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Targil 9 - unique factorization.
1. Consider ring Z i d s , where d is a positive integer.
(a) Prove that if d < 3 there is unique factorization property.
…
2
Solution. For any z, denote z  z  z  z . If z  a  ib d  Z i d s then
z  a 2  b 2 d is a nonnegative integer.
The proof is based on a sequence of lemmas, known as division with remainder
and Euclidean algorithm, most of which must look familiar.
The ring that we investigate, Z i s or Z i 2 s , will be denoted R.
Lemma 1. If z, w  R, w  0 , there exist q, r  R , such that z = qw + r and
w  r .
Lemma 2. For every z, w there exists a common divisor c  R such that any
common divisor of z and w in R is also divisor of c, and c = mz + nw, where
m, n  R . This c is unique up to multiplication by a unit and it is called greatest
common divisor.
Lemma 3. If p is irreducible, and st is divisible by p (while s, t , p  R ), then either
s or t is divisible by n.
After we prove the, life becomes easy. Indeed, take two different irreducible
factorizations of the same number: p1 p2  ...  pn  q1q2  ...  qm .
Both sides of the inequality are divisible by p1, so by lemma 3 one of the factors qi
in the right hand side is divisible by 1. If one irreducible number is divisible by
another, the quotient is a unit. So we may cancel out p1 with qi and a unit will
remain in RHS, but this unit can be hidden into another element of the product.
So we shall have a shorter identity, for which equivalence is already known by
induction (base of induction: if one side of identity had only one element of the
product, we get irreducible = something, so something is also irreducible, so both
factorization have 1 element and it is the same).
So, it remains to prove the lemmas:
Proof of lemma 1. Consider s = z / w , a complex number but not necessary in R.
Let q be the number in R closest to s. Since R geometrically is a rectangular
lattice, |s – q| is at most half the diagonal of a small rectangle, which is less than 1.
|s – q| < 1
|z – qw| = |sw – qw| = |s – q|·|w| < |w|
If we denote r = z – qw we got |r| < 1 and
z = qw + r
QED.
Proof of lemma 2. Using lemma 1, we build a sequence of identities:
z = qw + r1
w = q1r1 + r2
r1 = q2r2 + r3
r2 = q3r3 + r4
…
rn-2 = qn-1rn-1 + rn
The sequence rk is decreasing. A decreasing sequence of positive integers must
stop, so at some moment the remainder will be 0:
rn-1 = qnrn
If certain k in R divides both z and w, by first identity of the sequence it divides r1,
so by second identity it divides r2 and so on, hence by induction it divides rn. Also,
by the last identity rn divides rn-1 so by the identity before the last it divides rn-2 and
hence by the identity before that rn-3 and so on, and in the end we see that it divides
both z and w.
Since each common divisor k of z and w is a divisor of rn, so rn  kl  k  l ,
therefore rn  k . So rn is the longest among all common divisors of z and w.
Any other common divisor of the same length is divisible by rn, and the ration is of
length 1, so it is a unit, so any other divisor of maximal length is equivalent to rn.
From each identity we can express the rk+2 as a linear combination rk+1 and rk with
coefficient in r. We start with rn = rn-2 – qn-1rn-1.
Substitute rn-1 = rn-3 – qn-2rn-2, and we get rn = u1rn-3 + v1rn-2.
Substitute rn-2 = rn-4 – qn-3rn-3, and we get rn = u1rn-4 + v1rn-3.
This process continues until we get rn = uz + vw.
Remark. { uz + vw } is an ideal, spanned by z and w. All its elements are divisible
by all common divisors of z and w. What is more, with Euclidean algorithm we
found d, such that { kd } = { uz + vw }. Ideals for which such d exists are called
principal ideals.
The word “ideal” in ring theory historically appeared from thinking about rings
where these statements are wrong, so attempt to find one number generating the
ideal spanned by w and z fail completely, so they talked about “ideal numbers”,
which are not actual numbers we want them to be but subsets of the ring.
Proof of lemma 3. Assume that st is divisible by p but s and t separately are not.
The common divisor of p and anything is either a unit or equivalent to p, but s is
not divisible by p, so the greatest common divisor of s and p os 1, therefore
1 = as + bp
For similar reasons
1 = mt + np
Therefore
1 = (as + bp)(mt + np) = amst + (asn + bmt +bnp)p = Mst + Np = Kp
Hence p is a unit, which contradicts the assumption.
1. Consider ring Z i d s , where d is a positive integer.
…
(b) Prove that if d ≥ 3 there is no unique factorization property.
Solution
1  i d 1  i d   d  1
i d  i d   d
One of these two numbers is even and hence divisible by 2.
2 is irreducible: indeed, if 2 = kl, then k  l  4 , so either k or l is a unit or
k  l  2 but we don’t have numbers of length
2 in our ring.
So, if we would have unique decomposition, either 1  i d or 1  i d or i d or
i d would be divisible by 2, but they are not (you can divide them by 2 as
complex numbers, but you will get outside R).
2.* Find all integer solutions of equations:
(a) x2 + 4 = y3
…
Solution. (x + 2i)(x – 2i) = y3
The difference of between the two factors is 4i = – i(1+i)4.
1 + i is irreducible, so the greatest common divisor of x + 2i and x – 2i is (1+i)N,
where N is nonnegative integer not greater than 4.
Irreducible factorization of x + 2i is complex conjugate to irreducible factorization
of x – 2i, and 1 + i is equivalent to its complex conjugate, so 1 + i and its
equivalents appears in the same power in x + 2i and x – 2i.
The total power should be divisible by 3, therefore it is 0 or 3.
Apart from power of 1 + i the two factors x + 2i and x – 2i have no common
divisors, so both are cubes (because of unique factorization in the ring of Gaussian
numbers Z i s ).
So, x  2i   a  bi    a3  3ab2    3a 2b  b3  i .
3
 x  a 3  3ab 2   a 2  3b 2  a


