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  "body": "\n## Proof\n\nLet $\\nu_p$ be the [[p-adic valuation|04VCPS30ZE]]. Call $n$ a sum of two squares if $n = x^2 + y^2$ with integers $x, y$.\n\n**Lemma.** Let $p \\equiv 3 \\pmod 4$ be prime with $p \\mid x^2 + y^2$. Then $p \\mid x$ and $p \\mid y$.\n\n1. Suppose $p \\nmid y$. Then $\\gcd(y, p) = 1$, and [[Bézout's identity|0DQVHMYAGA]] gives a [[modular inverse|00GXTQMY8Z]] $w$ with $yw \\equiv 1 \\pmod p$.\n2. Multiplying $x^2 + y^2 \\equiv 0$ by $w^2$ gives $(xw)^2 \\equiv -1 \\pmod p$. So $-1$ is a [[quadratic residue|0SQDB87Z3T]] modulo $p$, and $\\left(\\frac{-1}{p}\\right) = 1$ as a [[Legendre symbol|01J8AMV9RS]].\n3. But $(p - 1)/2$ is odd, so the [[first supplement|0PHBS78HKT]] gives $\\left(\\frac{-1}{p}\\right) = -1$. This contradiction shows $p \\mid y$.\n4. Then $p \\mid x^2$, and $p \\mid x$ by [[Euclid's lemma|051VYKA1EX]].\n\n**If $n$ is a sum of two squares, then $\\nu_p(n)$ is even for each prime $p \\equiv 3 \\pmod 4$.**\n\n5. Fix such a prime $p$. We use strong [[induction|0ASJVEFA9J]] on $n \\ge 1$. Let $n = x^2 + y^2$.\n6. If $p \\nmid n$, then $\\nu_p(n) = 0$, which is even.\n7. If $p \\mid n$, the lemma gives $x = px'$ and $y = py'$ with integers $x', y'$. Then $n = p^2 n'$ with $n' = x'^2 + y'^2$, and $1 \\le n' \u003c n$.\n8. By the induction hypothesis, $\\nu_p(n')$ is even. By the [[valuation of a product|0850JJMJYZ]], $\\nu_p(n) = 2 + \\nu_p(n')$ is even.\n\n**If $\\nu_p(n)$ is even for each prime $p \\equiv 3 \\pmod 4$, then $n$ is a sum of two squares.**\n\n9. By [[the square times square-free decomposition|080Q3RPKY6]], $n = a^2 b$ with positive integers $a$, $b$ and $b$ [[square-free|004JW7A2N5]].\n10. Let $p \\equiv 3 \\pmod 4$ be a prime. If $p \\mid b$, then $\\nu_p(b) = 1$, as $p^2 \\nmid b$. Then $\\nu_p(n) = 2\\nu_p(a) + 1$ is odd, which is excluded. So $p \\nmid b$.\n11. By [[existence of prime factorization|020K0X7HVR]], $b = q_1 \\cdots q_r$ with primes $q_i$, where $r \\ge 0$. By step 10, each $q_i$ is $2$ or a prime $q_i \\equiv 1 \\pmod 4$.\n12. Each $q_i$ is a sum of two squares: $2 = 1^2 + 1^2$, and the others by [[Fermat's theorem on sums of two squares|043B3ENSTY]].\n13. By the [[Brahmagupta–Fibonacci identity|0K2DMACEZ4]], a product of two sums of two squares is a sum of two squares. By induction on $r$, $b$ is a sum of two squares. For $r = 0$, $b = 1 = 1^2 + 0^2$.\n14. Write $b = u^2 + v^2$. Then $n = a^2 b = (au)^2 + (av)^2$, a sum of two squares.\n\n## Idea\n\nA prime $p \\equiv 3 \\pmod 4$ cannot divide $x^2 + y^2$ without dividing both $x$ and $y$, because $-1$ is not a square modulo $p$. So such primes come off in pairs. Conversely, the remaining primes are sums of two squares, and the Brahmagupta–Fibonacci identity multiplies them together.\n",
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