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  "body": "\n## Statement\n\nLet $p$ be an odd prime.\nThen\n$$\\left(\\frac{2}{p}\\right) = (-1)^{(p^2-1)/8},$$\nwhere the left side is a [[Legendre symbol|01J8AMV9RS]].\n\n## Remarks\n\nSo $2$ is a quadratic residue modulo $p$ if $p \\equiv \\pm 1 \\pmod 8$, and a nonresidue if $p \\equiv \\pm 3 \\pmod 8$.\nFor example, $3^2 \\equiv 2 \\pmod 7$, but $2$ is not a square modulo $5$.\nOne proof uses [[Gauss's lemma|037ACYXB5Z]] with $a = 2$.\n\n## History\n\nEuler proved the second supplement.\n\n## Explanations\n\n### L1 human\n\nWhether $2$ is a square modulo $p$ depends only on the remainder of $p$ on division by $8$.\nThe remainders $1$ and $7$ give a square, and the remainders $3$ and $5$ do not.\n",
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  "name": "Second supplement to quadratic reciprocity",
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