a + ib)(s + it) = 1 + i(at + bs). Hence no
real prime can divide a + ib.
We now repeatedly use the fact that the Gaussian integers are a unique factorization
domain. Assume the Gaussian prime p + iq divides a + ib. Ruling out all other cases,
we show the only possibility is p + iq = 1 + i.
Because of the remark above, the case of a real prime p ≡ 3 mod 4, and q = 0, is
impossible. Hence we are left with the task to rule out the case of a Gaussian prime
p + iq with 5 ≤ p2 + q 2 ≡ 1 mod 4. In the 2l-th power of the given equation (3.3),
(a + ib)2l = (a2 + b2 )l
(3.4)
the two sides sides are both Gaussian integers as well as real, hence real positive integers.
Equation (3.4) implies that p + iq divides the natural number (a2 + b2 )l . Hence p − iq
divides this same number, too. Since the Gaussian prime p − iq divides (a + ib)2l , unique
factorization implies that p − iq divides a + ib, too. Since 5 ≤ p2 + q 2 ≡ 1 mod 4. the
two numbers p + iq and p − iq are relatively prime. Hence we conclude there product
(p + iq)(p − iq) = p2 + q 2 divides a + ib, too. But we have already shown that no real
prime divides a + ib. Hence the only possibilities left are
a + ib = iu (1 + i)s with u = 0, 1, 2, 3 and s = 0, 1.
Theorem 3.2 (The forgotten little theorem). A Gaussian integer does only have
an argument iπk
with integer k, l—an argument rational in degree measurement— if it
l
lies on the x-axis or on the y-axis, or on one of the two lines of slope ±1 through the
origin.
Proof. Assume that
(3.5)
z = a + ib =
√
a2 + b 2 e
iπk
l
with real integers a, b, k, l such that k and l are relatively prime. Let d := gcd (a, b) and
define a0 := ad and b0 := db . By the last problem, a0 + ib0 = iu (1 + i)s with u = 0, 1, 2, 3
and s = 0, 1. Multiplying both sides by d yields
a + ib = iu (1 + i)s d
with u = 0, 1, 2, 3 and s = 0, 1, and d any natural number. These are just the Gaussian
integers on the four lines Z, i · Z, (1 + i) · Z, and (1 − i) · Z, as to be shown.
29
Corollary. Let a + ib be a Gaussian integer with a 6= 0, b 6= 0.
Both real- and imaginary parts of (a + ib)l are nonzero if either l odd or |a| =
6 |b|. If
l > 0 is even, the following happens:
l
Re (a + ib)l
Im (a + ib)l
l ≡ 2 mod 4 Re (a + ib)l = 0 ⇔ |a| = |b|
Im (a + ib)l 6= 0
l
l ≡ 0 mod 4
Re (a + ib) 6= 0
Im (a + ib)l = 0 ⇔ |a| = |b|
10 Problem 3.4. Give the reason for Corollary 3.2.
Proof of Corollary 3.2. For any nonzero Gaussian integer a + ib, and integer l > 0, the
equation
p
(3.6)
(a + ib)l = iu (a2 + b2 )l
is equivalent to Re (a + ib)l = 0 for u odd, but equivalent to Im (a + ib)l = 0 for u even.
In all cases, the equation
(a + ib)4l = (a2 + b2 )2l
(3.7)
follows. Hence, as explained in the solution of Problem 3.3, we conclude that either
a = 0, or b = 0, or |a| = |b|. Since we have excluded the first two possibilities, we
conclude that a = ±b and hence equation 3.6 implies al (1 ± i)l = iu 2l/2 |a|l . Hence
l = 2m is even and (±i)m = iu . Hence either l ≡ 2 mod 4, and m, u both odd, and
Re (a + ib)l = 0— or l ≡ 0 mod 4, and m, u both even, and Im (a + ib)l = 0.
10 Problem 3.5.
(a + ib)l+2
a2 − b 2
(a + ib)l+4
Sl (a, b) := Im
2ab(a2 − b2 )
Rl (a, b) := Re
Calculate R0 , S0 , R4 , S4 . Prove that for all l ≡ 0 mod 4, Rl and Sl are integer homogenous and symmetric polynomials in a2 and b2 .
Answer.
(a + ib)2
a2 − b 2
=
=1
a2 − b 2
a2 − b 2
(a + ib)6
a6 − 15a4 b2 + 15a2 b4 − b6
R4 (a, b) = Re 2
=
= a4 − 14a2 b2 + b4
2
2
2
a −b
a −b
4a3 b − 4ab3
(a + ib)4
S0 (a, b) = Im
=
=2
2ab(a2 − b2 )
2ab(a2 − b2 )
(a + ib)8
8a7 b − 56a5 b3 + 56a3 b5 − 8ab7
S4 (a, b) = Im
=
= 4a4 − 24a2 b2 + 4b4
2
2
2
2
2ab(a − b )
2ab(a − b )
R0 (a, b) = Re
30
After some pain with the binomial formulas, one gets the general case with l = 4m:
m X
+m
X
s 4m + 2
a2m+2t b2m−2t
R4m (a, b) =
(−1)
2s
s=0 t=−m
m X
+m
X
(−1)s 4m + 4 2m+2t 2m−2t
S4m (a, b) =
a
b
2
2s
+
1
s=0 t=−m
Since the binomial coefficients even
are even, these are integer polynomials—the other
odd
assertions are easy to check, too.
10 Problem 3.6 (Calculation to the bones). Calculate R4m (1, 1) and S4m (1, 1),
confirming they are nonzero.
Answer.
2
2
Rl + (a + b )2abSl−4
Re (a + ib)l+2 + (a2 + b2 )Im (a + ib)l
=
a2 − b 2
l+2
Re (a + ib)
− Re i(a − ib)(a + ib)l+1
=
a2 − b 2
[a + ib − i(a − ib)](a + ib)l+1
= Re
a2 − b 2
(1 − i)(a + ib)l+1
= Re
a+b
With a = b = 1 and l = 4m ≥ 4, we get
R4m (1, 1) + 4S4m−4 (1, 1) = Re (1 − i)(1 + i)4m+1 /2 = (−4)m
With a = 1 + ε = z and b = −1, we get in the limit z → 1— using l’Hôpital’s rule
(1 + ε − i)4m+1
ε→0
ε
d (z − i)4m+1
= Re (1 − i)
dz
= Re (4m + 1)(1 − i)4m+1 = (4m + 1)(−4)m
R4m (1, −1) − 4S4m−4 (1, −1) = lim Re (1 − i)
Both formulas together imply S4m−4 = 2m(−4)m−1 . Actually, we check that for all
m≥0
R4m = (2m + 1)(−4)m , S4m = (2m + 2)(−4)m
3.3
Sums of two squares
From the decomposition of a Gaussian integer into Gaussian primes, one can quite easily
see how many solutions the integer equation m = a2 +b2 has for any given natural number
m.
31
10 Problem 3.7. We begin with some examples. For this illustration, I count
only the solutions of m = a2 + b2 with integers 0 ≤ a ≤ b, but give these solutions
completely. Give a list for m = 1 through 30.
m
list of a2 + b2
16
02 + 42
17
12 + 42
18
32 + 32
19
no solution
20
22 + 42
21
no solution
22
no solution
23
no solution
24
no solution
2
25 0 + 52 = 32 + 42
26
12 + 52
27
no solution
28
no solution
29
22 + 52
30
no solution
m list of a2 + b2
1
02 + 12
2
12 + 12
3
no solution
4
02 + 22
5
12 + 22
6
no solution
7
no solution
Answer.
8
22 + 22
9
02 + 32
10
12 + 32
11
no solution
12
no solution
13
22 + 32
14
02 + 22
15
12 + 22
It is still hard to find a pattern. Since the square of the absolute value of a complex
number is the sum of the squares of its real- and imaginary parts, we can use complex
arithmetic. Here are some further examples for illustration.
(a) |1 + 2i|2 = 12 + 22 = 5
(b) |3 + 2i|2 = 32 + 22 = 13
(c) The square of the absolute value of
(1 + 2i) · (−3 + 2i) = −7 − 4i,
and (1 + 2i) · (−3 − 2i) = 1 − 8i
implies 5 · 13 = 72 + 42 = 12 + 82 = 65.
(d) Squaring part (c) yields
(1 + 2i)2 · (−3 + 2i)2 = (−3 + 4i)(5 − 12i) = 33 + 56i
(1 + 2i)2 · (−3 − 2i)2 = (−3 + 4i)(5 + 12i) = −63 − 16i
The square of the absolute value yields
52 · 132 = 332 + 562 = 632 + 162 = 652
Are there more ways to solve a2 + b2 = 652 ? Indeed, there are still two further
solutions obtained as absolute squares of 13(−3 + 4i) = −39 + 52i and 5(5 − 12i) =
25 − 60i. Last not least, 652 is a perfect square.
392 + 522 = 252 + 602 = 02 + 652 = 652
32
(e) The absolute square of (1 + 2i)(1 + i) = −1 + 3i yields 10 = 12 + 32 .
10 Problem 3.8. Prove that a natural number is a perfect square if and only if
it has an odd number of divisors.
√
√
Answer. If d ∈ [1, m) is a divisor of m ∈ N, then md ∈ ( m, m]√is different one. In
that way, all divisors of m appear in pairs. The only exception is m in the case that
m is a perfect square.
Proposition 3.1. The integer equation m = a2 + b2 is solvable for a given natural
number m with integers a, b if and only if, in the prime decomposition of m, all primes
p ≡ 3 mod 4 appear with even multiplicity. In other words, m can be factored as
m = 2s g 2 c
where g has only prime factors congruent 3 mod 4, and
(3.8)
c=
N
Y
rνsν
ν=1
has only prime factors congruent 1 mod 4.
Theorem 3.3. The number of different ways one can decompose m = 2s g 2 c into squares
of integers a, b such that m = a2 + b2 is 4τ (c). Here
τ (c) =
N
Y
(1 + sν )
ν=1
is the number of divisors of c. The numbers a + ib from the solutions are exactly the
Gaussian integers
(3.9)
u
s
a + ib := i (1 + i) g
N
Y
(pν + iqν )σν (pν − iqν )τν
ν=1
All possible choices are obtained with u = 0, 1, 2, 3 and σν , τν ≥ 0 with σν + τν = sν for
ν = 1 . . . N.
For example m = 652 is decomposed as 652 = 52 · 132 and has 9 divisors. The possible
33
values of a + ib as solution of (3.9)—with unit factor iu = 1—are
(1 + 2i)2 (−3 + 2i)2
= (−3 + 4i)(5 − 12i) = 33 + 56i ,
2
(1 + 2i) (−3 + 2i)(−3 − 2i)
= (−3 + 4i) · 13
= −39 + 52i ,
2
2
(1 + 2i) (−3 − 2i)
= (−3 + 4i)(5 + 12i) = −63 − 16i ,
2
(1 + 2i)(1 − 2i)(−3 + 2i)
= 5 · (5 − 12i)
= 25 − 60i ,
(1 + 2i)(1 − 2i)(−3 + 2i)(−3 − 2i) = 5 · 13
= 65 ,
2
(1 + 2i)(1 − 2i)(−3 − 2i)
= 5 · (5 + 12i)
= 25 + 60i ,
2
2
(1 − 2i) (−3 + 2i)
= (−3 − 4i)(5 − 12i) = −63 + 16i ,
2
(1 − 2i) (−3 + 2i)(−3 − 2i)
= (−3 − 4i) · 13
= −39 − 52i ,
2
2
(1 − 2i) (−3 − 2i)
= (−3 − 4i)(5 + 12i) = 33 − 56i ,
As one sees there are still pairs of conjugate complex solutions, and one real solution
since 652 is a perfect square.
Remark. If m is not a perfect square, there are τ (c)
essentially different solutions of
2
m = a2 + b2 with 1 ≤ a ≤ b.
If m is a perfect square, there are τ (c)−1
essentially different solutions of m = a2 + b2
2
with 1 ≤ a ≤ b.
10 Problem 3.9. Prove that a Gaussian integer a + ib with prime decomposition (3.9) satisfies gcd (a, b) = 1 if and only if
s = 0 or s = 1;
g = 1;
either σν = 0 or τν = 0 for all ν = 1, . . . N .
Answer. A Gaussian integer has relatively prime real and imaginary parts if and only if
it is not divisible by any real prime. But the three conditions just imply:
a + ib is not divisible by 2;
a + ib is not divisible by any prime factor congruent to 3 mod 4;
a + ib is not divisible by any real prime congruent to 5 mod 4.
Proof of Proposition 3.1 and Theorem 3.3. For any given natural number m, let d be
the divisor of m assembling all prime factors congruent 3 mod 4. Let c be the divisor of m
assembling all prime factors congruent 1 mod 4. The factor c is completely decomposed
into its prime factors as in formula (3.8). In the ring of Gaussian integers, the usual
prime factoring of m will split further into irreducible factors
(3.10)
s
3s
2s
m = 2 dc = i (1 + i) d
N
Y
(pν + iqν )sν (pν − iqν )sν
ν=1
34
where the real factoring of d is not needed here.
Assume that m = a2 +b2 . Because of (a+ib)(a−ib) = a2 +b2 = m, we conclude that
a + ib is a divisor of m. The uniqueness of prime factorization implies equation (3.9)
holds for some values u ∈ {0, 1, 2, 3} and 0 ≤ σν ≤ sν , 0 ≤ τν ≤ sν for ν = 1 . . . N
(Why?). We now multiply equation (3.9) with the conjugate complex equation and get
2
2
s 2
a +b =2 g
N
Y
(pν + iqν )σν +τν · (pν − iqν )σν +τν
ν=1
We compare all factors with those in equation (3.10) and use the uniqueness of prime
factorization. Hence g 2 = d and σν + τν = sν for ν = 1 . . . N , as claimed. Too, we see
that no equation m = a2 + b2 can hold unless the factor d is a perfect square.
Too, the uniqueness of the prime factoring implies that all solutions given by equation (3.9) are different. It is straightforward to check that equation (3.9) yields solutions
of the equation m = a2 + b2 . Hence we get exactly
4τ (c) = 4
N
Y
(1 + sν )
ν=1
solutions.
10 Problem 3.10. Assemble different examples to illustrate Proposition 3.1 and
Theorem 3.3.
