Download Number Theory Review for Exam 1 ERRATA On Problem 3 on the

Number Theory
Review for Exam 1
ERRATA
On Problem 3 on the Pythagorean triples, the triple should be (z − 9, y, z)
On Problem 6 on the PPT’s, it should be a2 divided by 5.
Pythagorean Triples
The integer a divides the integer b provided there is a integer k with b = ak. We write this as a|b. One of the useful
properties about divisibility is the “2 out of 3” rule that says if an equation involves 3 integer terms (e.g. x2 + y 2 = z 2 ), and
the integer d divides 2 out of 3 of the terms (e.g. x2 and y 2 ), then d divides the third (e.g. z 2 ).
A rational is a number of the
√ which
√ form p/q where p and q are integers with q 6= 0. An irrational is a real number,
is not rational. For example, 3 is irrational (one can prove this as follows: Suppose to the contrary that 3 = p/q
where p and q are integers. Without loss of generality we may assume that p/q is reduced. Squaring and then
multiplying both sides by q 2 gives 3q 2 = p2 . So 3|p2 . This requires that 3|p. So now 9|p2 , and since 3q 2 = p2 ,
we have that 3|q 2 . So 3|q. This contradicts the assumption that p/q is reduced.).
A Pythagorean triple (or PT for short) is a triple (x, y, z) of positive integers such that x2 + y2 = z2 . A primitive
Pythagorean triple (or PPT for short) is a Pythagorean triple with the additional property that the only common divisors
of x, y, z are ±1. Examples of PT’s that are not PPT’s are: (6, 8, 10), (3 ∗ 12, 3 ∗ 5, 3 ∗ 13). and examples of PPT’s are:
(3, 4, 5), (5, 12, 13).
Note that if (x, y, z) is a Pythagorean triple, and k is a positive integer, then (kx, ky, kz) is a Pythagorean triple because
(kx)2 + (ky)2 = k2 (x2 + y2 ) = k2 (z2 ) = (kz)2 . Also, (x, y, z) is a PPT if and only if x and y have no common factors
other than ±1. We can argue this as follows: If d divides both x and y, then since x2 + y 2 = z 2 , the 2 out of 3 rule
implies that d divides z 2 . Since x, y, z have no common factors other than ±1, d = ±1.
Our first goal in the course was to give a complete description of PPT’s. In fact, we gave two complete descriptions.
Along the way to these descriptions we needed to establish some basic facts.
Basic fact 1: If (x, y, z) is a PPT, then exactly one of x and y is odd.
We argued this as follows: if both x and y are even, then z is even, and (x, y, z) is not primitive. If both x and y are odd,
say x = 2k + 1 and y = 2` + 1, then x2 + y 2 = 4(k2 + k + `2 + `) + 2. So when we divide z 2 by 4 we get a remainder of 2.
However, if z is even the remainder when z 2 is divided by 4 is 0, and if z is odd the remainder when z 2 is divided by 4 is 1.
So we are led to an impossible situation. So x and y can’t both be odd.
Basic fact 2: If (x, y, z) is a PPT, then z is odd. This can be argued (using the previous fact) as follows: One of x and
y is odd and the other is even. Since the square of an odd is odd, and the square of and even is even, x2 + y 2
is the sum of an odd number and an even number, which is odd. So z 2 is odd. This requires that z is odd.
Basic fact 3: If u and v are integers with no common divisors other than ±1, and uv is a perfect square, then both u
and v are perfect squares.
This can be argued as follows:We know that uv = k 2 for some integer k. So in the prime factorization of k 2 , every
prime p occurs and even number, say 2e, of times. Since and u and v have no common factors but ±1, either
p2e divides u or p2e divides v. So u and v are both products of primes to even powers. Thus, u and v are both
perfect squares.
Basic fact 4: If (x, y, z) is a PPT with x odd, then x2 = (z − y) ∗ (z + y).
Using basic facts 3 and 4, we see that there exist odd integers s and t so that z − y = s2 and z + y = t2 . We can now solve
for z and y to get z = (s2 + t2 )/2, and y = (t2 − s2 )/2. So x2 = z 2 − y 2 = s2 t2 , and it follows that x = st. This leads to:
Characterization of PPT’s: The PPT’s with first coordinate odd are precisely the triples of the form (st, (t2 −
2
s )/2, (s2 + t2 )/2) where s and t are positive odd integers with no common factors other than ±1 and s < t.
