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  "body": "\n## Proof\n\n1. Let $p$ be an odd prime and $a$ an integer. Let $\\left(\\frac{a}{p}\\right)$ be the [[Legendre symbol|01J8AMV9RS]]. Note that $(p-1)/2 \\ge 1$ is an integer.\n2. If $p \\mid a$, then $p \\mid a^{(p-1)/2}$. So $a^{(p-1)/2} \\equiv 0 = \\left(\\frac{a}{p}\\right) \\pmod p$. Assume from now on that $p \\nmid a$.\n3. Let $x = a^{(p-1)/2}$. By [[Fermat's little theorem|0M25PRN3WQ]], $x^2 = a^{p-1} \\equiv 1 \\pmod p$. So $p \\mid (x - 1)(x + 1)$.\n4. By [[Euclid's lemma|051VYKA1EX]], $x \\equiv 1$ or $x \\equiv -1 \\pmod p$. These two cases exclude each other, because $p \\nmid 2$.\n5. Suppose $a$ is a [[quadratic residue|0SQDB87Z3T]], so $a \\equiv y^2 \\pmod p$ for an integer $y$. Then $p \\nmid y$, since $p \\nmid a$.\n6. Then $x \\equiv (y^2)^{(p-1)/2} = y^{p-1} \\equiv 1 \\pmod p$, using [[powers of congruences|0J5VGJYYC3]] and Fermat's little theorem. So $x \\equiv 1 = \\left(\\frac{a}{p}\\right)$.\n7. Now we count residues. Let $Q$ be the set of nonzero classes in $\\mathbb{Z}/p\\mathbb{Z}$ that are squares. Each such class is $\\overline{y}^2$ with $1 \\le y \\le p - 1$.\n8. Since $(p - y)^2 \\equiv y^2 \\pmod p$, each class in $Q$ is $\\overline{y}^2$ for some $y$ with $1 \\le y \\le (p-1)/2$.\n9. Let $1 \\le y \u003c z \\le (p-1)/2$. If $y^2 \\equiv z^2$, then $p \\mid (z - y)(z + y)$. But $0 \u003c z - y \u003c p$ and $0 \u003c z + y \u003c p$, so this contradicts Euclid's lemma.\n10. So the classes $\\overline{y}^2$ with $1 \\le y \\le (p-1)/2$ are distinct, and $Q$ has exactly $(p-1)/2$ elements.\n11. By steps 5 and 6, every class in $Q$ is a root of the polynomial $X^{(p-1)/2} - 1$ over $\\mathbb{Z}/p\\mathbb{Z}$.\n12. The ring $\\mathbb{Z}/p\\mathbb{Z}$ [[is a field|0R8YXRC9AJ]]. So this polynomial has [[at most $(p-1)/2$ roots|0KJVQV3E0F]]. By steps 10 and 11, its roots are exactly the elements of $Q$.\n13. Suppose $a$ is a quadratic nonresidue. Then $\\overline{a} \\notin Q$, so $\\overline{a}$ is not a root, and $x \\not\\equiv 1 \\pmod p$.\n14. By step 4, $x \\equiv -1 = \\left(\\frac{a}{p}\\right) \\pmod p$.\n15. In all cases, $a^{(p-1)/2} \\equiv \\left(\\frac{a}{p}\\right) \\pmod p$.\n\n## Idea\n\nThe number $x = a^{(p-1)/2}$ squares to $1$, so it is $\\pm 1$ modulo $p$. Residues give $+1$ at once by Fermat. There are exactly $(p-1)/2$ nonzero residues, and they already fill all the roots of $X^{(p-1)/2} - 1$, so a nonresidue must give $-1$.\n",
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