Examples. For different values of m, here is a table listing the factors 2s , d, c, the prime
decomposition of c and the number of divisors τ (c). Furthermore, we list the possible
values of a + ib form solution (3.9). For abbreviation, I put the unit factor iu = 1, and
ignore the obvious conjugate complex ones.
m
2s d
c factor c τ (c) list of a + ib
5
1 1 5 5
2 −1 + 2i
8
8 1 1 1
1 −2 + 2i
9
1 9 1 1
1 −3
10
2 1 5 5
2 (1 + i)(1 + 2i) = −1 + 3i
26
2 1 13 13
2 (1 + i)(−3 + 2i) = −5 − i
20
4 1 5 5
2 −4 + 2i
36
4 9 1 1
1 6i
39
1 3 13 13
2 no solution
52
4 1 13 13
2 −4 − 6i
65
1 1 65 5 · 13
4 (1 + 2i)(−3 ± 2i) = −7 − 4i , 1 − 8i
130 2 1 65 5 · 13
4 (1 + i)(1 + 2i)(−3 ± 2i) = −3 − 11i , 9 − 7i
260 4 1 65 5 · 13
4 2i(1 + 2i)(−3 ± 2i) = 8 − 14i , 16 + 2i
35
3.4
Roots
Definition 3.2 (argument). The argument of a complex number z 6= 0 is the set of
angles which in polar coordinates yield z:
arg z = {θ ∈ R : z = |z|eiθ }
The principle value of the argument Arg z of a complex number z 6= 0 is the unique
value for the argument lying in the interval (−π, +π], including +π but excluding −π.
Remark. All values of the argument differ by integer multiples of 2π. Hence there always
exists a unique principle argument.
Nevertheless, the choice of a principle argument is just an arbitrary convention.
Especially, the set of arguments arg z depends continuously on z 6= 0, but the principle
argument does not depend continuously on z.
Lemma 3.1 (Principle value). The principle argument of
x + iy 6= 0 is given by the formula


arctan xy
if x > 0;



π


if x = 0 and y
2
(3.11)
Arg z = − π2
if x = 0 and y


y

π + arctan x
if x < 0 and y



−π + arctan y
if x < 0 and y
x
a complex number z =
> 0;
< 0;
> 0;
<0
Remark. Several computer languages provide Arg (x + iy) as a two variable function
arctan(x, y).
10 Problem 3.11. Give simple examples to illustrate all cases in Lemma 3.1.
Proof of Lemma 3.1 . The principle argument Arg z of any complex number z 6= 0 is
defined by restricting the argument of
z = |z| eiArg z
to the interval Arg z ∈ (−π, π]. Separating real- and imaginary parts yields
x = |z| cos Arg z , y = |z| sin Arg z
Hence xy = tan Arg z unless x = 0. But since both cos(θ ± π) = − cos θ and sin(θ ± π) =
− sin θ, the tangent function has period π. Hence the value of tan Arg z determines the
argument only modulo π—not yet modulo 2π.
The given cases result from comparing the signs of sin and cos with those of x and
y.
36
10 Problem 3.12. The angle α of the tangent to the graph of a differentiable
function y = f (x) with the x-axis is always given by
α = arctan f 0 (x) = arctan
dy
dx
Why does one not distinguish the cases occurring in Lemma 3.1?
Answer. The angle between the tangent and the x-axis is a directed angle between two
lines, and hence always lies in the interval (− π2 , π2 ).
But the principle argument of a complex number z is a directed angle between two
rays—the positive x-axis and the ray from the origin 0 to the point z. Hence it can take
any value in the interval (−π, π], including +π and excluding −π.
Definition 3.3 (right semi-plane). Define the right semi-plane as the set
RS = {z = x + iy ∈ C : x > 0 or (x = 0 and y ≥ 0) }
Remark. The right semi-plane is neither open nor closed. I use the word ”half plane”
only for an open set.
10 Problem 3.13. What is the exact interval to which the principle value of the
argument for any z 6= 0 in the right semi-plane is restricted.
Answer. For any z 6= 0 in the right semi-plane, the principle value of the argument lies
in the half open interval Arg z ∈ (− π2 , π2 ]. Conversely, Arg z ∈ (− π2 , π2 ] implies z ∈ RS.
Definition 3.4 (Principle value). The principle argument Arg z of any complex number z 6= 0 is defined by restricting the argument of
z = |z| eiArg z
to the interval Arg z ∈ (−π, π]. The principle n-th root is define as
p
√
iArg z
P.V. n z := n |z| e n
The principle logarithm of any complex number z 6= 0 is defined as
Ln z := ln |z| + iArg z
10 Problem 3.14. Prove that for all a, b ∈ RS and all n ≥ 2
Arg ab = Arg a + Arg b for a, b ∈ RS and a, b 6= 0
√
√
√
n
n
P.V. ab = P.V. n a · P.V. b
37
Answer. For the product of any nonzero a, b ∈ RS, the principal values Arg ab ∈ (−π, π],
and Arg a + Arg b ∈ (−π, π], too. Hence
Arg ab = Arg a + Arg b + 2πk
holds with k = 0, since both sides do not differ by an integer multiple of 2π. Now the
second assertion easily follows
p
p
p
p
√
Arg ab
Arg a+Arg b
Arg a
Arg b
n
n
P.V. ab = n |ab| ei n = n |ab| ei
= n |a| ei n · n |b| ei n
√
√
n
= P.V. n a · P.V. b
Here are some examples of principle third roots.
√
√
3+i
3
(a)
P.V. i =
√2
√
3−i
(b)
P.V. 3 −i =
2
s
−1 + i
1+i
(c)
P.V. 3 √
= √
2
2
√
√
1+i 3
3
(d)
P.V. −1 =
2
Note that the principal third root of any negative real, is not the negative real value.
10 Problem 3.15. In which examples does equality hold for the principle values.
Show enough of your calculations.
√
√ ?
√
(a) P.V. 3 i · P.V. 3 i = P.V. 3 −1
√
√
√
?
(b) P.V. 3 −i · P.V. 3 −i = P.V. 3 −1
rh
i
q
q
(c) P.V.
(d) P.V.
3
−1+i
√
2
q
3
1+i
√
2
· P.V.
· P.V.
3
q
3
?
−1+i
√
=
2
?
1+i
√ =
2
P.V.
P.V.
3
rh
3
−1+i
√
2
1+i
√
2
i2
38
2
Answer.
"√
#2
√
√
√
3+i
1+i 3
1+i 3
3
P.V. i · P.V. i =
= P.V. −1 =
=
2
2
2
#2
"√
√
√
√
√
√
3−i
1−i 3
1+i 3
3
3
3
P.V. −i · P.V. −i =
6= P.V. −1 =
=
2
2
2
s
s
2
−1 + i
−1 + i
1+i
3
3
√
√
P.V.
· P.V.
= √
= i 6=
2
2
2
s
√
2
√
−1 + i
3−i
3
3
√
= P.V. −i =
P.V.
2
2
√
3
(a)
(b)
(c)
√
3
(d)
s
P.V.
3
3.5
1+i
√ · P.V.
2
s
3
p
i2 h 2πi i2
3
πi
1+i h
2πi
√ = P.V. e 8
=
= e 24 = e 6
2
s
2
√
πi
1+i
3
3
√
P.V.
= P.V. i = e 6
2
The solution of the reduced cubic equation
Given is the equation
(3.12)
x3 + b x + c = 0
Since the quadratic term is lacking, it is called a reduced or depressed cubic. We want
b
to find all three solutions, including the complex ones. Putting x := u − 3u
leads to the
equation
3
b
3
(3.13)
u −
+c=0
3u
Hence z = u3 satisfies the quadratic resolvent equation
(3.14)
z 2 + cz −
b3
=0
27
The two solutions of equation (3.14) are
r
c
c2
b3
(3.15)
z1,2 = − ±
+
2
4
27
We get a resolvent equation with real or complex solutions, depending on whether the
discriminant
c2
b3
(3.16)
D=
+
4
27
39
is positive or negative. In the first case, the original cubic has only one or (exceptionally) two real solutions. The second case with complex solutions of the resolvent
equation (3.14) is called the casus irreducibilis. In that case, the original cubic has three
real solutions. Hence one is forced to go through some complex arithmetic —just in the
case where all final results are real. The solution procedure turns out to be different for
these two cases.
Case 1: D ≥ 0 The cubic has one real solution— or in the special case D = 0—two
real solutions. From Viëta’s formula
z1 z2 = −
(3.17)
b3
27
√
b
3
− √
z2
=
3 3 z1
Hence the real solution of the cubic (3.12) is
(3.18)
√
√
b
x1 = u −
= 3 z1 + 3 z2
s
s 3u r
r
3
2
3
3
b
b3
c
c
c
c2
+
+ − −
+
= − +
2
4
27
2
4
27
which is just Cardano’s formula. The two complex solutions are
√
√
x2 = 3 z1 ω + 3 z2 ω 2
(3.19)
√
√
x3 = 3 z1 ω 2 + 3 z2 ω
where
(3.20)
√
−1 + i 3
2π
2π
ω=
= cos
+ i sin
2
3
3
is the primitive third root of unity. Simplifying (3.18) and (3.19) yields
√
√
√
√
3 z + 3 z
3 z − 3 z √
1
2
1
2
(3.21)
x2,3 = −
±
3i
2
2
Case 2: D < 0 The cubic has three real solutions. In this case, the resolvent equation (3.14) has complex solutions. As already Rafael Bombelli discovered (1572),
this leads to a third root of a complex number in Cardano’s formula (3.18).
Actually, this case can only occur for b < 0. To cover the general case, extraction
of the third root has to be done via polar coordinates. Viëta’s formula (3.17) and
40
equation (3.15) imply
−b3
27
r
c
−b3 c2
=− ±i
−
2
27
4
r
−b3
(cos θ ± i sin θ)
=
27
|z1 |2 = z1 z2 =
(3.22)
z1,2
The argument θ can be calculated more simple from the real parts. One gets cos θ,
and finally θ:
r
−b3
c
− =
cos θ
2
27
(3.23)
−c
θ = arc cos q
3
2 −b
27
Hence equations (3.22) and (3.20) imply
r
√
−b
θ
θ
3
z1 =
cos + i sin
3
3
3
r
(3.24)
√
−b
θ + 2π
θ + 2π
3
+ i sin
z1 ω =
cos
3
3
3
and equations (3.17) and (3.18) imply that
r
−b
θ
x1 = 2
cos
3
r 3
−b
θ + 2π
(3.25)
x2 = 2
cos
3
r 3
−b
θ − 2π
x3 = 2
cos
3
3
are the three real solutions of the original cubic.
10 Problem 3.16. Calculate the roots in the limiting case D & 0+.
Answer.
r
c
x1
x1 = 2 3 − , x 2 = x3 = −
2
2
10 Problem 3.17. Calculate the roots in the limiting case D % 0− in terms of
c. Distinguish the cases c < 0 and c > 0. In both cases b < 0.
41
Answer. In the approach D % 0−, one gets
r
r
c2
b3 3 |c|
−b
=− ,
=
4
27 r 2
3
3
|c|
−b
=
2 r 27
−b3
c
cos θ
− =
2
27
Distinguish the cases c < 0 and c > 0.
Case c < 0. One gets cos θ = 1, and hence θ = 0. The result (3.25) implies
r
c
x1
x1 = 2 3 − , x 2 = x3 = −
2
2
Case c > 0. One gets cos θ = −1, and hence θ = π. The result (3.25) implies
r
x1
c
x2 = −2 3 − , x1 = x3 = −
2
2
10 Problem 3.18. Use Cardano’s formula and Bombelli’s method to get one
solutions of x3 − 30x − 36 = 0.
Answer. Cardano’s formula for a solution of x3 + bx + c = 0 yields in the present case
s
s
r
r
2
3
3
3
c
b
c2
b3
c
c
(3.26)
x1 = − +
+
+ − −
+
2
4
27
2
4
27
√
√
3
3
= 18 + 26i + 18 − 26i
√
To calculate the third root, let 3 18 + 26i = u+iv and 18+26i = (u+iv)3 . We compare
real- and imaginary parts and get the system
18 = u3 − 3uv 2 = u(u2 − 3v 2 )
26 = 3u2 v − v 3 = (3u2 − v 2 )v
With the help of the factoring, one finds the integer solution u + iv = 3 + i. Hence
Cardano’s formula gives x1 = 6.
10 Problem 3.19. Find the two other solutions of the equation x3 −30x−36 = 0.
Answer. The simplest way is division of the polynomial. One gets√(x3 − 30x − 36)/(x −
6) = x2 + 6x + 6. Hence the two other solutions are x2,3 = −3 ± 3.
42
Remark. Too,
we can find the two other roots by introducing the third root of unity
√
−1+i 3
into Cardano’s formula. This has to be done in a consistent way and
ω =
2
produces the solutions
√
√
√
√
√
−1 + i 3
−1 − i 3
3
2 3
x2 = ω 18 + 26i + ω 18 − 26i =
(3 + i) +
(3 − i) = −3 − 3
2 √
2 √
√
√
√
−1 − i 3
−1 + i 3
(3 + i) +
(3 − i) = −3 + 3
x3 = ω 2 3 18 + 26i + ω 3 18 − 26i =
2
2
which is the same answer as above.
3.6
The square root of a complex number
√
Proposition 3.2. A branch with Re z ≥ 0 for the square root of any complex number
z = x + iy is
sp
sp
2
2
p
x +y +x
x2 + y 2 − x
(3.27)
x + iy =
+ i sign (y)
2
2
√
The square root w = z of any complex number z = x + iy has two branches. The
second branch is w2 = −w.
√
Reason. Name the square root in question x + iy =: u + iv. Squaring yields x + iy =
(u + iv)2 . Separate the real- and imaginary part to get
x = u2 − v 2 , y = 2uv
The absolute value squared is
x2 + y 2 = |x + iy|2 = |u + iv|4 = (u2 + v 2 )2
Add and substrate
p
u2 + v 2 = x2 + y 2
u2 − v 2 = x
p
x2 + y 2 + x
u2 =
2
p
2
x + y2 − x
v2 =
2
Of the last two expression, one takes real square roots. u ≥ 0 has been assumed.
One needs still to determine the sign of v. But y = 2uv and u > 0 imply sign y =
(sign u)(sign v) = sign v. In the special case u = 0, both signs of v give a correct result
for the square root. This special case corresponds to y = 0 and x ≤ 0—the negative
numbers z ≤ 0.
43
10 Problem 3.20. Calculate the square roots:
√
√
√
(a)
−3 + 4i , −12 + 16i , −48 + 64i
√
√
√
(b)
−8 − 6i , −32 − 24i , −128 − 96i
√
√
√
(c)
−4 − 3i , −16 − 12i , −64 − 48
√
√
√
(d)
−6 + 8i , −24 + 32i , −96 + 128i
√
√
√
(e)
2i , 8i , 32i
A Gaussian integer that is the square of another Gaussian integer is called a perfect
square. Which of the numbers under the square roots are perfect squares, which are not?
3.7
Pythagorean triples
Any three integers such that a2 + b2 = c2 are called a Pythagorean triple.
10 Problem 3.21. Show that for any Gaussian integer p + iq, the real and
imaginary parts and the absolute value of its square are a Pythagorean triple.
Answer. Let the square be a+ib := (p+iq)2 . Its absolute value is c := |a+ib| = |p+iq|2 .
Clearly a2 + b2 = c2 holds.
10 Problem 3.22. Show that for any Gaussian integer a+ib 6= 0, it is impossible
that a + ib and b + ia are both perfect squares.
