2. Prove it by contradiction, assume none of them were a quadratic residue, then $\displaystyle \left( {\tfrac{a} {p}} \right) = \left( {\tfrac{b} {p}} \right) = \left( {\tfrac{{ab}} {p}} \right) = - 1$ however remember that $\displaystyle \left( {\tfrac{a} {p}} \right) \cdot \left( {\tfrac{b} {p}} \right) = \left( {\tfrac{{ab}} {p}} \right)$