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,
$$\left(\frac{a}{p}\right)\equiv {a}^{(p1)/2},\left(\frac{b}{p}\right)\equiv {b}^{(p1)/2},\text{and}\left(\frac{ab}{p}\right)\equiv {(ab)}^{(p1)/2}modp.$$ 
It is clear then that $(a/p)(b/p)\equiv (ab/p)modp$. 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 nonresidues modulo $p$ is a quadratic residue modulo $p$, which is not at all obvious.
Title  properties of the Legendre symbol 

Canonical name  PropertiesOfTheLegendreSymbol 
Date of creation  20130322 16:17:52 
Last modified on  20130322 16:17:52 
Owner  alozano (2414) 
Last modified by  alozano (2414) 
Numerical id  4 
Author  alozano (2414) 
Entry type  Theorem 
Classification  msc 1100 
Related topic  EulersCriterion 