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  "body": "\n## Proof\n\n1. Let $m$ and $n$ be coprime positive odd integers. For an odd integer $r$, put $\\delta(r) = (r - 1)/2$, an integer.\n2. Claim: for odd $r$ and $s$, $\\delta(rs) \\equiv \\delta(r) + \\delta(s) \\pmod 2$. Indeed $(rs - 1) - (r - 1) - (s - 1) = (r-1)(s-1)$. Both factors are even, so this is divisible by $4$, and $\\delta(rs) - \\delta(r) - \\delta(s)$ is even.\n3. By induction on $t$, $\\delta(r_1 \\cdots r_t) \\equiv \\sum_{i=1}^{t} \\delta(r_i) \\pmod 2$ for odd $r_1, \\dots, r_t$. For $t = 0$, both sides are $\\delta(1) = 0$.\n4. By the [[fundamental theorem of arithmetic|084HH1NDCB]], write $m = p_1 \\cdots p_s$ and $n = q_1 \\cdots q_t$ as products of primes, listed with repetition. All of them are odd, and $p_i \\ne q_j$ for all $i, j$ because $\\gcd(m, n) = 1$.\n5. By the definition of the [[Jacobi symbol|0XDB5AW7M1]], $\\left(\\frac{m}{n}\\right) = \\prod_{j=1}^{t} \\left(\\frac{m}{q_j}\\right)$, with [[Legendre symbols|01J8AMV9RS]] on the right.\n6. By the [[multiplicativity of the Legendre symbol|0NW1JP7TBR]], $\\left(\\frac{m}{q_j}\\right) = \\prod_{i=1}^{s} \\left(\\frac{p_i}{q_j}\\right)$. So $\\left(\\frac{m}{n}\\right) = \\prod_{i, j} \\left(\\frac{p_i}{q_j}\\right)$.\n7. In the same way, $\\left(\\frac{n}{m}\\right) = \\prod_{i, j} \\left(\\frac{q_j}{p_i}\\right)$.\n8. For each pair $(i, j)$, the [[law of quadratic reciprocity|085MJS6907]] gives $\\left(\\frac{p_i}{q_j}\\right) \\left(\\frac{q_j}{p_i}\\right) = (-1)^{\\delta(p_i) \\delta(q_j)}$.\n9. Multiplying over all pairs, $\\left(\\frac{m}{n}\\right) \\left(\\frac{n}{m}\\right) = (-1)^{\\sum_{i,j} \\delta(p_i) \\delta(q_j)} = (-1)^{A B}$, where $A = \\sum_i \\delta(p_i)$ and $B = \\sum_j \\delta(q_j)$.\n10. By step 3, $A \\equiv \\delta(m)$ and $B \\equiv \\delta(n) \\pmod 2$. So $AB \\equiv \\delta(m) \\delta(n) \\pmod 2$.\n11. Hence $\\left(\\frac{m}{n}\\right) \\left(\\frac{n}{m}\\right) = (-1)^{\\frac{m-1}{2} \\cdot \\frac{n-1}{2}}$.\n\n## Idea\n\nBoth Jacobi symbols break into Legendre symbols over pairs of prime factors, one from $m$ and one from $n$. Quadratic reciprocity handles each pair. The signs combine correctly because $(r-1)/2$ is additive modulo $2$ under multiplication of odd numbers.\n",
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