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A complex number a+bi is a Gaussian integer iff a and b are both integers. The collection of all Gaussian integers is normally denoted by Z[i]. This is a subring of C.

We are going to use the properties of the Gaussian integers to prove a result of Euler. This result is about the usual integers (the Gaussian integers are used in the proof but not in the statement).

Theorem Let p be an odd prime number. Then p is congruent to 1 mod 4 iff p can be written as a2+b2, for some integers a,b. Further, this expression as a sum of two squares is essentially unique.

A couple of remarks about the theorem

  • This theorem was first stated by Fermat in a letter to Mersenne but he didn't offer any proof. Pesky margins!
  • Saying that the expression is essentially unique means that any similar expression arises from the given one by sign changes of a and b or by swapping a and b.

Before we can prove this we need to find out some more about the Gaussian integers.

Lemma 1. The ring of Gaussian integers Z[i] is a unique factorization domain

For a proof see the UFD writeup.

Define a norm function on the Gaussian integers by d(a+bi)=a2+b2 (the square of the modulus). Note that this norm is always a non-negative integer and that d(xy)=d(x)d(y), for two Gaussian integers x,y.

Lemma 2

  • For a Gaussian integer u we have that u is a unit iff and only if d(u)=1.
  • The multiplicative group of the Gaussian integers is {1,-1,i,-i} (which is cyclic of order 4).

Proof if u is a unit then uu-1=1. Thus 1=d(1)=d(uu-1)=d(u)d(u-1). It follows that d(u)=1.

Write u=a+bi, for integers a,b. If d(u)=1 then a2+b2=1. Thus one of a or b must be 1 or -1 and the other one must be zero. This gives that the u with d(u)=1 are {1,-1,i,-i}. These are all units. This completes the proof of the lemma.

Lemma 3. Let u be a Gaussian integer. If d(u) is a prime number then u is a prime Gaussian integer.

Proof We have to show that u is irreducible. If u=ab for a,b non-units then d(u)=d(a)d(b). Since d(u) is a prime number, p say it must be that d(a)=p and d(b)=1 (or vice versa). But This is impossible, by Lemma 2, since b is assumed not to be a unit. This contradiction shows that u is irreducible and hence prime.

Let's prove the ==> direction of Euler's result. Suppose that p is an odd prime and suppose that p is congruent to 1 mod 4 (i.e. p-1 is divisible by 4).

Lemma 4 There is an integer c with c2 congruent to -1 mod p.

Proof There is an element of order 4 in the multiplicative group of the finite field Zp (the ring of integers mod p) for this group is cyclic (see the finite field writeup) and it's order is divisible by 4, by the assumption on p. Let t be this element. Then t2 has order 2. Now the polynomial x2-1 has roots 1 and -1 in Zp. Since a polynomial of degree 2 has at most two roots this tells us that -1 is the only element in the multiplicative group of Zp with order 2, so it must be that t2=-1. Letting c be an integer whose class mod p is t gives the result.

Back to ==> direction of Euler's result. We have p divides 1+c2, in Z by Lemma 4. This means that it also divides it in Z[i]. But now we can factorize 1+c2=(1+ci)(1-ci). So this means that p is not a prime element of Z[i] for we can't have p|(1+ci) or (1-ci). For, if p(x+yi)=1+ci for some integers x,y then equating real and imaginary parts we see that px=1 which is clearly impossible.

Since p is not prime in Z[i] it is not irreducible either. Thus we can write it p=(a+bi)(u+vi) as a product of nonunits in Z[i]. Taking d of both sides we get that p2= (a2+b2)(u2+v2). Since p is a prime number, the fundamental theorem of arithmetic tells us that either d(a+bi) and d(u+vi) both equal p or that one of them has norm 1. Since, neither is a unit, by assumption, the first possibility must occur (by Lemma 2) and p=d(a+ib)=a2+b2, is expressed as a sum of squares, as required.

Now for the uniqueness. Suppose that p=a2+b2=e2+f2. The argument at the end of the last paragraph shows that we have d(a+ib)=d(a-ib)=d(e+if)=d(e-if)=p. It follows from Lemma 3 that a+ib,a-ib,e+if,e-if are all prime Gaussian integers. Now by Lemma 1, since we have (a+ib)(a-ib)=(e+ib)(e-ib) uniqueness of factorisation says that either a+ib=(e+ib)u, or a+ib=(e-ib)u, for some unit u. By Lemma 2, u is one of 1,-1,i,-i. Substituting in these values we see that the difference between a,b and e,f is just given by sign changes or swapping,as is required.

Finally we prove the converse. Suppose that p=a2+a2. Since p is odd it must be that one of a and b is odd and the other is even. The square of an even integer is congruent to zero mod 4 and the square of an odd integer is congruent to 1 mod 4, hence the result.

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