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  "body": "\n## Statement\n\nLet $m$ and $n$ be coprime positive odd integers.\nThen\n$$\\left(\\frac{m}{n}\\right) \\left(\\frac{n}{m}\\right) = (-1)^{\\frac{m-1}{2} \\cdot \\frac{n-1}{2}},$$\nwhere the symbols are [[Jacobi symbols|0XDB5AW7M1]].\n\n## Remarks\n\nThe supplements also extend: for a positive odd $n$, $\\left(\\frac{-1}{n}\\right) = (-1)^{(n-1)/2}$ and $\\left(\\frac{2}{n}\\right) = (-1)^{(n^2-1)/8}$.\nThe proof uses the [[law of quadratic reciprocity|085MJS6907]] for each pair of prime factors.\nWith this law, one computes $\\left(\\frac{a}{n}\\right)$ by an algorithm like the [[Euclidean algorithm|0TRNT4X7J0]], with no factorization.\n\n## Explanations\n\n### L1 human\n\nThe Jacobi symbol obeys the same reciprocity law as the Legendre symbol.\nSo you can flip the symbol, reduce the top number modulo the bottom number, and repeat.\nThis gives a fast way to compute Legendre symbols for large primes.\n",
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  "msc": [
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  "name": "Quadratic reciprocity for the Jacobi symbol",
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