\\begin{thm}
Any prime equivalent to $1$ modulo $4$ is the sum of two squares.
\\end{thm}
\\begin{proof}
Let $p$ be a prime such that $p \\euqiv 1 \\pmod{4}$.
We already know that $-1$ is a quadratic residue modulo $p$, so for some $a$, $a^1+1^2 \\equiv 0 \\pmod{p}$.
Now let $m$ be the minimal natural number for which $mp$ is a sum of two squares, and write $a^2+b^2 = mp$. Take $a' \\equiv a \\pmod{m}$ and $b' \\equiv b \\pmod{b}$ such that $|a'|,|b'|\\leq m/2$.

Notice that module $m$, $aa'+bb' \\equiv a^2 + b^2 = mp \\equiv 0$, and $ab'-a'b \\equiv ab-ba = 0$. Thus, $x = \\frac{aa'+bb'}{m}$ and $y = \\frac{ab'-ba'}{m}$ are integers. Now $$x^2+y^2 = \\frac{(a^2+b^2)(a'^2+b'^2)}{m^2} = \\frac{a'^2+b'^2}{m}\\cdot p,$$
but $(a'^2+b'^2)/m \\leq 2(m/2)^2/m = m/2 < m$, contradicting the minimality of $m$, unless $x = y = 0$. But then $a' = b' = 0$ and $m\\divides a,b$, so $m^2 \\divides a^2+b^2 = mp$, so $m \\divides p$ and $m = 1$.
\\end{proof}