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Fix an even integer $n\geq 2$. Are there infinitely many primes of the form $a^n+1$, where $a$ is an integer?

Is there some theorem covering this, or is the problem still open for all even $n$?

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If $n=2\,k$ then $a^{2k}+1=(a^k)^2+1$. The existence of infinite primes of the form $x^2+1$ is an open problem

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There are definitely NOT infinitely many primes when $n>1$ is odd, since it'll factor as $(a+1)(a^{n-1}-a^{n-2}+\dots)$, where the latter factor is greater than $1$, whenever $a$ is greater than $1$.

Edit: Actually, for $n$ to be able to possibly satisfy such a condition, it must be a power of $2$, since if there was an odd prime $p$ dividing $n$,say $n=pk$, then again write $a^{pk}+1=(a^k)^p+1$ etc.

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  • $\begingroup$ Yes, sorry I meant even $n$ $\endgroup$ – python55 Jun 22 '15 at 6:52
  • $\begingroup$ @vgmath, What is the number of possible even values of $n$? Finite, finitely many, infinitely many ? $\endgroup$ – lab bhattacharjee Jun 22 '15 at 6:53

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