Gaussian Integers and Fermat's two square theorem

I was trying to explain to someone recently what I thought mathematics was all about and why it interests me. I found it difficult to put into words some of the ideas and complexities I have seen over the last few years and I had trouble explaining why mathematics is such never-ending treasure chest of ideas. Sometimes you do get the odd glimpse of a fundamental principle and the topic I have just read about Gaussian integers seems to exemplify that.

What is a Gaussian integer? These are a generalization of the ordinary set of integers Z and can be defined as the set of complex numbers

$$ \mathbb{Z} \left[i\right]=\left\{a+bi:a,b\in  \mathbb{Z}\right\}$$

Is it possible to find similarities between this set and the set of ordinary integers and if so what do these common properties tell us about the nature of integers in general? Well it turns out that these two sets do have properties that are in common. For example, it is possible to show that there are Gaussian prime numbers much as there are ordinary prime numbers and you can go on to define properties such as division, greatest common divisor, unique prime factorisation etc which is all very interesting.

So how does this come about? Well, firstly you define the norm of an element of Z[i] to be

$$norm(a+bi)=|a+bi|^{2}=a^{2}+b^{2}$$

and from the properties of the modulus we know that

$$|z_{1}z_{2}|=|z_{1}||z_{2}|$$

and so norms multiply, that is

$$norm\left(z_{1}z_{2}\right)=norm\left(z_{1}\right)norm\left(z_{2}\right)$$

Now because the norm is a real natural number we can define a Gaussian prime to be a Gaussian integer that is not the product of Gaussian integers of smaller norm. For example, 3+2i is a Gaussian prime because norm(3+2i)=13 which is prime in Z and so not the product of Gaussian integers of smaller norm. On the other hand 3+i is not a Gaussian prime because norm(3+i)=10 and since 2 and 5 divide 10 then we might expect 3+i to be factorisable, which it is since 3+i=(1-i)(1+2i) and the norm of 1-i is 2 and that of 1+2i is 5.

When it comes to deciding whether ordinary primes are also Gaussian primes we come to an interesting property. For example, 2 is not a Gaussian prime because 2=(1+i)(1-i) and so can be divided by 1+i or 1-i. On the other hand 3 is a Gaussian prime. This is because norm(3)=9 and there are no Gaussian integers with norm 3 which divide 9 (the smallest non-trivial norms are 2, 5, 8...). Now 2 is the sum of two squares, whilst 3 is not and you can go on to show that ordinary prime p is a Gaussian prime if and only if p is not the sum of two squares (or alternatively ordinary prime p is the sum of two squares if and only if p is not a Gaussian prime).

This leads on to the proof of what I think is a fascinating theorem. Fermat's two square theorem states that if p=4n+1 is prime, then p=a²+b² for some a,b in Z. Since for p>2 every prime is either of the form 4n+1 or 4n+3, then this characterises half the primes as being the sum of two squares. The proof comes from the use of Lagrange's Lemma which states that a prime p=4n+1 divides m²+1 for some integer m (this comes from Wilson's theorem). Since m²+1=(m-i)(m+i) in Z[i] although p divides m²+1 it does not divide m-i or m+i as m/p - i/p and m/p + i/p are not Gaussian integers. By the Gaussian prime divisor property if a Gaussian prime divides the product of Gaussian numbers (m-i)(m+i) then it should divide either m-i or m+i and, as it doesn't, then p can't be a Gaussian prime. As we have seen from above this means that p is the sum of two squares.

Why I find this fascinating is that although prime numbers appear to be a random sequence of numbers we do have this structured form for half the set. It makes me wonder if there is any construction for primes of the form 4n+3? Also, are the a,b that are used to construct the primes of the form 4n+1 drawn from all of Z or are there numbers which are missing? All this gets me thinking!