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  "body": "\n## Proof\n\nWrite $N(\\gamma) = \\gamma\\bar\\gamma$ for the [[norm|00HPT88ZX0]] of a [[Gaussian integer|0B268GVX95]] $\\gamma$. Then $N(\\gamma\\delta) = N(\\gamma) N(\\delta)$, since complex conjugation is multiplicative.\n\n1. A [[prime number|0SN0MAEKYX]] $p$ is either $2$, or odd with $p \\equiv 1 \\pmod 4$, or odd with $p \\equiv 3 \\pmod 4$. We treat the three cases.\n2. Case $p = 2$. We have $2 = (1 + i)(1 - i)$, so $2$ divides $(1 + i)(1 - i)$ in $\\mathbb{Z}[i]$.\n3. But $2 \\nmid 1 + i$ and $2 \\nmid 1 - i$, because $(1 \\pm i)/2$ has non-integer real part. So $2$ is not a [[Gaussian prime|0GTEHJQRFK]].\n4. Case $p \\equiv 1 \\pmod 4$. By [[Fermat's theorem on sums of two squares|043B3ENSTY]], $p = a^2 + b^2$ with integers $a$, $b$. So $p = (a + bi)(a - bi)$.\n5. If $p \\mid a + bi$ in $\\mathbb{Z}[i]$, then $p \\mid a$ and $p \\mid b$ in $\\Z$, so $p^2 \\mid a^2 + b^2 = p$. This is false. Likewise $p \\nmid a - bi$.\n6. So $p$ divides $(a + bi)(a - bi)$ but neither factor, and $p$ is not a Gaussian prime.\n7. Case $p \\equiv 3 \\pmod 4$. First, $p \\ne 0$, and $p$ is not a unit, since the [[units|06RH4XAGRY]] of $\\mathbb{Z}[i]$ are $\\pm 1$ and $\\pm i$.\n8. By the [[first supplement to quadratic reciprocity|0PHBS78HKT]], the [[Legendre symbol|01J8AMV9RS]] of $-1$ is $(-1)^{(p-1)/2} = -1$. So $-1$ is not a square modulo $p$.\n9. Claim: if $a$, $b$ are integers with $p \\mid a^2 + b^2$, then $p \\mid a$ and $p \\mid b$. Suppose $p \\nmid b$. By [[Bézout's identity|0DQVHMYAGA]], there is an integer $c$ with $bc \\equiv 1 \\pmod p$.\n10. Multiplying $a^2 + b^2 \\equiv 0$ by $c^2$ gives $(ac)^2 \\equiv -1 \\pmod p$, against step 8. So $p \\mid b$, and then $p \\mid a^2$, so $p \\mid a$ since $p$ is prime.\n11. Now let $\\alpha = a + bi$ and $\\beta$ be Gaussian integers with $p \\mid \\alpha\\beta$ and $p \\nmid \\alpha$. We show $p \\mid \\beta$.\n12. Since $p \\nmid \\alpha$, $p$ does not divide both $a$ and $b$. By the claim, $p \\nmid N(\\alpha) = a^2 + b^2$.\n13. As $p$ is prime, $\\gcd(N(\\alpha), p) = 1$. By Bézout's identity, $u N(\\alpha) + v p = 1$ for some integers $u$, $v$.\n14. Then $\\beta = u N(\\alpha) \\beta + v p \\beta = u \\bar\\alpha (\\alpha\\beta) + v p \\beta$. Both terms are divisible by $p$ in $\\mathbb{Z}[i]$, so $p \\mid \\beta$.\n15. By steps 7 and 14, $p$ is a Gaussian prime when $p \\equiv 3 \\pmod 4$.\n16. By steps 3, 6 and 15, $p$ is a Gaussian prime if and only if $p \\equiv 3 \\pmod 4$.\n\n## Idea\n\nFor $p = 2$ and $p \\equiv 1 \\pmod 4$, the prime is a product $(a + bi)(a - bi)$ and divides this product without dividing either factor.\nFor $p \\equiv 3 \\pmod 4$, the fact that $-1$ is not a square mod $p$ makes $N(\\alpha)$ prime to $p$ whenever $p \\nmid \\alpha$.\nThen a Bézout relation lets us cancel $\\alpha$ from $p \\mid \\alpha\\beta$, as for ordinary primes.\n",
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