Theorem 3.4 (Pythagorean triples). A nonzero Gaussian integer a + ib = (p + iq)2
is a perfect square if and only if the following three conditions hold
(i) the sum a2 + b2 is a perfect square c2 ;
(ii) gcd (a, b) =: g is either (a) a perfect square: g = d2 , or (b) twice a perfect square:
g = 2d2 ;
(iii) assuming 2t , but not 2t+1 divides d, we get two cases:
(iii a) if g = d2 , then 4−t a is odd and 4−t b is even—
(iii b) if g = 2d2 , then 2−2t−1 a is even and 2−2t−1 b is odd.
Corollary. Exactly one of the Gaussian integers a+ib = (p+iq)2 and b+ia = i(p−iq)2
is a perfect square if and only if
(i) the real and imaginary parts and the absolute value of a+ib are a Pythagorean triple;
(ii) the greatest common divisor gcd (a, b) is either a perfect square or twice a perfect
square.
44
Remark. It is always true that the imaginary part b of a perfect square is even, but this
condition is sufficient only in the special case gcd (a, b) = 1.
Remark. Condition (iii) in theorem 3.4 is only needed to determine which one of the
two numbers—either a + ib or b + ia—is a perfect square. In Problem 3.22, we have
already checked that these two numbers cannot be both perfect squares, unless they are
both zero.
Example 3.1. These are perfect squares, corresponding to t = 0, 1, 2 in (iii a) or (iii b):
√
√
√
(a)
−3 + 4i = 1 + 2i , −12 + 16i = 2 + 4i , −48 + 64i = 4 + 8i , . . .
√
√
√
(b)
−8 − 6i = −1 + 3i , −32 − 24i = −2 + 6i , −128 − 96i = −4 + 12i , . . .
but these are not perfect squares:
√
√ √
√
√
(−1 + 3i) 2 √
−4 − 3i =
, −16 − 12i = (−1 + 3i) 2 , −64 − 48 = (−2 + 6i) 2 , . . .
2 √ √
√ √
√
√
−6 + 8i = (1 + 2i) 2 , −24 + 32i = (2 + 4i) 2 , −96 + 128i = (4 + 8i) 2 , . . .
Here is the most simple example:
√
√
√
(a)
4 = 2 , 16 = 4 , 64 = 8 , . . .
√
√
√
(b)
2i = 1 + i , 8i = 2 + 2i , 32i = 4 + 4i , . . .
Proof of necessity in theorem 3.4 . Necessity is easier to prove. We assume that a+ib =
(p + iq)2 is a perfect square. Squaring the absolute values gives
c2 = a2 + b2 = |a + ib|2 = |p + iq|4 = (p2 + q 2 )2
Separating real- and imaginary parts yields the Pythagorean triple
a = p2 − q 2 , b = 2pq , c = p2 + q 2
Hence condition (i) holds.
To check (ii), let d := gcd (p, q). Clearly d2 divides both a and b, and hence d2
divides g := gcd (a, b). Let
p0 :=
q
a
b
c
p
, q 0 :=
, a0 := 2 , b0 := 2 , c0 := 2 ,
d
d
d
d
d
By these divisions we get
a0 = p02 − q 02 , b0 = 2p0 q 0 , c0 = p02 + q 02 , gcd (p0 , q 0 ) = 1
We show that gcd (a0 , b0 ) =: h can only be 1 or 2. Indeed, any odd prime dividing b0
needs to divides either p0 or q 0 , but not both. Hence it cannot divide a0 = p02 − q 02 .
45
Neither can 4 divide both a0 and b0 . Indeed, assuming 4 divides b0 , implies one number
among p0 and q 0 is even, the other one is odd. Hence a0 is odd.
Hence we see that h = 1 or h = 2 are the only possibilities left. Therefore gcd (a, b) =
2
d gcd (a0 , b0 ) = hd2 is either a perfect square or twice a perfect square, as claimed by
condition (ii).
Finally, we need to confirm that either case (iii a) or (iii b) occurs. Again, t ≥ 0 be
the integer for which 2t , but not 2t+1 divides d. These two cases can occur:
(iii a): g = d2 and h = 1 = gcd (a0 , b0 ) = gcd ((p0 − q 0 )(p0 + q 0 ), 2p0 q 0 ). Hence p0 ± q 0
are both odd, and hence one number among p0 and q 0 is odd, but the other one
is even. Hence a0 is odd, but 4 divides b0 . Similarly, 4−t a is odd and 4−t b is even,
since these are odd multiples of a0 and b0 .
(iii b): g = 2d2 and h = gcd (a0 , b0 ) = gcd ((p0 − q 0 )(p0 + q 0 ), 2p0 q 0 ) = 2. Hence p0 ± q 0
are both even, and—different from case (a)—p0 and q 0 are both odd.
Hence a0 is divisible by 4, but b0 ≡ 2 mod 4. Hence, 2−2t−1 a is even and 2−2t−1 b
is odd, since these are odd multiples of a0 and b0 .
Proof of sufficiency theorem 3.4 . We assume that the integers a and b satisfy the four
conditions (i) through (iii). Let g = gcd (a, b) = hd2 with h = 1 or h = 2 and as above
a0 :=
a
b
c
, b0 := 2 , c0 := 2
2
d
d
d
In cases (a) and (b) one gets
(iii a) if g = d2 , then gcd (a0 , b0 ) = 1, and a0 is odd, b0 is even. Hence c0 is odd, and
c0 ±a0
are both integers.
2
(iii b) if g = 2d2 , then gcd (a0 , b0 ) = 2,
a0
2
is even and
b0
2
is odd. Hence
c0
2
is odd.
As explained in Proposition 3.2, one obtains a square root of a complex number by the
formula
r
r
0 + a0
√
c
c 0 − a0
(3.27)
a0 + ib0 =
+ i sign (b)
2
2
0
0
0
In both cases (iii a) and (iii b), c ±a
and b2 are integers. Furthermore a2 + b2 = c2
2
implies
0 2
b
c 0 + a0 c 0 − a0
=
·
2
2
2
The two factors are relatively prime:
0
c + a0 c 0 − a0
gcd
,
=1
2
2
46
0
0
is a divisor of h = gcd (a0 , b0 ), which is
Indeed, the greatest common divisor of c ±a
2
0
0
either 1 or 2. But is the case (b) where h = 2, both numbers c ±a
turn out to be odd.
2
It is a (too easy) exercise to show that uniqueness of the prime factorization implies
Lemma 3.2. If the product of two relatively prime factors is a perfect square, then both
factors are perfect squares.
Hence we conclude that both roots
r
c 0 + a0
0
(E:doit)
p =
2
r
0
and q = sign (b)
c 0 − a0
2
√
are integers. In other word, we have√shown that a0 + ib0 = p0 + iq 0 is a Gaussian
integer. Multiplying by d, we get that a + ib = p + iq is a Gaussian integer, too, as to
be shown.
10 Problem 3.23. Generalize the Lemma 3.2 to the product of three or more
factors. Are the following conjectures true or false:
”If the product of three pairwise relatively prime factors is a perfect square,
then all three factors are perfect squares.”
”If the product of three relatively prime factors is a perfect square, then the
product of any two of them is a perfect square.”
Remark. Under the additional assumption that gcd (a, b) ≤ 2, indeed only the following
possibilities occur for perfect squares a + ib = (p + iq)2 :
p
odd
even
odd
3.8
q
even
odd
odd
a mod 4
1
3
0
b mod 4
0
0
2
c mod 4
1
1
2
h
1
1
2
Roots of unity
10
roots:
Problem 3.24. Find all different powers—exactly, expressed using square
√
−1 + i 3
(a) ω =
.
2
p
√
√
5 − 1 + i 10 + 2 5
(b) z =
.
4
1+i
(c) w = √ .
2
47
You need not show all calculations. Simplify your answers by using the conjugate complex. What is remarkable about these expressions?
Answer. It is remarkable that the different powers can be obtained simply by some sign
changes of some square roots:
√
√
−1 − i 3
−1 + i 3
2
(a)
, ω =
= ω , ω3 = 1
ω=
2
2
p
p
√
√
√
√
5 − 1 + i 10 + 2 5
− 5 − 1 + i 10 − 2 5
2
(b)
z=
, z =
,
4
4
z3 = z2 , z4 = z
−1 + i
1+i
(c)
, w4 = −1 ,
w = √ , w2 = i , w3 = √
2
2
5
6
3
w = w , w = −i , w7 = w
An exact trigonometric table of cos 3◦ k and sin 3◦ k for all k = 1, . . . , 15 can be
obtained using the three roots of unity ω, z and w. This trick goes in principle back to
Ptolemy!
(a) From the primitive third and eighth root of unity ω and w from Problem 3.24, we
get a product of argument 15◦ .
1
3
ω −1 w3 = e2πi[− 3 + 8 ] = e 24 = cos 15◦ + i sin 15◦
2πi
(b) We can calculate this product in terms of the exact square roots, obtained in
Problem 3.24.
√
√
√
1 + 3 + i( 3 − 1)
−1 − i 3 −1 + i
−1 3
√
· √
=
ω w =
2
2
2 2
(c) Finally, we compare the real- and imaginary parts obtained form (a) and (b), and
use the common denominator 4.
√
√
2+ 6
◦
cos 15 =
√ 4√
6− 2
sin 15◦ =
4
(d) In this case, we get the table immediately.
48
k 15◦
15◦
30◦
45◦
60◦
75◦
90◦
◦
cos√k 15
√
◦
sin√k 15
√
1
0
2+ 6
4√
3
√2
2
2
1
√ √2
6− 2
4
6− 2
4
1
√2
2
√2
3
√ √2
6+ 2
4
10 Problem 3.25. Calculate cos 15◦ + i sin 15◦ directly by formulas
r
r
θ
1 + cos θ
θ
1 − cos θ
cos =
, sin =
2
2
2
2
and confirm that you get the same values.
Answer.
◦
30
=
cos
2
r
1+
cos 30◦
2
s
=
p
√
√
2+ 3
2+ 3
=
4
2
√
◦
Similarly, one gets sin 15 =
√
2− 3
.
2
By squaring both sides, one checks that indeed
p
p
√
√
√
√
√
√
2+ 3
6+ 2
2− 3
6− 2
=
,
=
2
4
2
4
10 Problem 3.26. The following steps can be elaborated to obtain an exact
trigonometric table of cos 9◦ k and sin 9◦ k for all k = 1, . . . , 40.
(a) Find natural numbers p, q such that − p5 +
q
8
=
1
.
40
(b) With z, w from Problem 3.24, calculate the argument θ in
z −p wq = e2πiθ
(c) Hint: calculate
√
p
√
5 + 1 − i 10 − 2 5 1 + i
· √
4
2
(d) Calculate z −p wq with the exact square roots, obtained in Problem 3.24.
(e) Use Euler’s
2πi
e 40 = cos
(3.28)
2π
2π
+ i sin
40
40
and get exact expressions for cos 9◦ and sin 9◦ . Use the common denominator 8 in
your results.
Answer. (a) The natural numbers p, q have to satisfy −8p + 5q = 1. The smallest
solution is p = 3, q = 5.
(b) With z, w from Problem 3.24,
3
5
z −p wq = e2πi[− 5 + 8 ] = e 40
49
2πi
(d) With the same z, w from Problem 3.24, using z 5 = 1 and w2 = i, we can simplify
and get
p
√
√
5
+
1
−
i
10
−
2
5 1+i
· √
z −3 w5 = −z 2 w =
4
2
p
p
√
√
√
√
√
√
10 + 2 + 2 5 − 5
10 + 2 − 2 5 − 5
+i
=
8
8
(e) Comparing real- and imaginary part from parts (b) and (c) yields
p
√
√
√
10
+
2
+
2
5− 5
◦
cos 9 =
(3.29)
8 p
√
√
√
10
+
2
−
2
5
−
5
(3.30)
sin 9◦ =
8
10 Problem 3.27. Give the exact expression for the perimeter of a regular 20gon, inscribed into the unit circle. Compare with 2π. How large is the deviation?
Answer. The circumference of a regular 20-gon is
q
√
√
√
◦
40 sin 9 = 5
10 + 2 − 2 5 − 5 ≈ 6.257378602
Bur 2π ≈ 6.283185307 so only one decimal point. But the relative error of less than a
percent is a beginning.
In a similar way, we can use the twelfth and fifth root of unity, and get sin and cos for
the multiples of 6◦ .
(a) The natural numbers p, q have to satisfy 12p − 5q = ±1. The smallest solution is
p = 3, q = 7. Dividing by 60 yields
7
1
3
−
=
5 12
60
(b) With the twelfth and fifth roots of unity τ and z we get
3
7
z 3 τ −7 = e2πi[ 5 − 12 ] = e 60 = cos 6◦ + i sin 6◦
2πi
(d) We can calculate this product with the exact values of the square roots, simplify
and get
p
√
√
√
5
+
1
+
i
10
−
2
5
3−i
z 3 τ −7 = −z 3 τ −1 =
·
4
2
p
√
√
√
√
√ p
√
15 + 3 + 10 − 2 5
− 5 − 1 + 3 10 − 2 5
=
+i
8
8
50
(e) Comparing real- and imaginary part from parts (b) and (c) yields
p
√
√
√
15
+
3
+
10 − 2 5
◦
(3.31)
cos 6 =
8
√
√ p
√
−
5
−
1
+
3
10
−
2
5
sin 6◦ =
(3.32)
8
10 Problem 3.28. Give the exact expression for the perimeter of a regular 30gon, inscribed into the unit circle. Compare with 2π. How large is the deviation?
Answer. The circumference of a regular 30-gon is
q
√
√
√
15
◦
− 5 − 1 + 3 10 − 2 5 ≈ 6.271707796
60 sin 6 =
2
But 2π ≈ 6.283185307—so just one decimal point coincides.
10 Problem 3.29. Prove that
√ p
√
√
5 + 1 + 3 10 − 2 5
◦
(3.33)
cos 24 =
8p
√
√
√
15
+
3
−
10
−
2
5
(3.34)
sin 24◦ =
8
Here is a table assembling the information we have gathered so far.
θ
6◦
9◦
15◦
18◦
24◦
30◦
36◦
45◦
cos θ
√ √
√
15+ 3+ 10−2 5
√
√ 8√ √
10+ 2+2 5− 5
√ √ 8
2+ 6
q4 √
5+ 5
8
√
√ √
√
5+1+ 3 10−2 5
8
√
3
2
√
5+1
√4
2
2
√
sin θ
√
√ √
√
− 5−1+ 3 10−2 5
√
√8 √ √
10+ 2−2 5− 5
8 √ √
6− 2
4
√
5−1
4
√
√
√
√
15+ 3− 10−2 5
8
1
q √2
5− 5
8√
2
2
10 Problem 3.30. An exact trigonometric table of cos 3◦ k and sin 3◦ k for all
k = 1, . . . , 15 can be obtained with the same trick—which goes in principle back to
Ptolemy!
(a) Calculate
1 1 1
− − . With ω, z and w from Problem 3.24, get the argument θ in
3 5 8
ωz −1 w−1 = e2πiθ = cos 2πθ + i sin 2πθ
51
(b) Calculate ωz −1 w−1 with the exact square roots, obtained in Problem 3.24. Get a
readable result, do not distribute every term.
