It seems that $n^4+4$ and $(n+2)^4+4$ always have a prime factor other than $2$ or $5$. For example: $94^4+4$ and $96^4+4$ are both divisible by $4513$; $15^4+4$ and $17^4+4$ are both divisible by 257. I'm sure the reason is obvious, but I don't see it.
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3In fact both expressions have $n^2+2n+2$ as a nontrivial factor. – Peter Jun 21 '24 at 22:28
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@Peter Oh. Duh. – rogerl Jun 21 '24 at 22:31
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1$n^4+4 = (n^2+2n+2)(n^2-2n+2)$ and $(n+2)^4+4 =(n^2+2n+2)(n^2+6n+10)$ so they share a non trivial common factor. – Crostul Jun 21 '24 at 22:35
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2Cf. Sophie Germain’s identity – J. W. Tanner Jun 21 '24 at 22:46
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1Simpler: shift $,n\to n!-!1,$ then it's $\ n^2!+!1\mid (n \pm 1)^4!+4,$ by $,(i\pm!1)^4 = (\pm2 i)^2=-4\ \ $ – Bill Dubuque Jun 21 '24 at 23:40
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Or you could have brute-force computed their gcd by the Euclidean algoirthm $\ \ $ – Bill Dubuque Jun 21 '24 at 23:42