Solution: 0
This is equivalent to both discriminants $p^{2} \pm 4 c$ being squares. In other words, $p^{2}$ must be the average of two squares $a^{2}$ and $b^{2}$. Note that $a$ and $b$ must have the same parity, and that $\left(\frac{a+b}{2}\right)^{2}+\left(\frac{a-b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{2}=p^{2}$. Therefore, $p$ must be the hypotenuse in a Pythagorean triple. Such triples are parametrized by $k\left(m^{2}-n^{2}, 2 m n, m^{2}+n^{2}\right)$. But $p \equiv 3(\bmod 4)$ and is therefore not the sum of two squares. This implies that $p$ is not the hypotenuse of any Pythagorean triple, so the answer is 0 .