Another characterization of PPT’s is: The PPT’s with first coordinate even are precisely the triples of the form
(2k`, k 2 − `2 , k 2 + `2 ) where k and ` are positive integers, not both odd, no common factors other than ±1, and k > `.
Fermat’s last theorem states that the equation xn + y n = z n has no positive integer solutions for n ≥ 3. The theorem
was proven by Andrew Wiles in 1994. We proved Fermat’s last theorem in the special case that n = 4. In fact, we showed
that there do not exist positive integers x, y and z with x4 + y 4 = z 2 . Our idea was to show that if there exist positive
integers x, y, z with x4 + y 4 = z 2 , then we would be able to find other positive integers x0 , y 0 , and z 0 with x04 + y 04 = z 02 and
z 0 < z. This leads to a contradiction because we would be able to get smaller and smaller z 0 ’s among are solutions.
But since the postiive integers have a smallest element, we can’t continue forever. A simple instance of the above
1
is: Suppose that x, y and z are positive integers with x4 + y 4 = z 2 , 5|x and 5|z. Then x0 = x/5, y 0 = y/5 and z 0 = z/25 are
positive integers with x04 + y 04 = z 02 , and z 0 < z.
We also related the PPT’s with points on the unit circle. If (u, v, w) is a PPT, then (u/w, v/w) is a point on the unit
circle both of whose coordinates are both rational because u, v and w are positive integers. Conversely, if (p/q, r/q) is
a point on the unit circle both of whose coordinates are rationals (in reduced form), then (p, r, q) is a PPT. This is because
multiplying both sides of (p/q)2 + (r/q)2 = 1 by q 2 gives p2 + r2 = q 2 , and since p/q and r/q are reduced, (p, r, q)
is primitive.
Be able to:
• Find all PPT’s with a given x, a given y or a given z.
• Find all PT’s with a given z.
• Work with divisibility properties and PT’s.
• State Fermat’s last theorem.
• Be able to relate PPT’s to points on the unit circle.
Practice Problems
1. Show that there are only finitely many primes of the form n2 − 9 where n is a positive integer.
Note that n2 − 9 = (n − 3)(n + 3). So, if n2 − 9 is prime, then n − 3 is 1 and n + 3 is n2 − 9. I.e. n = 4. So if
n > 4, n2 − 9 is not prime.
2. Find all PPT’s of the form
(a) (15, y, z)
We need to find odd integers s and t with no common divisors so that st = 15 and s < t. The
possibilities are s = 1 and t = 15, or s = 3 and t = 5. This gives the PPT’s (15, (152 − 12 )/2 =
112, (152 + 12 )/2 = 113), and (15, (52 − 32 )/2 = 8, (52 + 32 )/2 = 17).
(b) (x, 80, z) We need to find integers k and ` not both odd and having no common divisor such that
k` = 80/2 = 40 and k < `. The possibilities for (k, `) are (1, 40), and (5, 8). Now x = `2 − k 2 , and
z = `2 + k 2 . So the PPT’s are (402 − 1 = 1599, 80, 402 + 1 = 1601)
(c) (x, y, 125) We need 125 = k 2 + `2 where k and ` are positive integers, not both odd, with no common
divisors, and ` > k. The possibilities are (2, 11) ((5, 10) doesn’t work because 5 and 10 have a common
divisor other than ±1)). So the PPT is (2k` = 44, `2 − k 2 = 117, 125)
3. Describe all PPT’s of the form (z − 9, y, z) In PPT’s, z is always odd. So z − 9 is even. We need to find
integers k and ` with 2k` = z − 9, and (k 2 + `2 ) = z. So we need 2k` = (k 2 + `2 ) − 9. This simplifies to
(` − k)2 = 9. So ` − k = 3, which simplifies to ` = k + 3. Thus the PPT has the form (2k` = 2k(k + 3), `2 − k 2 =
(k + 3)2 − k 2 = 6k + 9, `2 + k 2 = 2k 2 + 6k + 9). To be primitive, ` and k can’t have a common divisor. Since
` = k + 3, any divisor of ` and k divides 3. So we can’t have k be a multiple of 3.
