2. Since 4 is a perfect square: $\displaystyle \left( {\tfrac{{ - 4}} {p}} \right) = \left( {\tfrac{{ - 1}} {p}} \right)\left( {\tfrac{4} {p}} \right) = \left( {\tfrac{{ - 1}} {p}} \right) = {\left( { - 1} \right)^{\tfrac{{p - 1}} {2}}} = 1$
Let's denote $\displaystyle a=\left( {\tfrac{{p - 1}} {4}} \right)$. We have $\displaystyle \left( {\tfrac{a} {p}} \right) = \left( {\tfrac{a} {p}} \right)\left( {\tfrac{{ - 4}} {p}} \right) = \left( {\tfrac{{ - 4a}} {p}} \right) = \left( {\tfrac{{ - \left( {p - 1} \right)}} {p}} \right) = \left( {\tfrac{1} {p}} \right) = 1$