Typical problem on P with conditions I am looking for prime $p$ greater than or equal to $3$ such that $p|y^2 + 4$ as well as $4|p-3$. I need simple discussion to conclude the existence of $p$.
Thanks again.

If there were such a $y$, then the congruence $z^2\equiv -1\pmod{p}$ would have a solution. (Multiply $y$ by the inverse of $2$ modulo $p$.)
But it is a standard fact that if $4$ divides $p-3$, then $-1$ cannot be a quadratic residue of $p$.
