Coming back to this, consider the following.
[Warning: read like math ...]
Let p be a prime of the form 4k+1. It's easy to show that there exists u such that u^2=-1 (mod p). Indeed, simply using (2k)! works.
Consider all number of the form a+b.u (mod p) where a and b are in the range 0 <= a,b < sqrt(p). There are more than p of these, so by the pigeon-hole principle we have a0, b0, a1, b1 such that a0+b0.u = a1+b1.u (mod p).
Rearrange to give a0-a1 = (b1-b0).u (mod p) and square both sides. Remembering that u^2=-1 (mod p) we get
(a0-a1)^2 + (b1-b0)^2 = 0 (mod p)
A little checking shows that (a0-a1)^2 + (b1-b0)^2 < 2p, so that means
(a0-a1)^2 + (b1-b0)^2 = p
Thus we have shown that p is the sum of two squares.
I can reproduce this proof at will. I see how the proof works, I am convinced it's correct and valid. I still don't really understand why every prime of the form 4k+1 is the sum of two squares, especially since every argument someone tries to give to show somehow it's inevitable also works for 4k+3, where it's not true.