# properties of the Legendre symbol

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

###### Proposition.

The following properties are satisfied:

1. 1.

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

2. 2.

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

3. 3.

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

4. 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.

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.

Title properties of the Legendre symbol PropertiesOfTheLegendreSymbol 2013-03-22 16:17:52 2013-03-22 16:17:52 alozano (2414) alozano (2414) 4 alozano (2414) Theorem msc 11-00 EulersCriterion