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  "body": "\n## Proof\n\n1. Let $p$ be an odd prime and $a, b$ integers. Let $e = (p-1)/2$.\n2. By [[Euler's criterion|0M7DEDKSYB]], $\\left(\\frac{ab}{p}\\right) \\equiv (ab)^e \\pmod p$.\n3. We have $(ab)^e = a^e b^e$. By Euler's criterion for $a$ and for $b$, and [[products of congruences|072D5G4SK6]], $a^e b^e \\equiv \\left(\\frac{a}{p}\\right) \\left(\\frac{b}{p}\\right) \\pmod p$.\n4. So $u = \\left(\\frac{ab}{p}\\right)$ and $v = \\left(\\frac{a}{p}\\right) \\left(\\frac{b}{p}\\right)$ satisfy $u \\equiv v \\pmod p$.\n5. By the definition of the [[Legendre symbol|01J8AMV9RS]], $u$ and $v$ lie in $\\set{-1, 0, 1}$. So $\\abs{u - v} \\le 2$.\n6. Since $p \\mid u - v$ and $p \\ge 3 \u003e \\abs{u - v}$, we get $u - v = 0$.\n7. Hence $\\left(\\frac{ab}{p}\\right) = \\left(\\frac{a}{p}\\right) \\left(\\frac{b}{p}\\right)$.\n\n## Idea\n\nEuler's criterion turns the Legendre symbol into the power $a^{(p-1)/2}$ modulo $p$, and powers are multiplicative. Two numbers in $\\set{-1, 0, 1}$ that agree modulo an odd prime are equal.\n",
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