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  "body": "\n## Proof\n\n1. Let $p$ be an odd prime. Then $(p-1)/2$ is a natural number.\n2. By [[Euler's criterion|0M7DEDKSYB]] with $a = -1$, $\\left(\\frac{-1}{p}\\right) \\equiv (-1)^{(p-1)/2} \\pmod p$.\n3. Since $p \\nmid -1$, the [[Legendre symbol|01J8AMV9RS]] $\\left(\\frac{-1}{p}\\right)$ is $1$ or $-1$. Also $(-1)^{(p-1)/2}$ is $1$ or $-1$.\n4. So the two sides of step 2 differ by $0$ or by $\\pm 2$. Since $p \\ge 3$ divides the difference, the difference is $0$.\n5. Hence $\\left(\\frac{-1}{p}\\right) = (-1)^{(p-1)/2}$.\n\n## Idea\n\nEuler's criterion computes the symbol as $(-1)^{(p-1)/2}$ modulo $p$. Both sides are signs, and two signs that agree modulo an odd prime are equal.\n",
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