The existence of a solution of a quadratic congruence modulo $p$

Question

Let $p$ be an odd prime number and assume that $p\equiv 7(\textrm{mod}\,8)$. Since $(2\mid p)=1$, it follows that the congruence $2x^2\equiv1(\textrm{mod}\,p)$ is solvable. If $2a^2\equiv1(\textrm{mod}\,p)$, I want to prove that exactly one of the following congruences is solvable. $$ 2x^2-1\equiv a(\textrm{mod}\,p)\qquad ;\qquad 2x^2-1\equiv -a(\textrm{mod}\,p) $$ How to prove it? Thanks!