, 2016

Solution:
Clearly $p=2$ works with solutions $(0,0)$ and $(1,1)$ and not $(0,1)$ or $(1,0)$.

If $p \equiv 3 \pmod{4}$ then $-1$ is not a quadratic residue, so for $x^{3}+4 x \neq 0$, exactly one of $x^{3}+4 x$ and $-x^{3}-4 x$ is a square and gives two solutions (for positive and negative $y$), so there's exactly two solutions for each such pair $\{x,-x\}$. If $x$ is such that $x^{3}+4 x=0$, there's exactly one solution.

If $p \equiv 1 \pmod{4}$, let $i$ be a square root of $-1 \pmod{p}$. The right hand side factors as $x(x+2i)(x-2i)$. For $x=0,2i,-2i$ this is zero, there is one choice of $y$, namely zero. Otherwise, the right hand side is nonzero. For any fixed $x$, there are either $0$ or $2$ choices for $y$. Replacing $x$ by $-x$ negates the right hand side, again producing two choices for $y$ since $-1$ is a quadratic residue. So the total number of solutions $(x, y)$ is $3 \pmod{4}$, and thus there cannot be exactly $p$ solutions.