(c) Get exact expressions for cos 3◦ and sin 3◦ .
Solution. (a) Since
1 1 1
1
− − =
, we get the argument
3 5 8
120
1
1
1
ωz −1 w−1 = e2πi [ 3 − 5 − 8 ] = cos
2π
2π
+ i sin
120
120
(b) With the exact square roots from Problem 3.24,
"
p
√ #
√
√
−1
+
i
3
1
−
i
5
−
1
−
i
10 + 2 5
−1 −1
√
ωz w = ωwz =
2
4
2
p
√
√
√
√
3 − 1 + i( 3 + 1)
5 − 1 − i 10 + 2 5
√
=
4
2 2
p
p
√
√
√
√
√
√
√
√
( 3 + 1)( 5 − 1) − ( 3 − 1) 10 + 2 5
( 3 − 1)( 5 − 1) + ( 3 + 1) 10 + 2 5
√
√
+i
=
8 2
8 2
(e) We get exact expressions
p
√ √
√
√
√
2(
3
−
1)(
5
−
1)
+
2(
3
+
1)
5
+
5
cos 3◦ =
16
p
√ √
√
√
√
2(
3
+
1)(
5
−
1)
−
2(
3
−
1)
5+ 5
◦
sin 3 =
16
10 Problem 3.31. Calculate the circumference of a regular 60-gon. Which approximation of 2π do we get? How many decimal points are correct?
Answer. The circumference of an 60-gon is
q
√
√
√
15 √ √
◦
120 sin 3 =
2( 3 + 1)( 5 − 1) − 2( 3 − 1) 5 + 5 ≈ 6.280314749
2
2π ≈ 6.283185307
One gets two correct decimal points.
10 Problem 3.32. Gather the entire table with cos k3◦ and sin k3◦ for all k =
1 . . . 30. Go on as far as you get.
52
ω k z −k w−k = e
2πik
120
#k " √
"√
p
√
√ #k
3 − 1 + i( 3 + 1)
5 − 1 − i 10 + 2 5
√
=
4
2 2
# " √
" √
p
√ #
3
+
i
−
5
−
1
−
i
10
−
2
5
−
ω 2 z −2 w−2 = e =
2
4
p
√
√
√
√ p
√
√
15 + 3 + 10 − 2 5 − i 5 − i + i 3 10 − 2 5
=
8
4πi
120
p
√ # "√
√ #
1−i 3
5 − 1 + i 10 + 2 5
ω z w = e = −ωz =
2
4
p
√ p
√
√
√
√
√
5 − 1 + 3 10 + 2 5 − i 15 + i 3 + i 10 + 2 5
=
8
"
4 −4
−4
8πi
120
" √
# " √
p
√
√ #
−
3
−
1
+
i(
3
−
1)
5
−
1
−
i
10
−
2
5
−
√
= ωwz −2 =
ω 7 z −7 w−7 = e
4
2 2
p
p
√
√
√
√
√
√
√
√ √
√
2( 3 + 1)( 5 + 1) + 2( 3 − 1) 5 − 5
2(− 3 + 1)( 5 + 1) + 2( 3 + 1) 5 − 5
=
+i
16
16
14πi
120
Further values were obtained by trial and error for the signs of the roots.
53
θ
3◦
6◦
9◦
12◦
15◦
18◦
21◦
24◦
27◦
30◦
33◦
36◦
39◦
42◦
45◦
48◦
51◦
54◦
57◦
60◦
63◦
66◦
69◦
72◦
75◦
78◦
81◦
84◦
87◦
90◦
cos θ
√ √
√ √
√
√
2( 3−1)( 5−1)+2( 3+1) 5+ 5
16
√
√ √
√
15+ 3+ 10−2 5
√
√ 8√ √
10+ 2+2 5− 5
√
√ 8√
√
5−1+ 3 10+2 5
8
√ √
2+ 6
q4 √
5+ 5
8
√ √
√ √
√
√
2( 3+1)( 5+1)+2( 3−1) 5− 5
16
√
√ √
√
5+1+ 3 10−2 5
√
√ 8√ √
10− 2+2 5+ 5
8
√
3
2√ √
√ √
√
√
− 2( 3−1)( 5−1)+2( 3+1) 5+ 5
16
√
5+1
4
√ √
√
√
√
√
− 2(− 3+1)( 5+1)+2( 3+1) 5− 5
√
√ √
√16
15− 3+ 10+2 5
8
√
2
2√
√ √
√
− 5+1+ 3 10+2 5
8√
√ √
√ √
√
2( 3+1)( 5+1)−2( 3−1) 5− 5
16
q √
5− 5
8
√ √
√ √
√
√
2( 3+1)( 5−1)+2( 3−1) 5+ 5
16
1
2√
√ √
√
− 10+ 2+2 5+ 5
√
√ 8√
√
15+ 3− 10−2 5
√ √
√
√ 8 √
√
2(− 3+1)( 5+1)+2( 3+1) 5− 5
16
√
5−1
√4 √
6− 2
√4
√ √
√
− 15+ 3+ 10+2 5
8√ √
√
√
10+ 2−2 5− 5
8 √
√
√
√
− 5−1+ 3 10−2 5
8√
√ √
√ √
√
2( 3+1)( 5−1)−2( 3−1) 5+ 5
16
sin θ
√ √
√ √
√
√
2( 3+1)( 5−1)−2( 3−1) 5+ 5
√16
√ √
√
− 5−1+ 3 10−2 5
√
√8 √ √
10+ 2−2 5− 5
8
√
√ √
√
− 15+ 3+ 10+2 5
8
√ √
6− 2
4
√
5−1
4
√
√
√
√
√
√
2(− 3+1)( 5+1)+2( 3+1) 5− 5
16 √ √
√
√
15+ 3− 10−2 5
√
√ 8 √ √
− 10+ 2+2 5+ 5
8
1
√ √2
√ √
√
√
2( 3+1)( 5−1)+2( 3−1) 5+ 5
16
q √
5− 5
8
√
√ √
√
√
√
2( 3+1)( 5+1)−2( 3−1) 5− 5
√16
√ √
√
− 5+1+ 3 10+2 5
8
√
2
√
√ √
√2
15− 3+ 10+2 5
√
√
√
√ 8 √ √
− 2(− 3+1)( 5+1)+2( 3+1) 5− 5
16
√
5+1
4
√
√ √
√
√
√
− 2( 3−1)( 5−1)+2( 3+1) 5+ 5
16
√
3
√ √2
√
√
10− 2+2 5+ 5
8
√
√ √
√
5+1+ 3 10−2 5
√
√ 8 √ √
√ √
2( 3+1)( 5+1)+2( 3−1) 5− 5
16
q √
5+ 5
√ 8√
6+ 2
4√
√
√ √
5−1+ 3 10+2 5
√
√ 8 √ √
10+ 2+2 5− 5
8
√
√ √
√
15+ 3+ 10−2 5
√ √
√
√ 8 √ √
2( 3−1)( 5−1)+2( 3+1) 5+ 5
16
1
0
54
Part II
Algebra
1
Polynomials
1.1
Rational zeros of polynomials
Proposition 1.1. Assume x is a real or complex zero of a monic
5
integer polynomial
xr + ar−1 xr−1 + · · · + a1 x + a0 = 0
where ar−1 , . . . a1 , a0 are integers. If x is rational, then it is even an integer, and a
divisor of the constant term a0 , too.
Proof. Assume x =
yields
m
n
where m and n are relatively prime integers. Multiplying by nr
mr
mr−1
m
+
a
+
·
·
·
+
a
+ a0 = 0
r−1
1
nr
nr−1
n
mr + ar−1 mr−1 n + · · · + a1 mnr−1 + a0 nr = 0
mr = − ar−1 mr−1 + · · · + a1 mnr−2 + a0 nr−1 n
6
Now we argue the same way as in the proof of Proposition 1.9: Assume that any prime
p divides n. By the last line, the same prime divides mr . Hence, by Euclid’s Lemma
. Hence n = 1, and
p divides m. This cannot occur after cancellation of the fraction m
n
the rational root x is an integer.
To get the last claim, we reuse the formulas from above, now with n = 1, and get
mr + ar−1 mr−1 + · · · + a1 m + a0 = 0
m mr−1 + ar−1 mr−2 + · · · + a1 = −a0
Hence any zero m has to be a divisor of the constant coefficient a0 .
Proposition 1.2. Assume x is a real or complex zero of any polynomial
ar xr + ar−1 xr−1 + · · · + a1 x + a0 = 0
where ar 6= 0, ar−1 , . . . a1 , a0 are integers. If the root x = m
is rational with m and n
n
relatively prime, then m is a divisor of a0 and n is a divisor of ar .
5
6
A polynomial is called monic if the coefficient of the leading term xr is one.
Hence n is a divisor of mr and hence n = 1, since n and m are relatively prime.
55
Proof.
ar mr + ar−1 mr−1 n + · · · + a1 mnr−1 + a0 nr = 0
ar mr = − ar−1 mr−1 + · · · + a1 mnr−2 + a0 nr−1 n
− ar mr−1 + ar−1 mr−2 n + · · · + a1 nr−1 m = a0 nr
The second line implies that n is a divisor of ar mr . Since gcd (n, m) = 1, the Euclidean
Property stated in Proposition 1.4 implies that n is a divisor of ar .
The last line implies that m is a divisor of a0 nr . Since gcd (n, m) = 1, the Euclidean
Property stated in Proposition 1.4 implies that m is a divisor of a0 .
Rule: The rational solutions of an integer polynomial ar xr + · · · + a0 = 0, are among
the positive or negatives of the rational numbers top and bottom of which are divisors
of top and bottom of the solution of ar X + a0 = 0.
1.2
Gauss’ Lemma
Proposition 1.3 (Linear Gauss Lemma). Suppose that an integer polynomial has
a rational zero m
, with n, m are relatively prime. Then it can be factored into integer
n
polynomials
(1.1)
ar xr + · · · + a0 = (nx − m)(br−1 xr−1 + · · · + b0 )
If additionally gcd (ar , . . . , a0 ) = 1, then gcd (br−1 , . . . , b0 ) = 1, too.
Proposition 1.4 (Monic Linear Gauss Lemma). Suppose that a monic integer
polynomial has a rational zero. Then the root is an integer, and the polynomial can be
factored into monic integer polynomials
xr + · · · + a0 = (x − m)(xr−1 + · · · + b0 )
Proof. We may assume from the start that gcd (ar , . . . , a0 ) = 1. The factoring of equa, . . . , b0 = BB0 . We choose the
tion (1.1) holds with rational coefficients br−1 = Br−1
B
smallest possible denominator B. In that case, the assumption that gcd (ar , . . . , a0 ) = 1
implies that gcd (Br−1 , . . . , B0 ) = 1, too. 7 After multiplying with the least common
denominator B of the b0 , . . . br−1 , comparison of coefficients yields
Ba0
Ba1
Ba2
Bar−1
Bar
= −mB0
= nB0 − mB1
= nB1 − mB2
= nBr−2 − mBr−1
= nBr−1
7
Indeed, a common prime divisor of all Br−1 , . . . , B0 could not divide B and hence would be a
common divisor of all ar , . . . , a0 , what we just have ruled out.
56
Assume towards a contradiction that a prime p divides all Ba0 , . . . Bar . Either p does
not divide n or p does not divide m.
Take the second case. Successively, we conclude that p divides Ba0 = −mB0 , B0 , Ba1 −
nB0 = −mB1 , B1 , Ba2 −nB1 = −mB2 , B2 , . . . , Br−1 , hence all B0 , . . . Br−1 —contrary to
the assumption. Hence we conclude gcd (Ba0 , . . . Bar ) = 1. In the other case that p does
not divide n, we get the same conclusion, now going through the system in reversed order: Successively, we see that the prime p divides Bar = −nBr−1 , Br−1 , Bar−1 +mBr−1 =
nBr−2 , Br−2 , . . . , Ba1 + mB1 = nB0 , B0 .
In both cases, the conclusion gcd (Ba0 , . . . Bar ) = 1 holds. Hence B = 1, and in the
first place, the rational numbers b0 , . . . br−1 need to be integers.
Finally, we see that gcd (br−1 , . . . , b0 ) = 1. Indeed, since B = 1, a prime p dividing
all coefficients br−1 , . . . , b0 would divides all coefficients ar , . . . , a0 , too, contrary to the
assumption.
Theorem 1.1 (Gauss’ Lemma). An integer polynomial which factors over the rational
numbers into factors of lower degree, already factored into integer polynomials of lower
degree. The latter factoring is obtained from the former one by adjusting integer factors.
Especially, a monic integer polynomial that factors over the rational numbers, even
factors over the integers into monic integer polynomials.
Proof. Suppose that an integer polynomial of degree r ≥ 2 can be factored into rational polynomials of degree s ≥ 1 and t ≥ 1. Multiplying with the least common
denominators, we get the integer formula
(1.2)
B (ar xr + · · · + a0 ) = A (bs xs + · · · + b0 ) ct xt + · · · + c0
where ar xr + · · · + a0 is obtained from the given integer polynomial by dividing through
with the greatest common divisor of the coefficients. Because of using least common
denominators everywhere, and cancellation, we get
(1.3)
gcd (A, B) = 1
gcd (ar , . . . , a0 ) = 1
gcd (bs , . . . , b0 ) = 1
gcd (ct , . . . , c0 ) = 1
Question. Show that A = 1.
Answer. Assume any prime number p divides A, and derive a contradiction. By formula (1.2) this would imply that p divides all numbers Bar , . . . Ba0 .
Since gcd (ar , . . . , a0 ) = 1, we conclude that p divides B. But this contradicts the
assumption gcd (A, B) = 1.
Lemma 1.1. If a prime p divides all coefficients of the product of two polynomials with
integer coefficients
(1.4)
(bs xs + · · · + b0 ) ct xt + · · · + c0 ,
then p divides either gcd (bs , . . . , b0 ) or gcd (ct , . . . , c0 ).
57
Reason for the Lemma. Multiplying the terms of the product (1.4), we see that p divides
all the integers
b0 c 0
b0 c 1 + b1 c 0
b0 c 2 + b1 c 1 + b2 c 0
b0 ck + b1 ck−1 + · · · + bk c0
for all k ≥ 0. 8 We see that p divides either b0 or c0 . Now take the case that p divides
b0 . The case that p divides c0 can be deals with similarly. Because of the assumption
gcd (bs , . . . , b0 ) = 1, there exists an index 0 ≤ σ < s such that p divides b0 , b1 , . . . bσ but
not bσ+1 . But we have seen that p divides all the sums from above.
Successively, we put k = σ + 1, σ + 2 . . . and cancel the terms already known to be
divisible by p. Hence we see that p divides the sums
bσ+1 c0
bσ+1 c1 + bσ+2 c0
bσ+1 c2 + bσ+2 c1 + bσ+3 c0
bσ+1 cj + bσ+2 cj−1 + · · · + bσ+j+1 c0
and so on for all j ≥ 0. Hence we conclude that p divides c0 , c1 , c2 , . . . cj for all j ≥ 0.