4. Find all PT’s (not necessarily primitive) of the form (x, y, 225) First let’s find the PPT’s. We need to find k < `
with k 2 + `2 = 225. (k, `) = (9, 12) works. But 9 and 12 have a common divisor. So there are no PPT’s
with z = 225.
Next we find the PT’s. The possible multipliers are the divisors of 225: 3, 5, 9, 15, 25, 45, 75. So we
search for PPT’s of form (x, y, 75), (x, y, 45), (x, y, 25), (x, y, 15), (x, y, 9), (x, y, 5) and (x, y, 3).
This give the PPT’s (7, 24, 25), and (3, 4, 5).
We can make these have last coordinate 225 by scaling. So the PT’s are: (63, 216, 225) and (135, 180, 225)
5. In this problem we will consider the equation x4 − y 4 = z 2 where x, y and z are positive integers.
(a) Explain why if x, y, z have a common divisor d > 1, then there exist positive integers x0 , y 0 , and z 0 with x04 −y 04 = z 02
and z 0 < z.
We could use x0 = x/d, y 0 = y/d and z 0 = z/d2
2
(b) Assume now that x, y, z have no common divisors other than ±1. Show that (y 2 , z, x2 ) is a PPT We have
y 4 + z 2 = x4 so (y 2 )2 + z 2 = (x2 )2 , which means that (y 2 , z, x2 ) is a PPT.
(c) Argue that x is odd.
In a PPT the last coordinate is always odd. So x2 is odd. This requires that x is odd.
(d) Argue that z 2 = (x − y)(x + y)(x2 + y 2 )
z2 = x4 − y4 = (x2 − y2 )(x2 + y2 ) = (x − y)(x + y)(x2 + y2 )
(e) Show that if z is odd, then there exist positive integers k, ` and m so that x − y = k 2 , x + y = `2 and x2 + y 2 = m2 .
Note that z 2 = (x2 − y 2 )(x2 + y 2 ). If d divides x2 − y 2 and x2 + y 2 , then it divides their sum 2x2 ,
their difference 2y 2 , and z 2 . Since we are assuming that z is odd, d must be odd, and divided
x2 , y 2 , and z 2 . Since x, y and z have no common divisors but ±1, we conclude that d = ±1. Now
(x2 −y 2 ) = (x2 +y 2 ) = z 2 , and (x2 −y 2 ) and (x2 +y 2 ) have no common factors. So we know that x2 −y 2 = r2
from some integer r and x2 + y 2 = m2 for some integer m. We next look at (x − y)(x + y) = r2 , and
repeat the same kind of argument to show that x = k 2 and y = `2 for some integers k and `.
6. (a) Show that if a is an integer, then the remainder when a2 is divided by 5 is 0, 1, or 4.
a has the form 5q + r for some integers q and r with 0 ≤ q < 5. So a2 = (5q + r)2 = 5(5q 2 + 2qr) + r2 . So
the remainder when a2 is divided by 5 is the remainder when r2 is divided by 5. The possibilities
for r2 are 02 , 12 = 1, 22 = 4, 32 = 9, and 42 = 16, and the possible remainders are 0, 1, 4, 4, 1.
(b) Use (a) to show that if (x, y, z) is a PPT, then exactly one of x y, or z is a multiple of 5.
Since (x, y, z) is a PPT no two of x, y and z are multiples of 5. If x or y is a multiple of 5, we are
done. If neither x nor y is a multiple of 5, then when z 2 = x2 + y 2 is divided by 5 the remainder is
1 + 1, 1 + 4 or 3 (which comes from 4 + 4). The first and third are impossible–z 2 has remainder 0, 1
or 4. So 5 divides z, and we are done.
Base Representations
A geometric series is a series of the form a + ar + ar2 + · · ·. If |r| < 1, then the geometric series converges to a/(1 − r).
This can be argued as follows: the nth partial sum is sn = a + ar + · · · + arn . We multiply this equation by r, and subtract
the two equations to get (1 − r)sn = a − arn .
So sn = (a − arn )/(1 − r), and limn→∞ sn = a/(1 − r).
