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  "body": "\n## Statement\n\nLet $p$ and $q$ be distinct odd primes.\nThen\n$$\\left(\\frac{p}{q}\\right) \\left(\\frac{q}{p}\\right) = (-1)^{\\frac{p-1}{2} \\cdot \\frac{q-1}{2}},$$\nwhere the symbols are [[Legendre symbols|01J8AMV9RS]].\n\n## Remarks\n\nSo $\\left(\\frac{p}{q}\\right) = \\left(\\frac{q}{p}\\right)$ unless $p \\equiv q \\equiv 3 \\pmod 4$. In that case $\\left(\\frac{p}{q}\\right) = -\\left(\\frac{q}{p}\\right)$.\nFor example, $\\left(\\frac{7}{3}\\right) = \\left(\\frac{1}{3}\\right) = 1$, and $3 \\equiv 7 \\equiv 3 \\pmod 4$. So $\\left(\\frac{3}{7}\\right) = -1$.\nWith the [[first supplement|0PHBS78HKT]], the [[second supplement|0ZXW95S44M]] and [[multiplicativity|0NW1JP7TBR]], the law computes every Legendre symbol.\n\n## History\n\nEuler and Legendre conjectured the law.\nGauss first proved it in 1796, and he published proofs in his Disquisitiones Arithmeticae of 1801.\nGauss called it the fundamental theorem.\n\n## Explanations\n\n### L1 human\n\nThe law connects two questions: is $p$ a square modulo $q$, and is $q$ a square modulo $p$?\nThe answers are the same, except when both primes leave remainder $3$ on division by $4$. Then the answers are opposite.\nThis lets you swap and reduce the numbers, as in the [[Euclidean algorithm|0TRNT4X7J0]].\n",
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