Question. Use the Lemma to show that B = 1.
Answer. Assume any prime number p divides B, and derive a contradiction. By formula (1.2), and since A = 1, this would imply that p divides all coefficients of the
product (1.4). Hence by the Lemma, either p divides all coefficients bs , . . . , b0 or all
coefficients ct , . . . , c0 . This has been ruled out by assumption (1.3). Hence we conclude
that B = 1.
Now, since A = B = 1, the factoring (1.2) is indeed a factoring of the integer
polynomial ar xr + · · · + a0 into integer polynomials.
Since the integer polynomial ar xr + · · · + a0 was obtained from the given integer
polynomial by dividing through with the greatest common divisor of the coefficients, we
have factored the originally given polynomial, too.
10 Problem 1.1. Find all symmetric monic polynomials x4 + ax2 + bx2 + ax + 1,
that factor over the rationals and have
(i) Four rational zeros.
(ii) Exactly two rational zeros.
(iii) No rational zeros.
8
For simplicity, we put the coefficients of a polynomial equal zero if we exceed its degree.
58
Do the cases with one or three zeros occur? Why not?
Solution. The only possible rational zeros are +1 and −1. If x = 1 is a zero, it is a
double zero since
P (x) = x4 P (x−1 )
P 0 (x) = 4x3 P (x−1 ) − x2 P 0 (x−1 )
Hence P (x) = 0 and x2 = 1 imply P 0 (1) = −P 0 (1) = 0.
(i) Four rational zeros. Hence there are five choices with all zeros in the set {+1, −1}.
Three of them turn out to be symmetric:
(x + 1)4 , (x2 − 1)2 , (x − 1)4
(ii) Exactly two rational zeros. The cases with a double zero 1 are
(x − 1)2 (x2 + (a + 2)x + 1) = x4 + ax3 − 2(a + 1)x2 + ax + 1
where a is any integer a 6= 0, a 6= −4. The cases with a double zero −1 are
(x + 1)2 (x2 + (a − 2)x + 1) = x4 + ax3 + 2(a − 1)x2 + ax + 1
where a is any integer a 6= 0, a 6= +4. Too, we see that no solutions with a zero of
multiplicity three exist.
(iii) No rational zeros. The polynomial factors into two irreducible monic quadratics: Q and R. The symmetry implies
P (x) = Q(x)R(x) = x2 Q(x−1 )x2 R(x−1 ) = x4 P (x−1 )
Uniqueness of factorization into irreducible factors leads to four possibilities:
(a) x2 Q(x−1 ) = Q(x) and x2 R(x−1 ) = R(x).
(b) x2 Q(x−1 ) = −Q(x) and x2 R(x−1 ) = −R(x).
(c) x2 Q(x−1 ) = R(x) and x2 R(x−1 ) = Q(x).
(d) x2 Q(x−1 ) = −R(x) and x2 R(x−1 ) = −Q(x).
Here are the results:
(a) From x2 Q(x−1 ) = Q(x) = x2 + cx + 1 and x2 R(x−1 ) = R(x) = x2 + dx + 1,
we get
Q(x)R(x) = (x2 +cx+1)(x2 +dx+1) = x4 +(c+d)x3 +(cd+2)x2 +(c+d)x+1
with any integer c, d 6= ±2.
59
(b) x2 Q(x−1 ) = −Q(x) and x2 R(x−1 ) = −R(x) gives only the solution (x2 − 1)2
that we already have.
(c) x2 Q(x−1 ) = R(x) and x2 R(x−1 ) = Q(x) gives only the special case of (a) with
c = d, but nothing new.
(d) −x2 Q(x−1 ) = R(x) = x2 − cx − 1 and −x2 R(x−1 ) = Q(x) = x2 + cx − 1 gives
the extra solutions
Q(x)R(x) = (x2 + cx − 1)(x2 − cx − 1) = x4 − (c2 + 2)x2 + 1
These are new solution for any c 6= 0.
1.3
Eisenstein’s irreducibility criterium
Proposition 1.5 (The Eisenstein criterium). For a polynomial P to be irreducible
in the ring Z[z], it is sufficient that there exists a prime number p such that
1. the leading coefficient of the polynomial is not divisible by p;
2. all other coefficients except the leading one are divisible by p;
3. the constant coefficient is not divisible by p2 .
Proof. Assume toward a contradiction that a reducible polynomial would satisfy the
criterium. We would get the integer factorization
(1.5)
a0 + · · · + ar xr = (b0 + · · · + bs xs ) c0 + · · · + ct xt ,
where r = s + t and s, t, ≥ 1. Multiplying the terms of the product yields
a0
a1
a2
ak
=b0 c0
=b0 c1 + b1 c0
=b0 c2 + b1 c1 + b2 c0
=b0 ck + b1 ck−1 + · · · + bk c0
for all k ≥ 0. It is assumed that the prime p, but not p2 divides a0 = b0 c0 . Hence
p divides exactly one of the two numbers b0 and c0 . We may assume that p divides
c0 , but does not divide b0 . Furthermore, by assumption, p divides a0 , a1 , . . . ar−1 , but
not ar . Recursively, we conclude that p divides b0 c1 = a1 − b1 c0 and hence c1 , next
b0 c2 = a2 − b1 c1 − b2 c0 and hence c2 , . . . , cr−1 .
Hence p divides all coefficients of the factor ct xt +· · ·+c0 , since t ≤ r−1. This implies
that p divides all coefficients of the original polynomial ar xr + · · · + a0 , contradicting
the assumption that the prime p does not divide the leading coefficient ar .
From this contradiction, we see that no factoring of the polynomial into polynomials
of lower degree is possible, and hence it is irreducible.
60
q
√
10 Problem 1.2. Find the irreducible polynomial with zero z1 = 5−2 5 . Find
all its algebraic conjugates, and show that the number is a totally real algebraic integer.
Solution of Problem 1.2. Simple arithmetic shows that (2z 2 − 5)2 − 5 = 0 and hence
P (z) := z 4 − 5z 2 + 5 = 0
is the monic polynomial in the ring Z[z] with zero z1 . Hence z1 is an algebraic integer.
The Eisenstein criterium applies with p = 5. Hence its zeros are exactly the algebraic
conjugates of z. Obviously, these zeros are
s
s
s
s
√
√
√
√
5+ 5
5+ 5
5− 5
5− 5
, −
,
, −
2
2
2
2
which all four turn out to be real.
Lemma 1.2. For any prime p, the binomial coefficients
p
for k = 1 . . . p − 1
k
are divisible by p, but not by p2 .
Proposition 1.6. For any prime p, the polynomial
Φp (z) =
zp − 1
z−1
is irreducible over the integers.
Increasing powers. The assertion is true for p = 2. We assume now that p is an odd
prime. We substitute z = 1 + x and use the Eisenstein criterium to show that the
resulting polynomial P (x) = Φp (1 + x) is irreducible. The binomial formula implies
p (1 + x)p − 1 X p k−1
P (x) =
=
x
x
k
k=1
p
p 2
p
=p+
x+
x + ··· +
xp−3 + pxp−2 + xp−1
2
3
p−2
Because of Lemma 1.2, we see that all three assumptions for the Eisenstein criterium
(Proposition 1.5) are satisfied. Hence the polynomial P and hence Φp are irreducible
over the integers.
Lemma 1.3. For any odd prime p, the binomial coefficients
2
p
for k = 1 . . . p − 1
k
are divisible by p.
61
Proposition 1.7. For any prime p, the polynomial
2
zp − 1
Φ (z) = p
z −1
p2
is irreducible over the integers.
Proof. In the case p = 2, we get Φ4 = 1 + z 2 . which we can check to be irreducible. We
assume now that p is an odd prime. We substitute z = 1 + x and use the Eisenstein
criterium to show that the resulting polynomial
p(p−1)
Φp2 (1 + x) =
X
bi xi = p + · · · + xp(p−1)
i=0
is irreducible. The definition and the binomial formula imply
2
zp − 1
zp − 1
= Φp2 (z) ·
z−1
z−1
p2
(1 + x)p − 1
(1 + x) − 1
= Φp2 (1 + x) ·
x
x
p(p−1)
p2 2 p X p
X
X p
xl−1 =
bi x i ·
xk−1
l
k
i=0
l=1
k=1
We have obtained the integer factorization
a0 + · · · + ar xr = (b0 + · · · + bs xs ) c0 + · · · + ct xt ,
which is really possible in this context since it has been constructed above. The degrees
are now r = p2 − 1, s = deg Φp2 = p(p − 1) and t = deg Φp = p − 1. Multiplying the
polynomials and comparing the coefficients yields once more
a0
a1
a2
ak
=b0 c0
=b0 c1 + b1 c0
=b0 c2 + b1 c1 + b2 c0
=b0 ck + b1 ck−1 + · · · + bk c0
for all k ≥ 0. We know from Lemma 1.2 and 1.3 that the prime p divides all the
coefficients a0 , . . . , ap2 −1 , but ap2 = 1. That is enough to check inductively that p divides
all coefficients b0 , . . . , bp2 −p−1 . But the leading coefficient of Φp2 (1 + x) is bp2 −p = 1.
We see that the polynomial Φp2 (1+x) satisfies all three assumptions for the Eisenstein
criterium (Proposition 1.5). Hence Φp2 (1 + x) and hence Φp2 (x) are irreducible over the
integers.
62
Example p = 3, p2 = 9. We claim that
Φ9 (1 + x) =
6
X
bi x i
i=0
is irreducible. The definition and the binomial formula imply
z9 − 1
z3 − 1
= Φ9 (z) ·
z−1
z−1
9
(1 + x) − 1
(1 + x)3 − 1
= Φ9 (1 + x) ·
x
x
9
6
X 9
X
xl−1 =
bi xi · 3 + 3x + x2
l
i=0
l=1
We have obtained an integer factorization. Multiplying the polynomials and comparing
the coefficients yields once more
9
9
9
= b0 · 3 ,
= b 0 · 3 + b1 · 3 ,
= b0 · 1 + b1 · 3 + b2 · 3
1
2
3
9
9
9
= b1 · 1 + b2 · 3 + b3 · 3 ,
= b2 · 1 + b3 · 3 + b4 · 3 ,
= b3 · 1 + b4 · 3 + b5 · 3
4
5
6
9
9
9
= b4 · 1 + b5 · 3 + b6 · 3 ,
= b5 · 1 + b6 · 3 ,
= b6 · 1
7
8
9
We get the constant coefficients a0 = 9, b0 = 3 and c0 = 3 and the leading coefficient
b6 = 1. As in the general case, we to use the inductive argument to conclude that 3
is a divisor of the coefficients b0 , . . . , b5 . That is all we need. All assumptions in the
Eisenstein criterium are satisfied and hence Φ9 (1 + x) and Φ9 (x) are irreducible over the
integers.
Remark. The explicit result Φ9 (1 + x) = 3 + 9x + 18x2 + 21x3 + 15x4 + 6x5 + x6 is not
needed.
1.4
Descartes’ Rule of Signs
Definition 1.1 (Degree of a polynomial). We say that the polynomial P ∈ C[z] has
degree n provided that an 6= 0 is the highest nonzero coefficient, and write deg(P ) = n.
The constant nonzero polynomial has degree zero. The zero polynomial has degree ∞
or undefined. 9
Remark. For any two polynomials deg(P Q) = deg(P ) + deg(Q).
9
—depending on the taste of the author.
63
Proposition 1.8 (Descartes’ rule of signs). Let P (x) = an xn + an−1 xn−1 + · · · + a0
be a polynomial with real coefficients. The number of real positive zeros of P (x) is less
or equal the number of sign changes in the sequence an , an−1 , . . . , a0 ; with the zero terms
deleted.
The assertion holds even counting zeros with their multiplicities. Furthermore, the
difference between the number of sign changes and the number of positive zeros, counted
with multiplicities, is even.
10 Problem 1.3. Produce at least five polynomials of degree at least three, four,
five—with zeros you have chosen conveniently. Multiply out and check Descartes’ rule
of signs. Look for covering different possibilities, multiple zeros, and so on.
Examples. (i) (x − 1)4 = x4 − 4x3 + 6x2 − 4x + 1 has 4 sign changes and four positive
roots.
(ii) (x − 1)4 − 1 = x4 − 4x3 + 6x2 − 4x has 3 sign changes and one positive roots x1 = 2.
The other roots are 0, 1 ± i.
(iii) (x − 1)4 + 1 = x4 − 4x3 + 6x2 − 4x + 2 has 4 sign changes and no positive root.
√
.
The roots are 1 + (±1±i)
2
(iv) (x − 1)4 + 64 = x4 − 4x3 + 6x2 − 4x + 65 has 4 sign changes and no positive root.
The roots are −1 ± 2i, 3 ± 2i.
Proof of Descartes’ rule of signs. Prove the assertion by induction on the degree n. The
assertion is true for n = 0 and n = 1. For induction assumption, assume that the
assertion holds for all polynomials with degree less than n.
Given is a polynomial of degree n. We can assume that an > 0. and let c count the
number of sign changes in the sequence an , an−1 , . . . , a0 . Let 0 < x1 ≤ x2 ≤ · · · ≤ xN be
the increasing list of positive zeros of the polynomial P , counting multiplicities. Too,
we write N = N (P, (0, ∞)) for the count of zeros of P .
In the case that a0 = 0, the polynomial x−1 P (x) has degree n − 1, less than n, but
the same number of positive zeros and sign changes. Hence the assertion follows from
the induction assumption.
Assume a0 6= 0 and let ai 6= 0 be next nonzero coefficient. We need to distinguish
the cases a0 ai > 0 and a0 ai < 0.
(a) case a0 ai > 0. The divided derivative Q(x) := x−i+1 P 0 (x) has the same number c
of sign changes in the sequence nan , (n − 1)an−1 , . . . , iai of its coefficients.
Furthermore, Q(x) has at least one zero in each of the intervals
(1.6)
(0, x1 ), (x1 , x2 ), . . . , (xN −1 , xN )
64
hence at least N positive zeros. Hence N (P, (0, ∞)) ≤ N (Q, (0, ∞)). Note that a
zero of P 0 and hence Q exists in the open interval (0, x1 ), since |P | restricted to
[0, x1 ] does not achieve its maximum at 0, but in the open interval (0, x1 ). In the
case of a multiple zero a > 0, the interval (a, a) has to be replaced by the single
element set {a}. By this interpretation of the list (1.6), we see that the assertion
remains true counting multiplicities. Since deg(Q) < n, the induction assumption
yields N (P, (0, ∞)) ≤ N (Q, (0, ∞)) ≤ c, as to be shown.
