Proposition 1   The following properties are satisfied:

1. If $a\equiv b \mod p$ then $\displaystyle \left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$ .
2. If $a\neq 0 \mod p$ then $\displaystyle \left(\frac{a^2}{p}\right)=1$ .
3. If $a\neq 0 \mod p$ and $b\in \mathbb{Z}$ then $\displaystyle \left(\frac{a^2b}{p}\right)=\left(\frac{b}{p}\right)$ .
4. $\displaystyle \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\left(\frac{ab}{p}\right)$ .

Proof. The first three properties are immediate from the definition of the Legendre symbol. Remember that $(a/p)$ is $1$ if $x^2\equiv a \mod p$ has solutions, the value is $-1$ if there are no solutions, and equals $0$ if $a\equiv 0 \mod p$ .

The fourth property is a consequence of Euler's criterion. Indeed, $$\left(\frac{a}{p}\right)\equiv a^{(p-1)/2},\quad \left(\frac{b}{p}\right)\equiv b^{(p-1)/2},\quad \text{and } \left(\frac{ab}{p}\right)\equiv (ab)^{(p-1)/2} \mod p.$$ It is clear then that $(a/p)(b/p)\equiv (ab/p) \mod p$ . Since the numbers involved are all $\pm 1$ or $0$ , the congruence also holds with equality in $\mathbb{Z}$ .

Remark 1   Property (4) is somewhat surprising because, in particular, it says that the product of two quadratic non-residues modulo $p$ is a quadratic residue modulo $p$ , which is not at all obvious.