# Math Help - quadratic non residue

2. Assume not so let $n$ be the smallest QNR such that $n=ab$. Then $-1=\left(\frac{n}{p}\right)=\left(\frac{ab}{p}\righ t) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right) = (1)(1) = 1$ since $a,b < n$. Hence a contradiction.