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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]]. If $p = 2$, then $x = 1$ gives $x^2 = 1 \\equiv -1 \\pmod 2$. Also $2 \\not\\equiv 3 \\pmod 4$. So the statement holds for $p = 2$.\n2. Let $p$ be odd. Then $p \\equiv 1$ or $p \\equiv 3 \\pmod 4$.\n3. There is an integer $x$ with $x^2 \\equiv -1 \\pmod p$ exactly when $-1$ is a [[quadratic residue|0SQDB87Z3T]] modulo $p$.\n4. Since $p \\nmid -1$, this holds exactly when the [[Legendre symbol|01J8AMV9RS]] $\\left(\\frac{-1}{p}\\right)$ equals $1$.\n5. By the [[first supplement|0PHBS78HKT]], $\\left(\\frac{-1}{p}\\right) = (-1)^{(p-1)/2}$. This is $1$ exactly when $(p-1)/2$ is even, that is, when $4 \\mid p - 1$.\n6. So for odd $p$, a solution exists exactly when $p \\equiv 1 \\pmod 4$. By step 2, this means $p \\not\\equiv 3 \\pmod 4$.\n\n## Idea\n\nFor odd $p$, the first supplement gives the answer as the sign $(-1)^{(p-1)/2}$, which is $+1$ exactly for $p \\equiv 1 \\pmod 4$. The prime $2$ is checked by hand.\n",
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  "name": "Proof that −1 is a square modulo p if and only if p is not 3 modulo 4",
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