2
3
2
2
2  3a b  b   3a  b  b
The last equation implies b  1 or b  2 .
First option: b  1 .
3a 2  1  3a 2  b2  2
3a 2  1 or 3, so in previous equation it was + and b = 1
a2  1
We get x  2 , y = 0.
Second option: b  2 .
3a 2  4  3a 2  b2  1
3a 2  5 or 3, so in previous equation it was – and b = –2
a2  1
We get x   a 2  3b 2  a   1  12   11 , y = 5.
Verification shows that  2,0  ,  11,5 are indeed solutions.
2.* Find all integer solutions of equations:
…
(b) x2 + 2 = y3
 x  i 2  x  i 2   y
Solution.
3
In Z i 2 s we also have unique factorization.
The greatest common divisor of both brackets also divides their difference, which


3
 
k
is i 2 2  i 2 . So, it is i 2 , k not bigger then 3.
Irreducible factorizations of x  i 2, x  i 2 are complex conjugate and i 2 is
equivalent to its complex conjugate, so i 2 and its equivalents appears in the
same power in x  i 2, x  i 2 .
The total power of i 2 should be divisible by 3, and apart from that
x  i 2, x  i 2 have no common divisors, so both are cubes.

x  i 2  a  ib 2
  a
3
3
 6ab 2    3a 2b  2b3  i 2
 x  a 3  6ab 2   a 2  6b 2  a


2
3
2
2
1  3a b  2b   3a  2b  b
From the last equation:
b  1
 2
2
3a  2b  1
The last equation gives two options for 3a2: either 3 or –1.
It should be divisible by 3, so in both last equations sign is +, a =
a  1, b  1
x  a3  6ab2  5
Both options give a solution with y = 7.
3. Represent 1234321 and 123454321 as a2 + b2, where a, b are positive integers.
Solution:
1234321 = 11112 = (101·11)2
101 = 102 + 1
(a2 + b2)(c2 + d2) = (ac – bd)2 + (ad + bc)2
1012 = (102 + 1)(102 + 1) = 992 + 202
(101·11)2 = (992 + 202) ·112= 10892 + 2202
123454321 = 111112 = (41·271)2
41 = 25 + 16 = 52 + 42
412 = (52 + 42)(52 + 42) = 92 + 402
(41·271)2 = (92 + 402)·2712 = 25292 + 108402
Remarks.
I. In this solution, we used the lemma: sum of two squares times sum of two
squares is sum of two squares. This can come out of an ingenious algebraic trick,
(a2 + b2)(c2 + d2) = (ac – bd)2 + (ad + bc)2
or from a simple-minded observation: a  bi  c  di  x  iy .
II. In each case, we had two factors: one was 4k + 1, another 4m + 3.
There is no sense even to try decomposing q2 into sum of two positive squares, if q
is a prime number of type 4m + 3. Because the only solution of the equation
x2 + y2 = 0 (mod q)
is (0, 0). If we would have a different solution we would get:
(x / y)2 = x2 / y2 = –1 (mod q)
Then x/y is an element is of order 4 mod q, so q – 1 is divisible by 4 by Fermat
little theorem. Therefore, if a2 + b2 is divisible by 11 (or 271) then both a and b are
divisible by 11, so divide each of them by 11 and sum of squares by 11 2 and
continue from there. So, when we took 112 out of the brackets, it was the only
choice.
III. We could factorize each number in Gaussian integers. Decomposing natural
prime number only has a chance when it is 2 or of form 4k + 1.
So, for example
(a + bi)(a – bi) = 1234321 = 11112 = (101·11)2 = (10 + i)2(10 – i)2112
Factorization of a + bi is a unit times some factors from the right.
Norms of a + bi and a – bi are the same, so they should get the same number of
factors. Each of them should get 11 (since they are conjugate) and two from the
following 10 + i, 10 + i, 10 – i, 10 – i. That leaves us with only two options either
each factor gets 10 + i and 10 – i, or one takes 10 + i twice and another gets 10 – i
twice. So, there are only two ways to represent it as a sum of squares of integers.
The first way is 11112 + 02, and the second is the only answer to our question.