P∞
Fix b an integer with b ≥ 2. Given a real number γ with 0 ≤ γ < 1, there exist integers c1 , c2 , . . . , so that γ = k=1 ci /bi .
In fact, we can defined that ci recursively. We set γ0 = γ, and for i ≥ 0 we set γi+1 = bγi , ci+1 = bbγi+1 c.
We say that the base-b representation of the number γ is the expression .c1 c2 c3 c4 · · ·, and write (γ)b = .c1 c2 c3 c4 · · ·
The base-b representation of a number is not unique. In fact 1/8 has 2 base-8 representations: .1 (in base-8) and .07̄ (in
base-8). (Hint: 1/8 = 7/82 + 7/83 + 7/84 + · · ·).
If γ is rational, say γ = p/q where p and q are integers, then note that each i, γi has the form xi /q for some integer xi in
{0, 1, 2, . . . , q − 1}.
If γi = 0, then 0 = ci = ci+1 = ci+2 = · · ·, and the base-b representation of γ is .c1 c2 · · · ci−1
If γi = γj , then ci = cj , ci+1 = cj+1 , ci+2 = cj+2 , etc. Hence the base-b representation of γ repeats with repeating pattern
ci c2 · · · cj−1 .
Note that after q steps, either we’ve found a ci = 0, or i and j with ci = cj . So that after q steps, we’ve either terminated
or started to repeat.
So, we’ve shown that if γ = p/q is a rational number, then its base-b representation either terminates or
repeats.
Conversely, if the base-b representation of γ terminates, say it is .c1 c2 c3 · · · ck , then γ = p/q where p = c1 c2 · · · ck and q =
bk . If the base-b representation of γ repeats, say it is .c1 c2 . . . ck d1 d2 d3 · · · d` , then γ equals c1 /b1 +c2 /b2 +· · · ck /bk +a/(1−r)
where a = d1 /bk+1 + d2 bk+2 + · · · + d` /bk+` and r = 1/b` . Thus, γ is rational.
Be able to
• Find the base-b representation of a given fraction
• Given the base-b representation of a number, find the number
• Determine whether or not a number is rational based on its base-b representation
Practice Problems
3
1. (a) Find the base-7 representation of the number 9/49
b63/49c = 1
63/49 − 1 = 14/49.
b(14/49) ∗ 7c = b14/7c = 2
So, base-7 representation is .12
(b) Find the base-7 representation of the number 1/9.
b7/9c = 0
7/9 − 0 = 7/9
b49/9c = 5
49/9 − 5 = 4/9
b28/9c = 3
28/9 − 3 = 1/9
So the base-7 representation of 1/9 is .053.
(c) Find the number whose base-8 representation is .1234
It is 1/8 + 2/82 + 3/83 + 4/84 = (83 + 2 ∗ 82 + 3 ∗ 8 + 4)/84 = (512 + 128 + 24 + 4)/(4096) = 768/4096.
(d) Find the number whose base-8 representation is .12012
It is 1/8 + 2/64 + (0/83 + 1/84 + 2/85 ) + 1/83 ∗ (0/83 + 1/84 + 2/85 ) + · · · + The tale end is a a geometric series
with a = 10/85 and r = 1/83 . So that tale end is a/(1 − r) = (10/85 )/(1 − 1/83 ) = (10/85 )/(83 − 1)/83 =
10/(82 ∗ (83 − 1)) = 10/(64 ∗ 511). So the answer is: 1/8 + 2/64 + 10/(64 ∗ 511).
2. (a) Give the base-5 representations of two rational numbers.
.123, .4̄ work
(b) Give the base-5 representation of two irrational numbers. .10110111011110 · · · and .20220222022220 · · · work.
3. Which reduced fractions p/q have base-6 representation that terminates? Prove your answer. (Hint: Think about
what your answer was for base-10).
If p/q has base 6 representation then p/q = a1 /6d + a2 /6d+1 + · · · ak /6d+k = x/6d+k = x/(23+k 33+k ) for some integers
a1 , ldots, ak , x. When reduced, the denominator will be a multiple of a power of 2 and a power of 3. So answer is p/q
where q = 2e ∗ 3f for some positive integers e and f .
Continued Fractions
A finite continued fraction is a fraction of the form
a+0+
1
a1 +
1
a2 +
1
..