(b) case a0 ai < 0. The divided derivative Q(x) := x−i+1 P 0 (x) has only c − 1 of sign
changes in the sequence of its coefficients. Furthermore, Q(x) has at least one zero
in each of the intervals
(x1 , x2 ), . . . , (xN −1 , xN )
hence at least N − 1 positive zeros. Hence N (P, (0, ∞)) ≤ N (Q, (0, ∞)) + 1. Note
that the interval (0, x1 ) has to be left out in this case, since now |P | restricted
to [0, x1 ] can achieve its maximum at 0. With the interpretation of any interval
(a, a) with a > 0 as the single element set {a}, we see the assertion remains true
counting multiplicities. Since deg(Q) < n, the induction assumption yields
N (Q, (0, ∞)) ≤ c − 1
N (P, (0, ∞)) ≤ N (Q, (0, ∞)) + 1 ≤ c − 1 + 1 = c
as to be shown.
It remains to check the last assertion about the even difference. 10 The assertion clearly
holds for degree zero and one. Note that in any interval (a, b) between two consecutive
zeros P (a) = P (b) of a polynomials, the derivative P 0 has an odd number of zeros. This
yields an even number of extra zeros of P 0 in (0, ∞) beyond the ones claimed above.
Take the case (b): the difference N (Q, (0, ∞))+1−N (P, (0, ∞)) is even, Too, by the
induction assumption, the difference c − 1 − N (Q, (0, ∞)) is even. Hence the difference
c − N (P, (0, ∞)) is even, too.
In the case (a), the interval (0, x1 ) has to be included. Therefore, this time, the
difference N (Q, (0, ∞)) − N (P, (0, ∞)) is even. Too, by the induction assumption, the
difference c − N (Q, (0, ∞)) is even. Hence, in the end, the difference c − N (P, (0, ∞))
turns out to be even in all cases.
For any polynomial P (x) = an xn + an−1 xn−1 + · · · + a0 with real coefficients, we define
c+ : the number of sign changes in the sequence an , an−1 , . . . , a0 —with zero terms
deleted;
N+ : the number N (P, (0, ∞)) of positive zeros of P , counting their multiplicities;
10
This assertion has been added by Gauss.
65
c− : the number of sign changes in the sequence (−1)n an , (−1)n−1 an−1 , . . . , a0 —with
zero terms deleted;
N− : the number N (P, (−∞, 0)) of negative zeros of P , counting their multiplicities;
c0 : the multiplicity—also denoted by N (P, {0})— of z = 0 as a zero of polynomial P .
Hence ai = 0 for i < c0 , but ai 6= 0 for i = c0 .
Lemma 1.4 (Sum of sign changes). For any sequence an , an−1 , . . . , a0 with an 6= 0,
the sum c+ + c0 + c− is less or equal to n, and the difference n − (c+ + c0 + c− ) ≥ 0 is
even.
Furthermore, if n = c+ + c0 + c− , there cannot be more than one zero coefficient
between two nonzero coefficients; and for such the zero coefficient, the two nonzero
neighbors have different signs.
Proof. Prove the assertion by induction on the degree n. The assertion is true for
n ≤ 1. For induction assumption, assume that the assertion holds for all sequences
ak , ak−1 , . . . , a0 with k < n and ak 6= 0.
Given any sequence an , an−1 , . . . , a0 with an 6= 0. In the case an is the only nonzero
term, the assertion holds. Otherwise, let k < n be the maximal index with ak 6= 0. The
two sequences (±1)n an , (±1)n−1 an−1 , . . . , a0 have the added signs changes c+ + c0 + c− —
these are more than the added signs changes c0+ + c0 + c0− of the two shorter sequences
(±1)k ak , (±1)k−1 ak−1 , . . . , a0 . The difference are the additional sign changes occurring
between (±1)n an and (±1)k ak which are
(
1
if n − k is odd
c+ + c− − c0+ − c0− =
0 or 2
if n − k is even
By induction assumption, the difference k − (c0+ + c0 + c0− ) ≥ 0 is even. Hence the
difference n − (c+ + c0 + c− ) ≥ 0 is even, too.
Furthermore, if n = c+ + c0 + c− , then k = c0+ + c0 + c0− , too.
(
1=n−k
if n − k is odd
c+ + c− − c0+ − c0− =
2=n−k
if n − k is even
Thus there cannot be more than one zero coefficient between an 6= 0 and ak 6= 0. If
n − k = 2, then an ak < 0.
Corollary. A polynomial with real coefficients of degree n has at most n real zeros,
counting multiplicities.
Corollary. For a real polynomial of degree n with n real zeros, Descartes’ rule of signs
gives the exact number of positive and negative zeros.
Furthermore, there cannot be more than one zero coefficient between two nonzero
coefficients. For such the zero coefficient, the two nonzero neighbors have different signs.
66
Corollary. The even or odd coefficients of an even or odd real polynomial with only real
zeros have alternating signs and are nonzero—except for the lowest ones.
Proof. Put together, the rule of Descartes 1.8 and the Lemma 1.4 imply
N (P, (0, ∞)) + N (P, {0}) + N (P, (−∞)) ≤ c+ + c0 + c− ≤ n
(1.7)
In the case of n real zeros, equality holds everywhere, and hence
N (P, (0, ∞)) = c+ ,
N (P, (−∞)) = c− , and c+ + c0 + c− = n
Because of the last equality, as shown in the Lemma, the list of coefficients cannot have
more than one zero coefficient between two nonzero coefficients, and in that case the
two nonzero neighbors have different signs.
2
Tschebychev polynomials
Definition 2.1 (Tschebychev polynomials). The Tschebychev polynomials of first
and second kind are defined by
Tn (cos t) = cos nt and
Un (cos t) =
sin(n + 1)t
sin t
for integers n = 0, 1, 2, . . . .
Proposition 2.1. Both polynomials Tn and Un satisfy the same recursion formula
Tn+1 = 2xTn − Tn−1 Un+1 = 2xUn − Un−1
with the initial data T0 = U0 = 1 and T1 = x but U1 = 2x, respectively. Both Tn and Un
are integer polynomials of degree n. They are even for even n, odd for odd n. Moreover
Tn (1) = 1 and Un (1) = n + 1 for all n.
The Tn satisfy the composition formula Tn ◦ Tm = Tnm . The polynomials 2Tn (v/2)
are integer polynomials of the variable v.
n
0
1
2
3
4
5
6
7
8
9
Tn
1
x
−1 + 2x2
−3x + 4x3
1 − 8x2 + 8x4
5x − 20x3 + 16x5
−1 + 18x2 − 48x4 + 32x6
−7x + 56x3 − 112x5 + 64x7
1 − 32x2 + 160x4 − 256x6 + 128x8
9x − 120x3 + 432x5 − 576x7 + 256x9
67
Un
1
2x
−1 + 4x2
−4x + 8x3
1 − 12x2 + 16x4
6x − 32x3 + 32x5
−1 + 24x2 − 80x4 + 64x6
−8x + 80x3 − 192x5 + 128x7
1 − 40x2 + 240x4 − 448x6 + 256x8
10x − 160x3 + 672x5 − 1024x7 + 512x9
Lemma 2.1. Assume that n = 2m + 1 is odd. Let z = x + iy = eit and v = 2x = 2 cos t.
(2.1)
n
z −1
=
z−1
n−1
X
"
z k = (x + iy)m 1 + 2
k=0
m
X
#
"
cos kt = (x + iy)m 1 + 2
k=1
m
X
#
Tk (cos t)
k=1
(2.2)
m n−1 Y
2π k
2πi k
zn − 1 Y
m
x − cos
=
z − exp
= (2x + 2iy)
= (x + iy)m R(2x)
z−1
n
n
k=1
k=0
The polynomial
(2.3)
m m
Y
X
2π k
R2m+1 (2x) = 2
x − cos
=1+2
Tk (x)
2m + 1
k=1
k=1
m
is a monic integer polynomial of the variable v = 2x of degree m = (n − 1)/2.
Proof.
m
m
n−1
X
X
X
zn − 1
k
−m
k
(z k + z −k ) =: R2m+1 (2x)
z
=
1
+
=
z
z
=
(z − 1)z m
k=1
k=−m
k=0
To prove that these are integer polynomials of the variable v = 2x = z +z −1 , inductively
for all m, we use the binomial formulas for (z + z −1 )m . The example with n = 5, m = 3
has already been calculated in the formula (??) from the section about the heptagon.
Too, we can substitute (z k + z −k )/2 = cos kt and introduce the Tschebychev polynomials defined via Tk (cos t) = cos kt:
m
m
m
m
X
X
X
X
zn − 1
z k + z −k
k
=
z
=
1
+
2
=
1
+
2
cos
kt
=
1
+
2
Tk (cos t)
(z − 1)z m k=−m
2
k=1
k=1
k=1
with variable cos t = (z +z −1 )/2 = v/2. We see, again by induction on m, that 2Tm (v/2)
and hence Tm are integer polynomials. Combining the factors from conjugate complex
roots yields
(2.4)
Y
n−1 m zn − 1 Y
2πi k
2πi k
2πi k
=
z − exp
=
z − exp
z − exp −
z−1
n
n
n
k=0
k=1
m
m
Y
Y
z + z −1
2π k
2π k
2
m
=
z − 2z cos
+ 1 = (2z)
− cos
n
2
n
k=1
k=1
With x = (z + z −1 )/2 and v = 2x, the formula 2.2 is confirmed.
We need to factor the polynomial K ∈ Z[z] over the integers.
68
Proposition 2.2. Assume that n is odd. The polynomial K(z) ∈ Z[z] is irreducible
over the integers if and only if the polynomial R is irreducible over the integers.
Proof. Assume that there is an integer factoring K = P Q. The conjugate complex of
any root eiτ of P is a root e−iτ of P , too. Since n is odd, the real root −1 does not
occur, and the real root 1 has been divided out in the beginning. Hence the degree
of deg P = 2s is even and the factors with conjugate complex roots can be combined
as in formula (2.4). Thus we obtain an integer factoring of the polynomial R(v). The
converse is even easier to confirm.
The item k ≤ 8 of the sequence of formulas (2.3) are given.
m m
Y
X
2π k
2
Tk (x)
x − cos
=1+2
n
k=1
k=1
2π
2 x − cos
= 1 + 2T1
= 1 + 2x
3
4π
2π
x − cos
= 1 + 2T1 + 2T2
= −1 + 2x + 4x2
4 x − cos
5
5
3 3
Y
X
2π k
Tk
= −1 − 4x + 4x2 + 8x3
8
x − cos
=1+2
7
k=1
k=1
4
4
Y
X
2π k
16
x − cos
Tk = 1 − 4x − 12x2 + 8x3 + 16x4
=1+2
9
k=1
k=1
m
m m
Y
X
2π k
2
x − cos
=1+2
Tk (x)
n
k=1
k=1
5 Y
2π k
32
x − cos
= 1 + 6x − 12x2 − 32x3 + 16x4 + 32x5
11
k=1
6 Y
2π k
= −1 + 6x + 24x2 − 32x3 − 80x4 + 32x5 + 64x6
64
x − cos
13
k=1
7 Y
2π k
128
x − cos
= −1 − 8x + 24x2 + 80x3 − 80x4 − 192x5 + 64x6 + 128x7
15
k=1
8 Y
2π k
= 1 − 8x − 40x2 + 80x3 + 240x4 − 192x5 − 448x6 + 128x7 + 256x8
256
x − cos
17
k=1
m
69
One sees that the substitution v = 2x eliminates many powers of two.
2π
(2.5)
v − 2 cos
=1+v
3
2 Y
2π k
v − 2 cos
(2.6)
= −1 + v + v 2
5
k=1
4 Y
2π k
(2.7)
v − 2cos
= 1 − 2v − 3v 2 + v 3 + v 4
9
k=1
7 Y
2π k
(2.8)
v − 2 cos
= −1 − 4v + 6v 2 + 10v 3 − 5v 4 − 6v 5 + v 6 + v 7
15
k=1
8
Y
2π k
v − 2 cos
(2.9)
= 1 − 4v − 10v 2 + 10v 3 + 15v 4 − 6v 5 − 7v 6 + v 7 + v 8
17
k=1
The polynomial R9 has the divisor R3 , as expected. The irreducible factors are:
R9 = 1 − 2v − 3v 2 + v 3 + v 4 = (1 + v) · (1 − 3v + v 3 )
Moreover, we see that R15 has the divisor R3 · R5 . This is expected since
R15
=????????????????????
R3 · R5
The irreducible factors of R15 are:
−1 − 4v + 6v 2 + 10v 3 − 5v 4 − 6v 5 + v 6 + v 7 = (1 + v) · (−1 − 3v + 9v 2 + v 3 − 6v 4 + v 6 )
= (1 + v) · (−1 + v + v 2 ) · (1 + 4v − 4v 2 − v 3 + v 4 )
The Five Quadrable Lunes
Brian Shelburne
Faculty, Wittenberg University
Abstract Number: 5. A lune, the crescent shaped area formed by the intersection of
two circles, is quadrable if it is possible to construct a square with the same area using
only straight-edge and compass. Hippocrates of Chios (ca. 430 BCE) is credited with
discovering three such Lunes; two more were discovered in the 18th century. In the 20th
century Tschebatorev and Dorodnov proved there were only five. This talk will examine
the equation used to derive the five cases and present as examples the two cases which
are not normally seen.
70
3
3.1
Polynomials over the Complex Numbers
A proof of the Fundamental Theorem of Algebra
Given is any polynomial
P (z) =
n
X
ak z k
k=0
with complex coefficients ak ∈ C. We want to estimate its zeros.
10 Problem 3.1. Prove for any zero of a monic polynomial:
n
P (z) = z +
n−1
X
ak z k
k=0
"
|z| ≤ R0 := max 1,
n−1
X
#
|ak |
k=0
Answer. The assertion clearly holds for all zeros |z| ≤ 1. On the other hand, |z| > 1
implies
!
n−1
n
n
X
X
X
|ak |
|an−k ||z|n−1 = |z|n−1 |z| −
|an−k ||z|n−k ≥ |z|n −
|P (z)| ≥ |z|n −
k=1
k=1
k=0
Now P (z) = 0 and |z| > 1 imply that the right-hand side is non-positive, and hence
|z| ≤
n−1
X
|ak | ≤ R0
k=0
10
3.2. Choose any R > R0 . Show that the monic polynomials P (z) =
P Problem
k
z n + n−1
a
z
and
Q(z) = z n have the same number of zeros with absolute value
k=0 k
|z| ≤ R.
Solution. We use the Theorem of Rouché for the functions P and Q in the disk B(R, 0)—
of radius R around 0. On the boundary ∂B(R, 0), we get |z| = R > 1, and hence
n−1
n−1
X
X
k
|P (z) − Q(z)| = ak z ≤
|ak | ≤ R0 < R < Rn = |Q(z)|
k=0
k=0
Thus all assumptions of Rouché’s Theorem are satisfied. Hence the assertion follows.
71
Clearly, the result implies even that the polynomials P and Q have the same number
of zeros with absolute value |z| ≤ R0 —in the closed disk B(R0 , 0) = {z ∈ C : |z| ≤ R0 }.
