 Solution

For the odd prime $p$, the Fermat-Euler theorem gives $x^{p-1}\equiv 1(\mathrm{mod}p)$. Therefore the residue class $z=x^{(p-1)/2}$ obeys $(z-1)(z+1)\equiv 0(\mathrm{mod}p)$. Since a prime number dividing a product divides one of its factors,If $x$ is a quadratic residue, choose $y$ with $y^2\equiv x(\mathrm{mod}p)$. Then $y$ is nonzero modulo $p$, and Fermat's little theorem givesThis proves the quadratic residue direction of Euler's criterion.