Question:

Prove: If p = 1(mod 4) is a prime, then -4 and (p-1)/4 are both quadratic residues of p.

2. Since 4 is a perfect square: $\left( {\tfrac{{ - 4}}
{p}} \right) = \left( {\tfrac{{ - 1}}
{p}} \right)\left( {\tfrac{4}
{p}} \right) = \left( {\tfrac{{ - 1}}
{p}} \right) = {\left( { - 1} \right)^{\tfrac{{p - 1}}
{2}}} = 1$

Let's denote $a=\left( {\tfrac{{p - 1}}
{4}} \right)$
. We have $\left( {\tfrac{a}
{p}} \right) = \left( {\tfrac{a}
{p}} \right)\left( {\tfrac{{ - 4}}
{p}} \right) = \left( {\tfrac{{ - 4a}}
{p}} \right) = \left( {\tfrac{{ - \left( {p - 1} \right)}}
{p}} \right) = \left( {\tfrac{1}
{p}} \right) = 1$