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Use the given integer $u$ satisfying $u^2\equiv -1(\mathrm{mod}p)$. The Euclidean lattice under consideration has basis $(1,u),(0,p)$, because every vector in it is uniquely $(a,ua+bp)$ with $a,b\in \mathbb{Z}$. Its fundamental parallelogram therefore has areaWe use the following two-dimensional form of the Minkowski convex body theorem: a convex, centrally symmetric measurable subset of ${\mathbb{R}}^2$ of area strictly greater than four times the covolume of a full-rank Euclidean lattice contains a nonzero vector of that lattice. Apply it to the closed disk of squared radius $3p/2$. Its area is $3\pi p/2>4p$, so it contains a nonzero lattice vector $(a,b)$ satisfyingOn the other hand the defining modular congruence and $u^2\equiv -1(\mathrm{mod}p)$ implyThere is exactly one positive multiple of $p$ strictly below $2p$, so $\boxed{p=a^2+b^2}$. Neither coordinate can be zero, since a prime is not the square of an integer. This congruence lattice proof of the prime sum of two squares proves the required case of the sum of two squares theorem without assuming any conclusion at the critical area $4p$.

A positive integer $n$ is a sum of two squares of integers exactly when every prime factor congruent to three modulo four has an even prime exponent in its prime factorization. The zero-square convention permits $1=1^2+0^2$; zero itself separately has $0=0^2+0^2$.

For necessity, let $p\equiv 3(\mathrm{mod}4)$ divide $a^2+b^2$. If $p\nmid b$, then $(a{b}^{-1}{)}^2\equiv -1(\mathrm{mod}p)$, contradicting Euler's criterion for minus one at such a prime. Hence $p\mid b$, and then $p\mid a$ as well. Divide both coordinates by $p$ and repeat. Each removal subtracts two from the prime valuation of $n$, forcing it to be even.

For sufficiency, two is $1^2+1^2$. A prime number congruent to one modulo four has a square root of minus one by Euler's criterion, and the congruence lattice proof of the prime sum of two squares turns that root into a representation of the prime. Each even power of a prime congruent to three modulo four is already a perfect square, hence a sum of two squares with a zero term. Finally, the Brahmagupta-Fibonacci identity combines the representations across all prime powers in $n$.

For example, $45=3^2\cdot 5=6^2+3^2$ qualifies, while $21=3\cdot 7$ does not. The criterion examines prime exponents, rather than only the residue of the whole number modulo four.