From a necessary condition for an integer to be prime it follows that $n^{\frac{p-1}{2}}(p)\equiv\pm 1(p),$ since $n^{p-1}(p)\equiv 1(p)$ if $p\not\mid n$ and $p$ is a prime number. This motivates the following criterion, found by Leonhard Euler.
Let $p > 2$ be a prime number and let $n\in\mathbb Z$ be a given integer. The Legendre symbol modulo $p$ can be calculated using the formula
$$\left(\frac np\right)(p)\equiv n^{\frac{p-1}{2}}(p).$$
In particular: * $n$ is a quadratic residue modulo $p$ if and only if $p\not\mid n$ and $n^{\frac{p-1}{2}}(p)\equiv 1(p),$ * $n$ is a quadratic nonresidue modulo $p$ if and only if $p\not\mid n$ and $n^{\frac{p-1}{2}}(p)\equiv (p-1)(p)\equiv -1(p),$ * otherwise, we have $p\mid n$ and $n^{\frac{p-1}{2}}(p)\equiv 0(p).$
Proofs: 1
