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Find all prime numbers $p$ such that $5^p+ 4p^4$ is a perfect square.

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    $\begingroup$ $p=5$ is a solution. Did you really try? $\endgroup$ – Dietrich Burde Apr 20 at 16:36
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    $\begingroup$ How hard did you search? $p=5$ works. $\endgroup$ – lulu Apr 20 at 16:36
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    $\begingroup$ As a way to get started: Note that $5^p+4p^4=n^2\implies 5^p=(n-2p^2)(n+2p^2)$ so both of the factors on the right must be powers of $5$. $\endgroup$ – lulu Apr 20 at 16:38
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$5$ is the only such prime. Note that if $$5^p+4p^4=n^2$$ Then $$5^p=(n+2p^2)(n-2p^2)$$ Each factor must be a power of $5$, hence the difference of the factors is a multiple of $5$ unless $5^p=4p^2+1$. If this is not the case, then this difference is $4p^2$ and a multiple of $5$, so $p$ is a multiple of $5$. Since it is prime, $p=5$.

Otherwise, since $p^2>1$ we have $5p^2>5^p$, so $p^2>5^{p-1}$. By induction we can prove that this entails that $p<5$, so this is impossible to satisfy given the conditions.

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  • $\begingroup$ What if one of the factors equals $\pm1$? $\endgroup$ – Servaes Apr 23 at 7:36
  • $\begingroup$ @Serv It can't be $-1$, but it could be $1$. Let me think about that case. $\endgroup$ – Matt Samuel Apr 23 at 14:07
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    $\begingroup$ @Serva See my edit. $\endgroup$ – Matt Samuel Apr 23 at 14:27

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