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I stumbled upon this problem and thought I would try to finish it (it did not get answered in the previous post), but the part with $n$ a power of two seems rather tricky. The problem is as follows:

Let $a$ and $n$ be positive integers with $a > 1$. Prove that, if $a^n+1$ is a prime number, then $a$ is even and $n$ is a power of $2$.

Suppose $a$ is odd. Then $a=2k+1$ and therefore $a^n+1$ is even, hence not a prime. So $a$ must be even.

Now, by way of contradiction, suppose that $n$ is a power of $2$ but $a^n+1$ is composite, i.e. $$a^n+1=mk$$ for some integers $m,k$. Then $a^n=mk-1$, and since $a \mid a^n$, we must have $a \mid mk-1$. Hence $ak'=mk-1$, or $ak'+1 =mk$. I was hoping this would land a contradiction, but I can't seem to find it.

I also tried proving by contrapositive, i.e. assuming that $n$ is not a power of two and then showing that it wasn't a prime, but this did not seem to bring about much.

Any ideas?

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Numbersandsoon
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    Variants of this have been asked many times. You will perhaps find enough information here. If not, there are plenty of others. – André Nicolas Dec 14 '13 at 08:02
  • Gotcha. Didn't see how the argument with the $a$ odd applied, but letting $n=2^m k$ with $k$ odd and writing $(a^{2^m})^k$ does the trick. Very nice with $x+1 \mid x^a+1$. – Numbersandsoon Dec 14 '13 at 08:16

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No problem: if n is not a power of 2 then you write n = k*2^p, with k odd then repeat your previous argument and you can show in this instant that a^n + 1 is composite because it has a^(2^p) + 1 as a factor.

DeepSea
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