Prime numbers & perfect squares

Question

Find all prime numbers such that $2p^4-p^2+16$ is a perfect square.

$2p^4-p^2+16=n^2$

$16-n^2=p^2-2p^4$

$(4-n)(4+n)=p^2(1-2p^2)$

What should I do next?

Answer 1

Observe that for $p>3$ we have $$p \equiv 1\ \text {or}\ 2\ (\text {mod}\ 3).$$

Let for $p>3$ $\exists$ $k \in \Bbb N$ such that $$2p^4-p^2+16 = k^2.$$ But then $$2p^4-p^2+16 \equiv 2\ \not\equiv 0\ \text {or}\ 1 \equiv k^2\ (\text {mod}\ 3).$$

So $p=3$ is the only solution.