Similar with 10892 + 2202.
4. (a) How many ways are there to represent 2k p1i1 p2i2  ...  pnin  q1j1 q2j2  ...  qmjm as a
sum of 2 integer squares, if pi are prime numbers of type 4a + 1 and qj are prime
numbers of type 4b + 3 ?
Answer. 4(i1 + 1)(i2 + 1)…(in + 1) if all jk are even and 0 otherwise.
Solution. As we noticed in the remark II above, if qk divides a2 + b2 then it divides
both a and b, and we so we can say there are no decomposition into some of
squares if jk = 1 and if jk > 1 it can be reduced by 2 and we get an equivalent
problem. Repeating this, we get that if at least one jk is odd, then decomposition
into 2 squares doesn’t exist, and if all are even, the number of decompositions is
the same as for 2k p1i1 p2i2  ...  pnin .
The crucial fact is
Lemma. A prime number p of type 4k + 1 is representable as sum of integer
squares.
First proof of lemma. (with Gaussian integers and unique decomposition)
Consider the equation a2 + b2 = 0 (mod p).
It is equivalent to (a/b)2 = –1 (mod p) or to the statement a/b is of order 4 precisely
mod p and so it has solution (because primitive root exists mod p and p – 1 is 4k).
In other words, we have a2 + b2 = mp but neither a, nor b nor m is divisible by p.
Let z be the greatest common divisor of a  bi and p in the ring of Gaussian
integers, then z is a divisor of both a  bi  a 2  b 2 and p2, so it is 1, p or p2.
But z is a common divisor of a  bi and p, so zz is divisible by p and so z is not a
unit. Also z is not p2 otherwise a2 + b2 would be divisible by p2. So z  p .
Second proof of lemma. (Elementary)
As before, we argue that there is “ 1 ” mod p, i. e. a number c s. t.
c2 = –1 (mod p).
Let S be a set of nonzero integer numbers in 0, p . Then S   p  .

Denote cS = {cx | x  S}. Mark points on a circle of length p corresponding to the
points of cS. You get a circle of length p divided into  p  arcs. So one of the

arcs is shorter than
p . So, we have 0  k , l  p such that ck – cl = b (mod p)
and 0  b  p . Take a = k – l and we get a  p , a  0 , and ca = b (mod p).
Therefore 0 < a2 + b2 < p + p = 2p and a2 + b2 = a2 (1 + c2) = 0 (mod p).
So, a2 + b2 = p. QED
From the first approach we see that this decomposition is in fact unique: indeed if
(a + bi)(a – bi) = p then a  bi  a  bi  p is prime so a + bi and a – bi are
irreducible. So if c 2 + d 2 = p then c + di is equivalent to either c + di or c – di,
under multiplication by units.
Notice that a + bi and a – bi are not equivalent unless the angle between them is a
multiple of 90°, that happens only when p = 2. Therefore for 2 we have 4
representations as a2 + b2, and for prime p = 4k + 1 we have 8 representations.
Now consider a composite number 2k p1i1 p2i2  ...  pnin .
Assume that z  2k p1i1 p2i2  ...  pnin , where z is Gaussian integer.
Up to multiplication by a unit, z is defined by its factorization, so we must count
number of possible factorizations and multiply by 4. The only irreducible number
of
= 2 up to equivalence is 1+ i, so we should have it in power k. Othe
irreducible factors should have
= pt. For each t, we should have precisely it of
those, and there are two types (up to equivalence), any we can have 0, 1, 2, … , it
of the first type and the rest of the second type. So, for each t we have it + 1
choices and these choices are independent, so there are (i1 + 1)(i2 + 1)… (in + 1)
factorization. Now multiply by the number of units to get the answer.
4. (b)* For R > 0, show that number if integer points in the disc { x2 + y2 ≤ R } is


