Irish Mathematical Olympiad

We have
$$\begin{aligned} 0 &= p(h+z) - p(h-z) \\ &= a((h+z)^4 - (h-z)^4) + b((h+z)^3 - (h-z)^3) \\ &\quad + c((h+z)^2 - (h-z)^2) + d((h+z) - (h-z)) \\ &= 8a(h^3z + hz^3) + 2b(3h^2z + z^3) + 4chz + 2dz \\ &= z(8ah^3 + 6bh^2 + 4ch + 2d) + z^3(8ah + 2b). \end{aligned}$$
This is satisfied for all $z \in \mathbb{C}$ iff
$$8ah^3 + 6bh^2 + 4ch + 2d = 0 \quad \text{and} \quad 8ah + 2b = 0.$$
The second equation implies $h = -\frac{b}{4a}$. Substituting this into the first of these equations, we obtain, after multiplying by $4a^2$,
$$b^3 - 4abc + 8a^2d = 0.$$
Thus, this condition is necessary. Conversely, if it holds, then, with $h = -b/(4a)$ we see that $8ah^3 + 6bh^2 + 4ch + 2d = 0$ and $8ah + 2b = 0$ hold.