Number Theory or Geometry?

First prove that every prime has the form $p\equiv1\pmod4$ can be expressed as the sum of squares:

Relevant Wiki: Legendre Symbol

Because of Legendre Symbol: $(-1/p)=1,$ if $p\equiv1\pmod4$ , we can find an integer $a$ which is coprime to $p$ , satisfies $a^2\equiv-1\pmod p$ .

Let's first look at this case: $ax\equiv y\pmod p$ , there is at least a solution $x_0,y_0$ which $ax_0\equiv y_0\pmod p$ , satisfies $0<\lvert x_0\rvert<\sqrt p,0<\lvert y_0\rvert<\sqrt p$ (see the proof below). Squaring both side gives $x^2+y^2\equiv0\pmod p$ , but $0<x_0^2+y_0^2<2p$ forced $x^2+y^2=p$ .

Next step, by letting $q=2xy,r=x^2-y^2$ (WLOG, let $x>y$ ), we can yields a triplet of $q,r,p$ which are three sides of a right angle triangle with, integer values of side lengths.

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The proof:

Consider the set $S=\{ax-y|0\leq x<\sqrt p,0\leq y<\sqrt p\}$ , there are exactly $(\lfloor p\rfloor+1)^2>\sqrt p^2=p$ elements inside that, by Pigeonhole Theorem, at least a pair of integers in the set are congruent modulo $p$ , says $(x_1,y_1),(x_2,y_2)$ , so $\begin{aligned}ax_1-y_1&\equiv ax_2-y_2\pmod p\\\\a(x_1-x_2)&\equiv y_1-y_2\pmod p\end{aligned}$ By letting $x_0=x_1-x_2,y_0=y_1-y_2$ , we conclude that this is the same as $ax_0\equiv y_0\pmod p$ , but if $x_0=0$ will cause $y_0=0$ , this says $x_1=x_2,y_1=y_2$ , exactly the same element in the set $S$ , contradict the assumption, this completes the proof.