R R R R  R   R 
1  4   R                     ...  .
 3   5   7   9  11 13 


Solution. Let us consider again the question that we have solved in 4(a): how
many ways are there to represent integer N > 0 as a2 + b2, where a and b are
integers (or, in other words, how many integer points does a circle with center at
the origin and radius N have). We had an answer in terms of factorization, but
there is a nicer way to formulate the answer.
Notations. Let Z(N) be number of pairs of integers (a, b) such that N as a2 + b2,
A(N) number of all divisors of N of type 4k + 1 (not necessary prime),
B(N) number of all divisors of N of type 4k + 3 (not necessary prime).
Claim. For N > 0, Z(N) = 4(A(N) – B(N)).
Proof. We shall conclude it from the answer of 4(a) which uses factorization
N = 2k p1i1 p2i2  ...  pnin  q1j1 q2j2  ...  qmjm .
First of all, both answers disregard even divisors, so it is enough to prove it for odd
N, i. e. k = 0.
Now there is a matching of divisors of N related to q1, each couple has a ratio q1
precisely: for any d which is a divisor of N, if of q1 in the decomposition of d has
the same parity as j1 we divide by q1, otherwise we multiply by d.
In each pair of this matching one divisor is of type 4k + 1, another of type 4k + 3.
If j1 is odd then the matching is perfect and A(N) – B(N) = 0, and so is Z(N).
If j1 is even then the matching is not perfect: the things that aren’t cancelled out by
this matching are precisely divisors of N’ = N / q1j1 . But we saw that Z(N) = Z(N’)
so it is enough to prove for N’.
Doing the same for every qs we either prove the statement or reduce it to the same
statement for number M = p1i1 p2i2  ...  pnin . All its divisors are of type 4k + 1, hence
A(M) = (i1 + 1)(i2 + 1)…(in + 1) , B(M) = 0, and the rest of it follows form 4(a).
Now that we have proved the claim, we shall prove 4(b).
Number of points in the disc is
 R
 R
 R
N 0
N 1
N 1
 Z  N   Z  0   Z  N   1  4  A N   B  N 
Notation. D(d, N) = 1 if d divides n
0 otherwise.
 R
 R
N 1
N 1


  A  N   B  N       D  4k  1, N    D  4l  3, N   

 R

 k 0


l 0

 R

 R    R 
  D  4k  1, N    D  4l  3, N    
    4l  3 
k 0 N 1
l 0 N 1
k 0  4k  1 
l 0
QED.
Remarks.
1. No convergence issues here when interchanging brackets because all summands
except finite number of them are zeroes.
2. I know 4(b) from a wonderful book of Hilbert and Con-Vossen “Anschauliche
Geometrie” (it has English and Russian translations).
3. Of course, in the limit when we neglect the [], we get the famous Gregory 1 1

Leibnitz formula   4arctan 1  4 1    ...  . It makes sense, since for large
 3 5

circles the discrete area (number of integer points) is close enough to the normal
area (the difference of this two is bounded by the number of tiles chopped by the
boundary, which is bounded by constant times length of circle, which is O(radius)
while area is O(radius2).
4. Area can be roughly measured also by triangular lattice, so formula of 4(c) also
produces a formula for  when applied to large discs, but this time there’s an ugly
irrational factor related to area of regular hexagonal tile. This is no great wonder
1 
1
1 1 1
x  x2
3 k 1
3k 2
dx , and that is an
since 1     ...     x  x dx  
3
2 4 5
1

x
0 k 0
0
1
elementary integral of type 
dt so it comes to an expression with  .
1 t2
5. The question “when a number is sum of two squares” is known as Fermat
theorem, it has many proofs (thanks Shahar for the link
http://en.wikipedia.org/wiki/Proofs_of_Fermat's_theorem_on_sums_of_two_squar
es#Zagier.27s_.22one-sentence_proof.22 )
4. (c)* Consider “triangular” lattice, formed by points of Z   , where   1,
 3  1. Show that number of points of this lattice in the disc { x2 + y2 ≤ R } is


