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In Wikipedia there's a list of quite non-trivial and beautiful proofs of Fermat's two squares theorem. Actually I'm a bit surprised because this fact belongs to a very small set of mathematical fact that I, well, "see" and understand.

Here's how I'm approaching it: Let's assume that $a^2 + b^2 = p$, where $p$ is prime. For $p>2$ if $a^2$ is even, than $b^2$ is odd - and vice versa. A square can be even only if the original number is even, the same about an odd square. That actually means that we can rewrite as sum as $4n^2 + (2m + 1)^2 = p$.

This could be rewritten as $4(n^2 + m^2 + m) + 1 = p$. We actually hadn't use the fact that p is prime so far, only the fact that prime numbers are odd. Also we can notice that n should be even as well, i.e the even factor is divisible by four but not by any bigger power of two.

I do realize that it could not be that simple so I'm requesting for pointing out what atually I'm missing.

shabunc
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    You aren't missing anything. The square of any integer is either of the form $4k$ or $4k+1$. If you add two together then you can only hope to get integers of the form $4k$, $4k+1$, or $4k+2$. You can never get an integer of the form $4k+3$. Of course, the hard part of the theorem is to show that any prime of the form $4k+1$ actually can be expressed as the sum of two squares. – lulu Sep 03 '15 at 14:36
  • @lulu oh! I see, I should proof that there always such sum existing. Thanks! – shabunc Sep 03 '15 at 14:40
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    As you'll see from the wikipedia page you linked to, that part is really not easy at all. But the proofs are beautiful and the techniques (like infinite descent) are very important in number theory. Enjoy! – lulu Sep 03 '15 at 14:44

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