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  "body": "\n## Statement\n\nLet $p$ be an odd prime and let $a$ be an integer.\nThen\n$$a^{(p-1)/2} \\equiv \\left(\\frac{a}{p}\\right) \\pmod p,$$\nwhere $\\left(\\frac{a}{p}\\right)$ is the [[Legendre symbol|01J8AMV9RS]].\n\n## Remarks\n\nSo for $p \\nmid a$, the integer $a$ is a quadratic residue modulo $p$ if and only if $a^{(p-1)/2} \\equiv 1 \\pmod p$.\nFor example, $3^3 = 27 \\equiv -1 \\pmod 7$, so $3$ is a quadratic nonresidue modulo $7$.\nThe proof uses [[Fermat's little theorem|0M25PRN3WQ]]: $x = a^{(p-1)/2}$ satisfies $x^2 \\equiv 1$, so $x \\equiv \\pm 1 \\pmod p$.\nA consequence is the [[multiplicativity of the Legendre symbol|0NW1JP7TBR]].\n\n## History\n\nLeonhard Euler found the criterion.\n\n## Explanations\n\n### L1 human\n\nTo test whether $a$ is a square modulo $p$, raise $a$ to the power $(p-1)/2$.\nThe result is $1$ for a square and $-1$ for a non-square, modulo $p$.\nWith [[fast modular exponentiation|04DR5PM25N]], this test is quick even for large $p$.\n",
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