Choose $n = m^2$ and we have that $am^4 + bm^2 + c$ is a quadratic residue modulo $p$ for every prime and hence a perfect square by a well-known result that $k$ is a quadratic residue modulo every prime $\Longleftrightarrow k$ is a perfect square.
However, by another well-known result (and not that hard to prove) that a polynomial is a perfect square at every integer iff it is a perfect square in $\mathbb{Z}[x]$, we have $am^4 + bm^2 + c = P(m)^2$ for some integer polynomial $P$.
Remark that then $P$ must be of the form $P(x) = dx^2 + e$, hence $(a,b,c) = (d^2, 2de, e^2)$ for some selection of $d,e$. Clearly the problem is true in this case, and hence we are done.