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  "body": "\n## Proof\n\n1. Let $p$ be an odd prime and $h = (p-1)/2$. Since $p \\nmid 2$, [[Gauss's lemma|037ACYXB5Z]] gives $\\left(\\frac{2}{p}\\right) = (-1)^n$. Here $n$ counts the $k \\in \\set{1, \\dots, h}$ whose residue of $2k$ exceeds $p/2$.\n2. For $1 \\le k \\le h$, we have $2 \\le 2k \\le p - 1$. So the least positive residue of $2k$ is $2k$ itself.\n3. So $n$ counts the integers $k$ with $p/4 \u003c k \\le h$. An integer $k$ satisfies $k \\le p/4$ exactly when $k \\le \\lfloor p/4 \\rfloor$, by the definition of the [[floor function|0YVRNEDSVG]].\n4. Since $\\lfloor p/4 \\rfloor \\le p/4 \u003c h$ for $p \\ge 3$, this gives $n = h - \\lfloor p/4 \\rfloor$.\n5. Write $p = 8m + c$ with $m \\ge 0$ and $c \\in \\set{1, 3, 5, 7}$. We compute $n$ and $N = (p^2 - 1)/8$ in each case.\n6. If $c = 1$: $h = 4m$ and $\\lfloor p/4 \\rfloor = 2m$, so $n = 2m$. Also $N = 8m^2 + 2m$. Both are even.\n7. If $c = 3$: $h = 4m + 1$ and $\\lfloor p/4 \\rfloor = 2m$, so $n = 2m + 1$. Also $N = 8m^2 + 6m + 1$. Both are odd.\n8. If $c = 5$: $h = 4m + 2$ and $\\lfloor p/4 \\rfloor = 2m + 1$, so $n = 2m + 1$. Also $N = 8m^2 + 10m + 3$. Both are odd.\n9. If $c = 7$: $h = 4m + 3$ and $\\lfloor p/4 \\rfloor = 2m + 1$, so $n = 2m + 2$. Also $N = 8m^2 + 14m + 6$. Both are even.\n10. In every case $n$ and $(p^2 - 1)/8$ have the same parity. So the [[Legendre symbol|01J8AMV9RS]] satisfies $\\left(\\frac{2}{p}\\right) = (-1)^n = (-1)^{(p^2-1)/8}$.\n\n## Idea\n\nFor $a = 2$ the multiples $2, 4, \\dots, p - 1$ are already reduced, so Gauss's lemma only asks how many even numbers below $p$ exceed $p/2$. That count is $h - \\lfloor p/4 \\rfloor$, and its parity depends only on $p$ modulo $8$, just like $(p^2 - 1)/8$.\n",
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