properties of the Legendre symbol

Let $p$ be an odd prime and let $a$ be an arbitrary integer. Let $\left({\displaystyle \frac{a}{p}}\right)$ be the Legendre symbol of $a$ modulo $p$. Then:

Proposition.

The following properties are satisfied:

1. If $a\equiv bmodp$ then $\left({\displaystyle \frac{a}{p}}\right)=\left({\displaystyle \frac{b}{p}}\right)$.

2. If $a\ne 0modp$ then $\left({\displaystyle \frac{a^2}{p}}\right)=1$.

3. If $a\ne 0modp$ and $b\in \mathbb{Z}$ then $\left({\displaystyle \frac{a^{2}b}{p}}\right)=\left({\displaystyle \frac{b}{p}}\right)$.

4. $\left({\displaystyle \frac{a}{p}}\right)\left({\displaystyle \frac{b}{p}}\right)=\left({\displaystyle \frac{ab}{p}}\right)$.

Proof.

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

The fourth property is a consequence of Euler’s criterion. Indeed,

$$\left(\frac{a}{p}\right)\equiv a^{(p-1)/2},\left(\frac{b}{p}\right)\equiv b^{(p-1)/2},\text{and }\left(\frac{ab}{p}\right)\equiv {(ab)}^{(p-1)/2}modp.$$

It is clear then that $(a/p)(b/p)\equiv (ab/p)modp$. Sincethe numbers involved are all $\pm 1$ or $0$, the congruence alsoholds with equality in $\mathbb{Z}$.∎

Remark.

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.