.
where a0 is an integer, and a1 , a2 , . . . , an are positive integers. For shorthand, we write this fraction as [a0 ; a1 , a2 , . . . , an ].
We can show that the finite continued fraction [a0 ; a1 , a2 , . . . , an ] is rational as follows: The proof is by induction
on n. If n = 0, then [a0 ] = a0 which is rational. If n = 1, then [a0 ; a1 ] = a0 + 1/a1 which is rational because a sum of
rationals is rational.
1
Assume that [a0 ; a1 , . . . , ak ] is rational. Now [a0 ; a1 , . . . , ak , ak+1 ] = a0 + [a1 ;a2 ,...,a
. By induction, [a1 ; a2 , . . . , ak+1 ] is
k+1 ]
rational. Since the recripricol of a nonzero rational is rational, and the sum of rationals is rational, we see that [a0 ; a1 , . . . , ak+1 ]
is rational.
Also, we know that every rational number is equal to some finite continued fraction. Indeed, we have an algorithm. Let
α = a/b where a, b are integers with b > 0. We divide b into a to get a = bq + r for some integer r with 0 ≤ r < b. Then
a/b = q + r/b = q + 1/(b/r). We next divide r into b to get b = rq2 + r2 for some integer r2 with 0 ≤ r1 < r2 . This gives
1
b/r = q2 + r2 /r. So that a/b = q + q2 +1/(r/r
. We just keep finding quotients and remainders until we get a remainder of 0.
2)
Note we will eventually do this because at each stage we are remainders are getting smaller.
Note for an integer m ≥ 2, we have that m/1 = (m − 1) + 1/1. For that the finite continued fractions [m] and [m − 1; 1]
are equal. More generally, the finite continued fractions [a0 ; a1 , a2 , . . . , an ] and [a0 ; a1 , a2 , . . . , an−1 , an − 1, 1] are equal when
an ≥ 2.
4
Given a finite continued fraction [a0 ; a1 , a2 , . . . , an ] its kth convergent is the finite continued fraction Ck = [a0 ; a1 , a2 , . . . , ak ].
So C0 = [a0 ] = a0 , C1 = [a0 ; a1 ] = a0 a1 + 1/a1 , C2 = a0 + a2 /(a1 a2 + 1) = (a0 a1 a2 + a0 + a2 )/(a1 a2 + 1).
Calculating convergents from their definitions is a royal pain. Fortunately, we can calculate convergents using a recurrence
relation. We let Ck = pk /qk . The recurrences are given by:
pk+1
p0 = a0
q0 = 1
p1 = a0 ∗ a1 + 1
q1 = a1
= ak+1 ∗ pk + pk−1 qk+1 = ak+1 ∗ qk + qk−1
Some important properties of the recurrence are:
pk qk−1 − pk−1 qk = (−1)k−1
pk and qk have no common factors other than ±1
Ck − Ck−1 =
(−1)k−1
qk qk−1
Ck − Ck−2 =
−1k ak
qk qk−2
Most importantly, these say that the even convergents are increasing i.e. C0 < C2 < C4 < · · ·, the odd convergents are
decreasing C1 > C3 > C5 > · · ·, every odd convergent is larger than every even convergent, and the value of the continued
fraction lies between the even and odd convergents.
An infinite continued fraction is a fraction of the form, where a0 is an integer, and each ai (i ≥ 1) is a postive integers.
We use the shorthand [a0 ; a1 , a2 , . . . , ] for this continued fraction. We defined the kth convergent, Ck , as before. Since the
all even convergents and the odd convergents get closer and closer to each other (see property (3) above), limk→∞ Ck exists.
This means that every infinite continued fraction equals some real number γ.
We also showed that every irrational number is the value of some infinite continued fraction. In fact, we had an algorithm.
Let γ be an irrational number. Set γ0 = γ, and a0 = bγ0 c, and for i ≥ 1 set γi = 1/(γi−1 − ai−1 ), and ai = bγi c.
In fact, we showed that the value of an infinite continued fraction must be irrational. So, finite continued fractions
represent rational numbers, and infinite continued fractions represent irrational numbers.
Continued fractions are very useful in many different ways. First they are good approximations to an irrational number.