R R R R R  R   R 
1  6   R                        ...  .
 2   4   5   7   8  10  11


Solution. Open discs of radius 1 centered at points of Z   cover the plane, from
here we get in Z   the logical chain we got in problem 1(a):
Division with remainder => Euclidean algorithm => unique factorization.
Notice, that there are 6 units.
Next we come to a question, which circles of radius N have points on integer
lattice and how many. It is the same as representing N in the form a2 – ab + b2 (or,
which is an equivalent problem, in the form a2 + ab + b2).
The first answer, similar to 4(a), is: if N = 3k p1i1 p2i2  ...  pnin  q1j1 q2j2  ...  qmjm , where pi
are of type 3s+1, and qj are of form 3s+2, then it depends on the parity of jl.
If at least one of them is odd, then it is impossible; if all of them are even, then the
number of representations is 6(i1 + 1)(i2 + 1)…(in + 1).
Of course, 6 here is the number of units in our ring.
The proof is very similar to 4(b), so we shall say only about the differences.
Firstly, the ring of Gaussian numbers is replaced by Z   .
Secondly, the modular arithmetic is slightly different: in the previous example, we
were reducing a2 + b2 = 0 (mod p) to (a/b)2 = –1 (mod p); in a similar way it is
possible to reduce a2 + ab + b2 = 0 (mod p) to (a/b) = –3 (mod p).
So this time we have to prove that –3 is quadratic mod p if and only if p isn’t 3k–1.
This is known from quadratic reciprocity for example.
After that, we reformulate the answer:
Notations. Let T(N) be number of pairs of integers (a, b) such that N as a2 +ab+ b2,
U(N) number of all divisors of N of type 3k+1 (not necessary prime),
V(N) number of all divisors of N of type 3k+2 (not necessary prime).
Claim. For N > 0, T(N) = 6(U(N) – V(N)).
The proof and the rest of it is similar to 4(b), I won’t rewrite it (I could’ve cut and
paste it, but it wouldn’t be fair to the readers).
k

1
 k  1 
5. Suppose n > m > 0 are integers,   arctan  m n  , prove    
 .
2
2


 
 m n 
({x} denotes fractional part of x, which is a number in [0,1) equivalent to x mod 1)
Solution.
Notation. Given a real number a and a positive number b we shall denote a % b a
number x in [0, b) such that x – a = nb, where n is integer. It is quite obvious that x
is uniquely defined.
k


1
The statement of the problem can be rewritten as  k  %  
 .
2
2
 m n 
We shall need the following lemma.
Lemma. Let z = x + iy be a Gaussian integer. Let B be the union of four lines,
coordinate and diagonal B = {x = 0} U {y = 0} U {x + y = 0} U {x = y}.
If zk is not in B, then z is not in B.
We shall prove the lemma in the end. Denote w = n + im. Then  k  % is the
angle between horizontal axis and wk, measured counterclockwise. By lemma, it is
nonzero. So, sin   k  %  
Im  wk 
wk

 k  %  sin   k  %  
Im  wk 
w
k
Im  wk 
w
k
.


1
 k 

2
2
w
 m n 
1
k
QED, if we prove the lemma.
First proof of lemma. In the previous questions we have actually classified all
irreducible Gaussian numbers:
(a) there are 4 things equivalent to 1+i.
(b) There are real or imaginary things, equivalent to natural prime numbers of
type 4k+3.
(c) For each natural prime p of type 4k+1 there are 4 Gaussian irreducible
numbers equivalent to a + bi and 4 other prime equivalent to a – bi; here
a2 + b2 = p.
If zk is in B, then z4k is real. Therefore, for each irreducible factor a + bi in zk of
type (c) we have the complex conjugate factor. So, the same is true for z. So, the
factorization of z has complex conjugate pairs of type (c) which give a real number
in the product times real factors of type (a) times units and factors of type (b)
which preserve the British cross B.
Second proof of lemma. Consider the formula cos(2x) = 1 – 2cos2x
It means that if cos x is a rational number of denominator more than 2, then the
sequence cos(2x), cos(4x), cos(8x), …, cos(2nx), … has growing denominators and
can’t have finite number of states, so x is not rational number degrees.
Cosine of a Gaussian number is a square root of a rational number.
So if z  rei and the argument of zn is rational in degrees, then the argument of z is
rational in degrees, and cos  2  is also a rational number, so its denominator is 2
1
at most. This leaves finite number of possibilities: cos  2   0, 1,  .
2
Some of these   30 , 60 , 120 , 150  have irrational tan which cannot
appear in Gaussian numbers; others are the British cross.