By the last Problem (3.1), all zeros z of polynomial P lie in the disk B(R0 , 0). On the
other hand, the polynomial Q(z) = z n obviously has n zeros, counting multiplicities,
and they all lie in the disk B(R0 , 0). Hence we get the celebrated result:
Main Theorem (The Fundamental Theorem of Algebra). Every polynomial of
degree n has exactly n complex zeros, counting their multiplicities.
3.2
Meditation on Descartes’ rule of signs
Which insight does one gain by combining the Fundamental Theorem of algebra 3.1
with Descartes’ rule of signs? From Descartes’ rule of signs and Lemma 1.7, we get for
the number of complex zeros of a real polynomial P :
(3.1)
N (P, C \ R) = n − N (P, (0, ∞)) − N (P, {0}) − N (P, (−∞))
n − c+ − c0 − c− c+ − N+ c− − N−
+
+
=2
2
2
2
where all three fractions are nonnegative integers.
Corollary. A polynomial with real coefficients of degree n has at least n − c+ − c0 − c−
complex zeros. They occur in pairs of conjugate complex ones.
10 Problem 3.3 (The sleepless Descartes). Give eight or more examples of
polynomials which show that all three fractions in equation (3.1) can be zero or nonzero
in all combinations. Polynomials of degree up to six are needed!
P
only real zeros
x2 + 1
(x + 2)((x − 1)2 + 1)
(x − 2)((x + 1)2 + 1)
x2 (x − 1)2 + 1
x2 (x + 1)2 + 1
(x2 − 1)2 + 1
(x2 + 1)(x4 − 2x2 + 2)
distributed n − c+ − c0 − c−
0
x2 + 1
2
3
x − 2x + 4
0
x3 − 2x − 4
0
4
3
2
x − 2x + x + 1
2
4
3
2
x + 2x + x + 1
2
x4 − 2x2 + 2
0
6
4
x −x +2
2
c+ − N+
0
0
2
0
2
0
2
2
2
c− − N−
0
0
0
2
0
2
2
2
Definition 3.1. For any integers a and b, the notation
”b ≥ a”
means b ≥ a and b − a is even. Similar notation is used with other comparison signs.
Similarly, for any integer m ≥ 1, the notation
means b ≤ a and b − a is divisible by m.
72
m
”b ≤ a”
In the following I want to count the zeros of a real polynomial in the right- and left
half plane.
H+ : Let H+ := N (P, {x + iy : x > 0} denote the number of zeros in the open right
half plane;
H− : Let H− := N (P, {x + iy : x < 0} denote the number of zeros in the open right
half plane;
H0 : denote the number of zeros on the imaginary axis.
10 Problem 3.4. Prove that a real polynomial with only imaginary roots is always
even or odd. Furthermore, there are no sign changes in its sequence of coefficients, and
all even respectively odd coefficients between the lowest nonzero one and the leading one
are nonzero.
Proof. Since P (z) = P (z) for all complex z, the zeros occur in conjugate complex pairs.
Because all roots are assumed to be purely imaginary, the roots occur in pairs of zi and
zi = −zi , with the same multiplicity. Hence the polynomial is even or odd.
P (x) = xp [ap + ap+2 x2 + · · · + an xn−p ] = xp Q(x2 )
(3.2)
where n − p is even and deg Q = (n − p)/2. The polynomial Q has only real negative
zeros. Hence by Corollary 1.4 to the rule of Descartes, its sequence of coefficients has
no sign changes.
Proposition 3.1. An odd or even real polynomial (3.2) has at least as many roots in
the right half plane as there are sign changes in the sequence
ap , ap+2 , ap+4 , . . . an
of its coefficients. On the other hand, the number of roots on the positive imaginary axis
is at most the number q− of sign changes in the alternated sequence
ap , −ap+2 , ap+4 , · · · ± an
of its coefficients. In both assertions, the differences are even. Put together, we even get
(3.3)
2
2
2
2
c+ ≤ m − q− ≤ N (P, {z : Re z > 0}) ≤ N (P, {z : Re z ≥ 0}) ≤ c0 + m + q− ≤ n − c+
4
(3.4)
N (P, {z : Re z = 0}) ≤ c0 + 2q−
Proof. Let n = c0 + 2m and p = c0√and R(z) := z −p P (z). Since the polynomial R(z)
is even, the polynomial Q(z) := R( z) is well defined, and obviously Q(0) = R(0) 6= 0
and deg Q = m.
73
The mapping z → w = z 2 is a bijection of the right half plane {x + iy : x > 0}
onto the slitted plane C \ (−∞, 0]. Hence the zeros of polynomial R(z) in the right half
plane are mapped to the zeros of w 7→ Q(w) = R(z) in the slitted plane.
We use—for polynomial Q—the Fundamental Theorem of Algebra, and Descartes’
rule of signs, and Lemma 1.4 on the sum of sign changes. Hence we conclude
N (P, {z : Re z > 0}) = N (R, {z : Re z > 0}) = N (Q, C \ (−∞, 0]})
2
2
= m − N (Q, (0, ∞)) ≥ m − q− ≥ q+ = c+
The mapping z → w = z 2 is a bijection of the positive imaginary axis to the negative
real axis. Again counting zeros of polynomial z → R(z) and zeros of w 7→ Q(w) = R(z),
we get
2
N (P, {iy : y > 0}) = N (R, {iy : y > 0}) = N (Q, (−∞, 0)}) ≤ q−
4
N (P, {z : Re z = 0} = c0 + 2N (Q, (−∞, 0)}) ≤ c0 + 2q−
10 Problem 3.5. Is it true or not.
For a real polynomial, the number of real or complex zeros with positive real part
is less or equal the number of sign changes in the sequence an , an−1 , . . . , a0 ;
with the zero terms deleted.
Remark. This assertion is wrong. Here are examples, which show:
The number of real or complex zeros of P (x) with positive real part can be
less, equal or even greater the number of sign changes in the sequence
an , an−1 , . . . , a0 ; with the zero terms deleted. In each case, the difference
is even.
P
(x + 1)4 + 64
(x − 1)4 + 64
(x − 1)4 − 1
(x + 2)((x − 1)2 + 1)
distributed
zeros
x + 4x + 6x + 4x + 65 1 ± 2i, −3 ± 2i
x4 − 4x3 + 6x2 − 4x + 65 3 ± 2i, −1 ± 2i
x4 − 4x3 + 6x2 − 4x
0, 2, 1 ± i
3
x − 2x + 4
1 ± i, −2
4
3
2
H+
2
2
3
2
c+
0
4
3
2
10 Problem 3.6. Is it true or not.
For a real polynomial, the number of real or complex zeros with positive real part
is at least half the number of sign changes in the sequence an , an−1 , . . . , a0 ;
with the zero terms deleted.
Remark. This assertion is wrong. Here are examples, which show:
74
The number of real or complex zeros of P (x) with positive real part can
be less, equal, or greater than half the number of sign changes in the
sequence an , an−1 , . . . , a0 ; with the zero terms deleted.
P
(x − 3)((x + 1)2 + 9)
(x − 1)(x2 + 1)
(x − 1)4 + 64
(x + 1)4 + 64
distributed
zeros
x3 − x2 + 4x − 30
3, −3 ± i
x3 − x2 + x − 1
1, ±i
4
3
2
x − 4x + 6x − 4x + 65 3 ± 2i, −1 ± 2i
x4 + 4x3 + 6x2 + 4x + 65 1 ± 2i, −3 ± 2i
H+
1
1
2
2
c+
3
3
4
0
10 Problem 3.7. List all examples mentioned up to now—and some more interesting ones.
75
P
P distributed
only real zeros
zeros
only imaginary zeros
x2 + 1
x2 + 1
even or odd polynomial
±i
(x2 + 1)2 − 2
x4 + 2x2 − 1
(x2 + 1)(x4 − 6x2 + 25)
x6 − 5x4 + 19x2 + 25
(x2 − 1)2 + 1
x4 − 2x2 + 2
(x2 + 1)(x4 − 2x2 + 2)
x6 − x4 + 2
(x − 3)((x + 1)2 + 9)
x3 − x2 + 4x − 30
(x − 1)(x2 + 1)
x3 − x2 + x − 1
(x − 1)(x − 2)(x2 + 1)3
x8 − 3x7 + 5x6 − 9x5 + 9x4 − 9x3 + 7x2 − 3x + 2
(x + 2)((x − 1)2 + 1)
x3 − 2x + 4
((x − 2)2 + 2)((x + 1)2 + 2)
x4 − 2x3 + x2 + 18
(x + 1)4 + 64
x4 + 4x3 + 6x2 + 4x + 65
(x − 1)4 − 1
x4 − 4x3 + 6x2 − 4x
(x − 1)4 + 1
x4 − 4x3 + 6x2 − 4x + 2
(x − 1)4 + 4
x4 − 4x3 + 6x2 − 4x + 5
((x − 1)4 + 4)2
x8 − 8x7 + 28x6 − 56x5 + 78x4 − 88x3 + 76x2 − 40x + 18
x2 (x − 2)2 + 1
x4 − 4x3 + 4x2 + 1
76
±
p√
p√
2 − 1, ±i
2+1
±2 ± i, ±i
√
± 1±i
√
±i, ± 1 ± i
3, −3 ± i
1, ±i
1, 2, (±i)3
1 ± i, −2
√
√
2 ± i 2, −1 ± i 2
1 ± 2i, −3 ± 2i
2, 1 ± i, 0
1+
(±1±i)
√
2
2 ± i, ±i
(2 ± i)2 , (±i)2
1±
√
1±i
H+
c+
c+
c+
0
0
0
0
≥ c+
c+
1
1
2
2
2
2
2
2
1
3
1
3
2
8
2
2
2
2
2
0
3
3
4
4
2
4
4
8
2
2
H−
c−
c+
c−
0
0
0
0
≥ c+
c+
1
1
0
2
2
2
2
2
2
0
0
0
0
0
1
1
2
0
2
4
0
0
0
0
0
0
4
0
2
0
Lemma 3.1. The number of positive imaginary zeros of a real polynomial is at most the
number of the sign changes q− (odd) and q− (even) in the nonempty alternating sequences
a0 , −a2 , a4 , −a6 , . . .
a1 , −a3 , a5 , −a7 , . . .
of all nonzero coefficients of one parity, either odd or even.
Proof. Let P (x) + σP (−x) = xp Qσ (x2 ) with parity σ = ±1. If P (iy) = 0 with y > 0,
then P (−iy) = 0, and hence Qσ (−y 2 ) = 0. Hence Descartes rule implies
N (P, (0, i∞)) ≤ N (Q, (−∞, 0)) ≤ q− (σ)
as long as Q is not identically zero. This happens for σ = −1 and P even, as well as
for σ = 1 and P odd. Otherwise, one can use both parities. In general, one gets two
inequalities for both parities.
10 Problem 3.8. Is it true or not.
For a real polynomial, the number of real or complex zeros with positive or negative real part
is at least twice the minimum of the number of sign changes in the two
sequence (±1)n an , (±1)n−1 an−1 , . . . , a0 ; with the zero terms deleted.
Conjecture 1 (Sleepy Descartes Conjecture). The nonnegative integer numbers
c0 , c+ , c− , N+ , N− , H+ , H− characterizing a real polynomial of degree deg P can take
any values compatible with the restrictions
2
2
(3.5)
N+ ≤ c+ , , N− ≤ c−
(3.6)
N+ ≤ H+ , N− ≤ H− ,
(3.7)
c+ + c− + c0 ≤ deg P ,
(3.8)
H+ + H− + c0 ≤ deg P
(3.9)
2 min(c+ , c− ) ≤ H+ + H−
2
2
2
2
2
Proof of the sleepy Descartes conjecture.
3.3
Estimation of zeros with Rouché’s Theorem
10 Problem 3.9. Assume that
(3.10)
|an−1 | >
n−2
X
|ak | + |an |
k=0
P
k
Use the Theorem of Rouché to show that the polynomial P (z) = n−1
k=0 ak z has exactly
n − 1 zeros of absolute value less than one. Under the additional assumption an 6= 0,
the polynomial has exactly one zero with absolute value greater than one.
77
Solution. We use the Theorem of Rouché for the functions P and Q(z) = an−1 z n−1 in
the unit disk B(1, 0) = {z ∈ C : |z| ≤ 1}. Indeed, on the boundary of this disk, we get
|z| = 1, and
n−2
X
X
|P (z) − Q(z)| = ak z k ≤
|ak | + |an | < |an−1 | = |Q(z)|
k6=n−1
k=0
All assumptions of Rouché’s Theorem are satisfied, and the first assertion follows.
The second assertion can for example be shown by applying the Theorem of Rouché
to the functions
f (z) = z n P (z −1 ) = a0 z n + · · · + an−1 z + an
—also called the mirror polynomial of P — and the function g(z) = an−1 z, again in the
unit disk B(1, 0). Since again |f (z) − g(z)| < |g(z)| on the boundary ∂B(1, 0), both
functions f and g have exactly one zero inside the unit disk. Under the assumption
an 6= 0, we get z 0 6= 0 from f (z 0 ) = 0. Hence P (z 0−1 ) = z 0−n f (z 0 ) = 0 and |z 0−1 | > 1
yields the n-th zero outside the unit disk.
3.4
Perron’s irreducibility criterium
Proposition 3.2 (The Perron criterium). For a monic polynomial to be irreducible
in the ring Z[z], it is sufficient that all its constant coefficient a0 6= 0 and the second
highest coefficient dominates
(3.11)
|an−1 | >
n−2
X
|ak | + 1
k=0
Corollary. An an integer monic polynomial with a0 6= 0 and exactly one root of absolute
value greater than one and no root of absolute value equal to one, is irreducible.
The Proposition follows from the Corollary. Indeed, by Problem (3.2), the polynomial has n − 1 zeros of absolute value in 0 < |z| < 1 and one zero of absolute value
larger one.
Proof of the Corollary. Assume toward a contradiction that there would exist a rational
factorization. By Gauss’ Lemma, we would get a factorization even into monic integer
polynomials P (z) = B(z)C(z) with deg(B) = s ≤ n − 1 and deg(C) = t ≤ n − 1. Furthermore, we can choose C to be the irreducible factor with the unique zero of absolute
value larger one. Hence B(z) would have only zeros z1 , z2 , . . . zs of absolute value in
(0, 1). Since P is assumed to be monic, B and C are both monic, too. Furthermore
a0 = b0 c0 , and Viëta’s formula implies
(3.12)
0 6= |b0 | = |z1 | · |z2 | · · · |zs | < 1
78
which cannot be true for any integer polynomial.
The only way out of this contradiction is that B = 1 and deg(B) = 0. Hence the
polynomial P is irreducible.
10 Problem 3.10. Try to generalize Perron’s criterium to polynomials which are
not monic. 11
3.5
The limiting case
This subsection investigates zeros and reducibility for the polynomial P (z) =
in the limiting case
(3.13)
|an−1 | =
n−2
X
Pn
k=0
ak z k
|ak | + |an |
k=0
Here is a main result.
Proposition 3.3. Assume the equality (3.13), and additionally 0 6= |an | ≤ |a0 | for
n ≥ 3, or 0 6= |an | < |a0 | for n =P
1 or 2.
