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  "body": "\n## Proof\n\n1. Let $p$ and $q$ be distinct odd primes, $h = (p-1)/2$ and $k = (q-1)/2$. Put $S(q, p) = \\sum_{x=1}^{h} \\lfloor qx/p \\rfloor$ and $S(p, q) = \\sum_{y=1}^{k} \\lfloor py/q \\rfloor$.\n2. Since $p \\nmid q$, [[Eisenstein's lemma|0XSDBCP4RZ]] gives $\\left(\\frac{q}{p}\\right) = (-1)^T$ with $T = \\sum_u \\lfloor qu/p \\rfloor$, over the even $u$ with $2 \\le u \\le p - 1$.\n3. For $1 \\le x \\le h$, the number $qx/p$ is not an integer, by [[Euclid's lemma|051VYKA1EX]]. For a non-integer $z$ and an integer $N$, $\\lfloor N - z \\rfloor = N - 1 - \\lfloor z \\rfloor$, by the definition of the [[floor function|0YVRNEDSVG]].\n4. Let $u$ be even with $u \u003e h$, and put $x = p - u$. Then $x$ is odd and $1 \\le x \\le h$. By step 3, $\\lfloor qu/p \\rfloor = \\lfloor q - qx/p \\rfloor = q - 1 - \\lfloor qx/p \\rfloor$.\n5. Since $q - 1$ is even, $\\lfloor qu/p \\rfloor \\equiv \\lfloor qx/p \\rfloor \\pmod 2$.\n6. Every $x \\in \\set{1, \\dots, h}$ arises exactly once: an even $x$ from $u = x \\le h$, and an odd $x$ from the even number $u = p - x \\ge h + 1$.\n7. So $T \\equiv S(q, p) \\pmod 2$, and $\\left(\\frac{q}{p}\\right) = (-1)^{S(q, p)}$. Exchanging the roles of $p$ and $q$ gives $\\left(\\frac{p}{q}\\right) = (-1)^{S(p, q)}$.\n8. Let $R$ be the set of integer points $(x, y)$ with $1 \\le x \\le h$ and $1 \\le y \\le k$. It has $hk$ points.\n9. No point of $R$ satisfies $qx = py$. Otherwise $p \\mid qx$, so $p \\mid x$ by Euclid's lemma, which is impossible for $1 \\le x \u003c p$.\n10. Fix $x$ with $1 \\le x \\le h$. Since $qx/p$ is not an integer, the integers $y \\ge 1$ with $y \u003c qx/p$ are $1, \\dots, \\lfloor qx/p \\rfloor$.\n11. Also $qx/p \\le qh/p \u003c q/2$, so $\\lfloor qx/p \\rfloor \\le k$. Hence the points of $R$ with $py \u003c qx$ number $\\sum_{x=1}^{h} \\lfloor qx/p \\rfloor = S(q, p)$.\n12. In the same way, the points of $R$ with $qx \u003c py$ number $S(p, q)$.\n13. By steps 9, 11 and 12, $hk = S(q, p) + S(p, q)$.\n14. So the [[Legendre symbols|01J8AMV9RS]] satisfy $\\left(\\frac{p}{q}\\right) \\left(\\frac{q}{p}\\right) = (-1)^{S(p,q) + S(q,p)} = (-1)^{\\frac{p-1}{2} \\cdot \\frac{q-1}{2}}$.\n\n## Idea\n\nEisenstein's lemma writes $\\left(\\frac{q}{p}\\right)$ as $(-1)$ to a sum of floors, which counts lattice points under the line $y = qx/p$. Swapping $p$ and $q$ counts the points on the other side of the same line. Together they fill a rectangle with $\\frac{p-1}{2} \\cdot \\frac{q-1}{2}$ points, since no point lies on the line.\n",
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