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?
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1Either $p \mid 4+n$ or $p \mid 4-n.$ But $p$:cannot divide both of them. For otherwise $p \mid 8 \implies p=2.$ But then $2p^4 - p^2 + 16 = 44,$ which is not a perfect square. This shows that either $p^2 \mid 4+n$ or $p^2 \mid 4-n.$ – little o Apr 20 '19 at 18:00
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See also this question, or this one. – Dietrich Burde Apr 20 '19 at 18:05
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1$p=3$ so that $2p^4-p^2+16=13^2$. – Dietrich Burde Apr 20 '19 at 18:13
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@Dietrich Burde can you please check my solution below? – little o Apr 20 '19 at 19:06
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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.
little o
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How did you conclude that 2p^4−p^2+16≡2(mod 3)? Did you just check for the next few primes? – Tapi Apr 20 '19 at 19:08
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3Because if $p \equiv 1\ \text {or}\ 2\ (\text {mod}\ 3)$ then $p^2 \equiv 1\ (\text {mod}\ 3)$ and hence $p^4 \equiv 1\ (\text {mod}\ 3).$ So we have $$\begin{align} 2p^4-p^2 + 16 & \equiv 2.1 - 1 + 16\ (\text {mod}\ 3). \ & \equiv 17\ (\text {mod}\ 3). \ & \equiv 2\ (\text {mod}\ 3). \end{align}$$ Hope it will help. – little o Apr 20 '19 at 19:13