Then the polynomial P (z) = nk=0 ak z k has exactly one zero of absolute value greater
than one and at most one simple zero of absolute value equal one.
10 Problem 3.11. Find an integer polynomial satisfying equality (3.13), and
additionally |an | > |a0 | =
6 0 and n ≥ 3, which has no zero of absolute value greater than
one. Can it be even reducible?
10 Problem 3.12. Assume the equality (3.13). Use the logarithmic residue Theorem to show that the polynomial P has at most n − 1 zeros of absolute value less
one—even counting multiplicities.
Why can we not claim existence of exactly n − 1 zeros of absolute values less one?
Show there is at most one zero with absolute value greater one.
Solution of Problem 3.12. Let ε > 0 be small enough such that the circle ∂B(1 − ε, 0)
incloses all zeros P (z) = 0 with |z| < 1. Choose α > 0 small enough such that
α|Q(z)| < |P (z)| on ∂B(1−ε, 0). The Theorem of Rouché 4.2 implies N (P, B(1−ε, 0)) =
N (P + αQ, B(1 − ε, 0))— equal number of zeros for the functions P and P + αQ inside
the disk B(1 − ε, 0).
Since the coefficients of the polynomial P + αQ satisfy the strict inequality (3.9), we
know from the solution of Problem 3.9 that this polynomial has n − 1 zeros of absolute
value less than one. Hence N (P + αQ, B(1 − ε, 0)) ≤ n − 1.
Together, we conclude that N (P, B(1 − ε, 0)) = N (P + αQ, B(1 − ε, 0)) ≤ n − 1, as
claimed. We cannot claim existence of exactly n − 1 zeros of P with absolute values less
11
I did not have any success.
79
one. Indeed, in the limit α → 0, some of the zeros of P + αQ can move from inside to
the boundary ∂B(1, 0) of the unit disk.
We show there is at most one zero with absolute value greater one. Let ε > 0 be small
enough and R > 1 large enough such that annulus A := B(R, 0) \ B(1 + ε, 0) contains
all zeros P (z) = 0 with |z| > 1. Choose α > 0 small enough such that α|Q(z)| < |P (z)|
on its boundary ∂A. The Theorem of Rouché 4.2 and the result of Problem 3.9 imply
N (P, A) = N (P + αQ, A)) ≤ 1, as claimed.
10 Problem 3.13. Assume once more the equality (3.13), and additionally
Pn a0 6= 0k
and an 6= 0. Show by elementary considerations that the polynomial P (z) = k=0 ak z
has at most one zero of absolute value one. Indeed, if all coefficients ak are real, this
zero can possibly only be either 1 or −1.
Proof. Assume there exists a zero with |z| = 1. The estimate
X
X
X
n−1
k
|ak | = |an−1 |
|ak z k | =
|an−1 | = |an−1 z | = ak z ≤
k6=n−1
k6=n−1
k6=n−1
contains all equalities. Equality in the triangle inequality implies that all summands are
nonnegative multiples of any nonzero term. Hence
|ak |
a0 for all k 6= n − 1
|a0 |
X
a0 X
ak z k = −
an−1 z n−1 = −
|ak |
|a
0|
k6=n−1
k6=n−1
a
0
an−1 z n−1 = −
|an−1 |
|a0 |
a0
|an |
an z n =
|a0 |
ak z k =
which rules out—as possible zero—all but one value
(3.14)
z1 = −
|an | an−1
an |an−1 |
In the case of real coefficients, this leaves only either +1 or −1 for a possible zero on
the unit circle.
Rather obviously (z − 1)2 = z 2 − 2z + 1 is a polynomial the coefficients of which
satisfy equality (3.13), but has a double zero.
10 Problem 3.14. Assume once more the equality (3.13), and additionally 0 6=
|an | ≤ |a0 | and n ≥ 3, or 0 6= |an | < |a0 | for n = 1 orP2. Use the previous Problems 3.12
n
k
and 3.13, and show that the polynomial P (z) =
k=0 ak z has exactly one zero of
absolute value greater than one.
80
Solution of Problem 3.14. Assume—towards a contradiction—|z| ≤ 1 for all zeros. Viëta’s
formula, and the assumption would imply
|a0 | = |an | |z1 | · |z2 | · · · |zn | ≤ |an | ≤ |a0 |
Hence there is equality everywhere. All zeros would have absolute values one. By
Problem 3.13, they need all to be equal to the only one possible value z1 , given by
formula (3.14). Hence
P (z) = an (z − z1 )n
n
ak = an
(−z1 )k for all k = 0, . . . , n
k
n
X
2|an−1 | = 2n|an | =
|ak | = |an |2n
k=0
Hence n = 1 or n = 2 and P (z) = an (z − z1 )n , which we ruled out by assumption.
With these steps, we have proved Proposition 3.3. The Corollary 3.4 immediately implies
Proposition 3.4. A monic integer polynomial P ∈ Z[z] satisfying the weak inequality
(3.15)
|an−1 | =
n−2
X
|ak | + 1
k=0
and additionally 0 6= a0 can either
(i) be irreducible,
(ii) or be equal to P = (x + 1)2 or P = (x − 1)2 ,
(iii) or have a factorization P = QR with Q = x + 1 or Q = x − 1, the second factor
R being irreducible.
10 Problem 3.15. Assume that the polynomial P ∈ R[z] satisfies equality (3.13),
and additionally that ak > 0 for all k = 0, 1, . . . , n. Let n ≥ 2. How many positive zeros
does P have. What follows from Descartes’ rule of signs about the negative zeros?
Solution. The sequences of coefficients a0 , a1 , . . . an has 2 sign changes. By the rule of
Descartes number of positive zeros can only be 0 or 2. Since P (1) = 0, there are exactly
two positive zeros, counting multiplicity: N (P, (0, ∞)) = 2.
The sequences of coefficients a0 , −a1 , . . . (−1)n an has n − 2 sign changes. By the rule
of Descartes number of negative zeros can only be N (P, (−∞, 0)) ≤ n − 2, where the
missing ones give pairs of conjugate complex zeros.
10 Problem 3.16. Assume that the polynomial P ∈ R[z] satisfies equality (3.13),
(z)
and additionally that an > 0, and P (1) = 0. Divide Pz−1
and calculate P 0 (1).
81
Proof. All ak ≥ 0 for k 6= n − 1, but an−1 < 0. Hence
R(z) :=
n−2 X
k
X
P (z)
= an z n−1 −
aj z k
z−1
k=0 j=0
R(1) = an −
n−2 X
k
X
n−2
X
aj = an −
(n − 1 − k)ak
k=0 j=0
=
n
X
k=0
kak = P 0 (1)
k=0
P
Putting rk := kj=0 aj for k = 0, . . . , n − 2, and rn−1 := an —and replacing n − 1 by
n yields the following result:
Corollary. Any integer polynomial
n
R(z) = rn z −
n−1
X
rk z k
with 1 ≤ rn ≤ r0 ≤ r1 ≤ r2 ≤ · · · ≤ rn−1
k=0
has exactly one zero of value greater than one, and n − 1 zeros of absolute value less
than one. Furthermore, if rn = 1, then R is irreducible.
Proof. The result is clearly true for n = 1. Let n ≥ 2, and define P (z) = (z − 1)R(z).
The polynomial P ∈ Z[z] satisfies 1 ≤ an+1 ≤ a0 and has degree deg(P ) = n + 1 ≥ 3. It
is left to the reader to check that equality (3.13) holds, now with n replaced by n + 1.
Hence Proposition 3.3 applies.
As a conclusion, we see once more that P has exactly one zero of absolute value
greater than one, and one simple zero of absolute value equal one. Hence the polynomial
R has one zero of absolute value greater than one, and n − 1 zeros of absolute value less
than one. By Corollary 3.4, R is irreducible, provided that it is monic.
4
4.1
The Residue Theorem and its Consequences
The residue theorem and the logarithmic residue theorem
Main Theorem (The residue theorem). Let Ω ⊆ C be any domain and γ be a closed
piecewise smooth path in Ω. Let f be a function that is analytic in Ω except for isolated
singularities.
Assume that the path γ is contractible to a point inside Ω, and assume that the
function f has no singularities on the path γ. Then
I
X
1
(4.1)
f (z) dz =
n(γ, aj )Res (f, aj )
2πi γ
j
82
where n(γ, aj ) are the winding numbers of the path γ around the singularities aj of the
function f and Res (f, aj ) are the residues at them.
Corollary (The residue theorem in the simple form). Let γ be a simple closed
piecewise smooth curve, and let Ω ⊆ C be a domain that contains both γ and the interior
domain D surrounded by γ.
Let f be a function that is analytic in Ω except for isolated singularities. Assume
that the function f has no singularities on the path γ. Then
I
X
1
f (z) dz =
Res (f, aj )
(4.2)
2πi γ
j
where the sum runs over the singularities aj of the function f in the interior domain D
of γ.
Definition 4.1 (Multiplicity of singularity). For any meromorphic function f with
either an isolated singularity or zero at z0 , the multiplicity µ(f, z0 ) is defined to be the
multiplicity of the zero, or the negative multiplicity of the pole, respectively. In other
words, the Laurent expansion of f at z0 has a leading term a(z − z0 )µ(f,z0 ) with a 6= 0.
Theorem 4.1 (The argument principle (logarithmic residue theorem)). Let
Ω ⊆ C be any domain and γ be a closed piecewise smooth path in Ω. Let f be a nonzero
meromorphic function in Ω.
Assume that the path γ is contractible to a point inside Ω, and assume that the
function f has no zeros or poles on the path γ. Then
I 0
X
X
f (z)
1
dz =
n(γ, aj )µ(f, aj ) −
n(γ, pk )µ(g, bk )
(4.3)
2πi γ f (z)
j
k
Here n(γ, aj ) is the winding number of the path γ around the singularity aj , whereas
µ(f, aj ) is the multiplicity of the singularity or zero.
Remark. In the two sums, zeros and poles are counted according to their multiplicities,
which is negative for the poles.
Reason for the logarithmic residue theorem. If f has a pole or zero at z0 , the Laurent
expansions in a neighborhood with small |z − z0 | are
f (z) = a(z − z0 )µ + b(z − z0 )µ+1 + . . .
f 0 (z) = aµ(z − z0 )µ−1 + b(µ + 1)(z − z0 )µ + . . .
f 0 (z)
µ
b
=
+ + ...
f (z)
z − z0 a
0
Hence the residue Res ( ff , z0 ) = µ = µ(f, z0 ) is just the multiplicity of the zero or pole
of f . Now the result follows from the residue Theorem.
83
Corollary (The logarithmic residue theorem in its simple form). Let γ be a
simple closed piecewise smooth curve, and let Ω ⊆ C be a domain that contains both γ
and the interior domain D surrounded by γ. Let f be a meromorphic function in Ω.
Assume that the function f has no zeros or poles on the path γ. Then the integral
I 0
f (z)
1
dz = N − P
(4.4)
2πi γ f (z)
equals the number N of zeros of f minus the number P of poles in the interior domain
of γ. Both zeros and poles have to be counted according to their multiplicities.
4.2
Comparing two functions
Proposition 4.1 (Comparison at the zeros of two functions). Let Ω ⊆ C be any
domain and γ be a closed piecewise smooth path in Ω. Let f, g be any meromorphic, and
h be any holomorphic functions in Ω.
Assume that the path γ is contractible to a point inside Ω, and assume that the
functions f and g have no zeros or singularities on the path γ. With the quotient
function F := fg , we get
(4.5)
1
2πi
I
h(z)
γ
X
X
F 0 (z)
dz =
h(aj )n(γ, aj )µ(f, aj ) −
h(bk )n(γ, bk )µ(g, bk )
F (z)
j
k
Here n(γ, aj ) is the winding number of the path γ around the singularity aj , whereas
µ(f, aj ) is the multiplicity of the singularity or zero.
Corollary (Comparison at the zeros in the interior domain). Let γ be a simple closed
piecewise smooth curve, and let Ω ⊆ C be a domain that contains both γ and the interior
domain D surrounded by γ. Let f and g be two holomorphic functions in Ω, and assume
that f and g have no zeros on the path γ. Put F := fg .
(4.6)
1
2πi
I
γ
F 0 (z)
dz = N (f, D) − N (g, D)
F (z)
is the number N (f, D) of zeros of f in the interior domain of γ, counted according to
their multiplicities—minus N (g, D), the number of zeros of the denominator g.
Main Theorem (Theorem of Rouché). Let γ be a simple closed piecewise smooth
curve, and let Ω ⊆ C be a domain that contains both γ and the interior domain D
surrounded by γ.
Let the two holomorphic functions f and g be defined in Ω, and assume that they
satisfy the strict inequality |f (z) − g(z)| < |f (z)| + |g(z)| for all z ∈ γ on the path.
Then the equality N (f, D) = N (g, D) holds for the number of zeros of f , and g,
respectively, in the interior domain of γ, counted according to their multiplicities.
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Remark. The more traditional form of the theorem makes the stronger assumption
|f (z) − g(z)| < |f (z)| on the path z ∈ γ.
Theorem 4.2 (Theorem of Rouché in general form). Let Ω ⊆ C be any domain
and γ be a closed piecewise smooth path in Ω. Assume that the path γ is contractible to
a point inside Ω.
Let the two meromorphic functions f and g be defined in Ω, and assume that they
have no poles and satisfy the strict inequality |f (z) − g(z)| < |f (z)| + |g(z)| for all z ∈ γ
on the path. Then the equality
X
X
(4.7)
n(γ, aj )µ(f, aj ) =
n(γ, bk )µ(g, bk )
j
k
holds. Here n(γ, aj ) is the winding number of the path γ around the singularity aj ,
whereas µ(f, aj ) is the multiplicity of the singularity or zero.
The proof depends on the following more general Proposition.
Proposition 4.2. As in the general Theorem of Rouché 4.2, let Ω be any domain, f, g
be any nonzero meromorphic functions in Ω with no poles or zeros on γ. Let γ be a
closed piecewise smooth path which is in Ω contractible to a point.
Put F := fg and assume that the path F (γ) is contractible to a point in the domain
C \ {0}. In other words, we assume that the winding number n(F (γ), 0) = 0.
Then the same equality (4.7), as claimed in the general Theorem of Rouché 4.2 holds.
Completing the proof for the Theorem of Rouché. On the path γ, the quotient F :=
f
satisfies |F (z) − 1| < |F (z)| + 1. Hence F (z) ∈ C \ (−∞, 0], which is a simply
g
connected domain. Hence the image path F (γ) is contractible to a point inside the
domain C \ {0} ⊃ C \ (−∞, 0] ⊃ F (γ). Hence the winding number is n(F (γ), 0) = 0.
Hence the substitution rule yields
I 0
I
1
F (z)
1
dw
N (f, D) − N (g, D) =
dz =
=0
2πi γ F (z)
2πi F (γ) w
The equality N (f, D) = N (g, D) shows that f and g have the same number of zeros in
the interior domain of γ, counting their multiplicities.
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