Find all prime numbers $p$ such that $5^p+ 4p^4$ is a perfect square.
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Tapi
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6$p=5$ is a solution. Did you really try? – Dietrich Burde Apr 20 '19 at 16:36
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6How hard did you search? $p=5$ works. – lulu Apr 20 '19 at 16:36
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1As 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 $5$. – lulu Apr 20 '19 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.
Matt Samuel
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@Serv It can't be $-1$, but it could be $1$. Let me think about that case. – Matt Samuel Apr 23 '19 at 14:07
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