# How can we summarize Legendre symbol $(\frac{11}{p})$

We note that $11\equiv 3 \pmod 4$. So by using the law of quadratic reciprocity to get $(\frac{p}{11})$, we need to discuss the residue of $p\pmod 4$.

I'm wondering how to give a specific formula like $(\frac{2}{p})=1$ if $p\equiv \pm1\pmod 8$ and $(\frac{2}{p})=-1$ if $p\equiv \pm 3\pmod 8$. By the way, if we need to find the fomula for $(\frac{2}{p})$ at first, why should we discuss $8k\pm1$ and $8k\pm3$?

## 1 Answer

$f(n) = (\frac{n}{q})=\prod_{p^k \| q} (\frac{n}{p})^k$ is a Dirichlet character (completely multiplicative and $q$-periodic). So is $g(n) = (\frac{q}{n})= \prod_{p^k \| n} (\frac{q}{p})^k$, and this is really the statement of quadratic reciprocity. Here you can show it from $i \sqrt{11} = \sum_{n=1}^{11} \zeta_{11}^{n^2}$, yielding $$\sqrt{11} (\frac{11}{p}) \equiv \sqrt{11} \ 11^{(p-1)/2} \equiv \sqrt{11}^p \equiv(-i\sum_{n=1}^{11} \zeta_{11}^{n^2})^p \equiv(-i)^p \sum_{n=1}^{11} \zeta_{11}^{p n^2} \bmod p$$ which depends only on $p n^2 \bmod 11$ and $p \bmod 4$, thus on $p \bmod 44$.