Let $p > 2$ be a prime number. The Legendre symbols are completely multiplicative, i.e. for any two integers $n,m\in\mathbb Z$ we have $$\left(\frac{nm}{p}\right)=\left(\frac{n}{p}\right)\cdot\left(\frac{m}{p}\right).$$
In other words: * The congruence $x^2(p)\equiv nm(p)$ is solvable if and only if the congruences $x^2(p)\equiv n(p)$ and $x^2(p)\equiv m(p)$ are solvable.
Yet in other words: * If $n$ and $m$ are both quadratic residues (respectively both quadratic nonresidues) modulo $p$, then their product $nm$ is a quadratic residue modulo $p.$ * If one of the numbers $n$ and $m$ is a quadratic residue modulo $p$ and the other a quadratic nonresidue modulo $p$, then their product $nm$ is a quadratic nonresidue modulo $p.$
In general, if $p > 2$, $r\ge 2,$ and $n_1,\ldots,n_r$ are integers, then $$\left(\frac{n_1\cdots n_r}{p}\right)=\left(\frac{n_1}{p}\right)\cdots\left(\frac{n_r}{p}\right).$$
Proofs: 1
