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  "body": "\n## Statement\n\nLet $p$ be an odd prime and let $a$ be an integer with $p \\nmid a$.\nReduce each of the integers $a, 2a, \\dots, \\frac{p-1}{2} a$ to its least positive residue modulo $p$.\nLet $n$ be the number of these residues that are greater than $p/2$.\nThen $\\left(\\frac{a}{p}\\right) = (-1)^n$, where the left side is the [[Legendre symbol|01J8AMV9RS]].\n\n## Remarks\n\nFor example, take $p = 7$ and $a = 3$. The residues of $3, 6, 9$ are $3, 6, 2$, and only $6$ is greater than $7/2$. So $\\left(\\frac{3}{7}\\right) = -1$.\nWith $a = 2$, the lemma gives the [[second supplement to quadratic reciprocity|0ZXW95S44M]].\nGauss used the lemma in one of his proofs of the [[law of quadratic reciprocity|085MJS6907]].\n\n## Explanations\n\n### L1 human\n\nWrite the first half of the multiples of $a$ modulo $p$, and count how many land in the upper half of $1, \\dots, p - 1$.\nAn even count means that $a$ is a square modulo $p$, and an odd count means that it is not.\n",
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