We proved the following 3 results about an irrational number α, a rational approximation r/s, and a convergent Ck = pk /qk
of the continued fraction expansion of α:
Convergents are the best relative approximants If |sα − r| < |qk α − pk |, then s ≥ qk+1
Convergents are the best approximants If |α − r/s| < |α − Ck |, then s ≥ qk .
The only good approximants are convergents If |α − r/s| < 1/(2s2 ), then r/s = Ck for some k.
Continued fractions are
√ also very useful in solving certain kinds of equations. Let d be a positive integer which is not
a perfect square. Then d is irrational.√ Pell’s equation is x2 − dy 2 = 1. We showed that if x and y are positive integers
2
that solve Pell’s equation, then
√ |x/y − d| < 1/2y , (here’s how) and hence (be the third approximant property), x/y is
necessarily
a convergent of d. The upshot of this is: to find a solution to Pell’s equation, we hunt among the convergents
√
of d.
Be able to
• Find finite continued fractions of a number
• Find convergents of a continued fraction
• Know and use properties of the convergents of a continued fraction
• Know and use approximation properties of convergents
• Be able to use convergents to find a solution to Pell’s equation.
1. Find the value of the finite continued fraction [−2; 1, 2, 1, 3, 4].
−81/64.
5
2. Find two finite continued fractions equal to 12/7. 12/7 = [1; 1, 2, 2] = [1; 1, 2, 1, 1]
3. The value of the finite continued fraction [a0 ; a1 , a2 , a3 , . . . , ] is x.
(a) Which is bigger x or [a0 ; a1 , a2 ]? Justify you answer. x, because [a0 ; a1 , a2 ] = C2 , and value is always bigger than
even convergents.
(b) Which is bigger x or [a0 ; a1 , a2 , a3 ]?
[a0 ; a1 , a2 , a3 ] = C3 > x, as odd convergents are always bigger than the value.
(c) Explain why |x − [a0 ; a1 , a2 , a3 | < 1/6
|x − C3 | < 1/q3 q4 , and q3 ≥ 3, q4 ≥ 4. So x − C3 | < 1/12 < 1//6.
4. Give 2 continued fractions whose value equals a rational number.
Any finite continued fraction works.
5. Give 2 continued fractions whose value equals an irrational number.
Any infinite continued fraction works.
6. Find the best approximation to ln 2 by a rational with denominator less than or equal to 100.
ln 2 is approximately .69315
So a0 = 0. 1/(ln 2) ≈ 1.4427. So a1 = 1.
1/.4427 ≈ 2.2589. So a2 = 2.
1/2589 ≈ 3.826. So a3 = 3
1/.826 ≈ 1.1593. So a4 = 1
1/.1593 ≈ 6.2793. So a5 = 6.
1/.2793 ≈ 3.5804. So a6 = 3.
Now we compute C6 = 61/88 and C7 = 192/279
Answer is 192/279.
7. It is known that e = [2; 1, 1, 2, 1, 1, 4, 1, 1, 6, . . . , ]
(a) Find C4 , and C5 .
C5 = 87/32, C4 = 19/7
(b) I have a fraction r/s with s < q5 , what can you say about |es − r|?
It is at least q6 = 39.
(c) I have a fraction r/s with s < q5 , what can you say about |e − r/s|?
It is at least q5 = 32.
√
8. (a) Find the infinite continued fraction expansion of 7 (use exact calculations-not a calculator).
√
(b) Find
√ the first 5 convergents of 7.
b 7c = 2,
so a0 = 2
√
7+2
√1
=
3 , so a1 = 1.
7−2
√3
7−1
√2
7−1
√3
7−2
√1 =
7
√
=
=
√
7+1
2 ,
7+1
3 ,
√
so a2 = 1.
so a3 = 1.
= 7 + 2, so a4 = 4.
2 repeats, so a5 = a1 , ots.
√
Answer: 7 = [2; 1, 1, 1, 4]
(c) Use the answers√to (b), to find a solution to Pell’s equation x2 − 7y 2 = 1.
Convergents of 7 are 2/1, 3/1, 5/2 8/3, . . ..
So possible solutions are (x, y) = (2, 1), (3, 1), (5, 2), (8, 3) etc. x = 8, y = 3 is a solution because 82 − 7 ∗ (3)2 =
64 − 63 = 1.
6