For example: We know for $b=12$ and $n=1$, $12^1+1=13$ is prime. A quick check with WolframAlpha shows that there no other primes of the form $12^n+1$ up to $n=1000$. Are the any other primes for this base? Also what about numbers of the form $b^n+1$ where $b$ is some fixed even integer?

My thoughts:

  • for $b=2^k$ $$b^n+1=2^{kn}+1$$ so if we have the primes of the form $2^n+1$, we have those of $(2^k)^n+1$ aswell.

  • for $b=k^2$ $$b^n+1=({k^n})^{2}+1$$ and it is an open question, as these numbers are a subset of numbers of the form $m^2+1$.

Now let $b=12$ for example, then $b$ is not a square number and no power of $2$. What can we say about this base?

  • $\begingroup$ Related topic $\endgroup$ – Jakobian Nov 2 '18 at 11:32
  • $\begingroup$ For $n$ odd and greater than $1$, we have $b^n+1 = (b + 1)(b^{n-1}-b^{n-2}+\cdots + 1)$, so at the very least, $n$ needs to be even. In fact, any odd divisor of $n$ gives a similar decomposition, so $n$ must be a power of $2$. $\endgroup$ – Arthur Nov 2 '18 at 11:33
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    $\begingroup$ If $b\ge 2$, $b^n+1$ prime implies that $n$ is a power of $2$ $\endgroup$ – ajotatxe Nov 2 '18 at 11:33
  • $\begingroup$ $b$ can't be odd, it would mean that $2|b^n+1$ (unless $b=1$, but that's irrelevant) $\endgroup$ – Jakobian Nov 2 '18 at 11:34
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    $\begingroup$ I doubt that this is known for any base. People have long searched for the next Fermat prime and, in failing to find one, have come to suspect that there are no more...but there is no proof either way. $\endgroup$ – lulu Nov 2 '18 at 11:36

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