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  "body": "\n## Proof\n\nLet $p$ be an odd [[prime|0SN0MAEKYX]]. Congruences are [[modulo|0TPW25CBN6]] the stated modulus.\n\n**If $p = x^2 + y^2$, then $p \\equiv 1 \\pmod 4$.**\n\n1. An even integer $2k$ has square $4k^2 \\equiv 0 \\pmod 4$. An odd integer $2k + 1$ has square $4k^2 + 4k + 1 \\equiv 1 \\pmod 4$.\n2. So $x^2 + y^2$ is $0$, $1$ or $2$ modulo $4$.\n3. Since $p$ is odd, $p$ is $1$ or $3$ modulo $4$. With $p = x^2 + y^2$, step 2 leaves only $p \\equiv 1 \\pmod 4$.\n\n**If $p \\equiv 1 \\pmod 4$, then $p = x^2 + y^2$.**\n\n4. Since $(p - 1)/2$ is even, the [[first supplement to quadratic reciprocity|0PHBS78HKT]] gives $\\left(\\frac{-1}{p}\\right) = 1$, a [[Legendre symbol|01J8AMV9RS]].\n5. So $-1$ is a [[quadratic residue|0SQDB87Z3T]] modulo $p$: there is an integer $a$ with $a^2 \\equiv -1 \\pmod p$.\n6. Let $k$ be the largest integer with $k^2 \\le p$. Since $p$ is prime, it is not a perfect square, so $k \u003c \\sqrt p \u003c k + 1$ by [[irrationality of square roots of non-squares|0N10YRNNWV]].\n7. Let $S$ be the set of pairs $(u, v)$ of integers with $0 \\le u, v \\le k$. It has $(k + 1)^2 \u003e p$ elements.\n8. Map $(u, v)$ to the residue of $u - av$ modulo $p$. There are only $p$ residues. By the [[pigeonhole principle|0HGVCWQGY3]], two distinct pairs $(u, v) \\ne (u', v')$ have $u - av \\equiv u' - av'$.\n9. Put $x = u - u'$ and $y = v - v'$. Then $x \\equiv ay \\pmod p$, the pair $(x, y)$ is not $(0, 0)$, and $\\abs{x}, \\abs{y} \\le k$.\n10. By steps 5 and 9, $x^2 + y^2 \\equiv a^2y^2 + y^2 = (a^2 + 1)y^2 \\equiv 0 \\pmod p$.\n11. By steps 6 and 9, $0 \u003c x^2 + y^2 \\le 2k^2 \u003c 2p$.\n12. The only multiple of $p$ strictly between $0$ and $2p$ is $p$. So $p = x^2 + y^2$.\n\n## Idea\n\nSquares are $0$ or $1$ modulo $4$, which gives one direction. For the other, $-1$ is a square $a^2$ modulo $p$. The pigeonhole principle finds small $x$ and $y$ with $x \\equiv ay$, so $p$ divides $x^2 + y^2$, and smallness forces $x^2 + y^2 = p$.\n",
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