$a^n+1$ prime $\Rightarrow n = 2^k$ is a power of $2$

Question

I'm trying to prove that $a^n+1$ can only be prime if $n$ is a power of $2$. Is there a general factorization of $a^n+1$?

Answer 1

Hint: $\ a^{\large k}+1\mid a^{\large n}+1\,$ if $\ k\mid n\ $ & $ $ quotient $\,{\color{#0af}{{\large \frac{n}k}=m}}\,$ is $\rm\color{#0af}{odd},\,$ by

$\bmod\, a^{\large k}+1\!:\,\ \color{#c00}{a^{\large k}\equiv -1} \,\Rightarrow\,\ a^{\large n} =\, (\color{#c00}{a^k})^{\large \color{#0af}{\frac{n}k}}\equiv (\color{#c00}{-1})^{\large \color{#0af}m}\equiv\:\! {-}1$

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Or $ $ put $\ x = a^{\large k}\ $ in $\ x+1\mid x^{\large\color{#0af}m} + 1,\, $ via Factor Theorem: $\ x-c\, \mid\, f(x)-f(c)\ $ in $\ \mathbb Z[x].$

Note that this is the special case $\,\color{#0af}m\,$ odd, $\ x\to -x\ $ in this prior question today.

Answer 2

For some non-negative integers $b$ and $k$, we have $n=2^b(2k+1)$, in which case

$$a^n+1 = (a^{2^b}+1)(a^{2k\cdot2^b}-a^{(2k-1)2^b}+a^{(2k-2)2^b}-\cdots -a^{3\cdot2^b}+a^{2\cdot2^b}-a^{2^b}+1)$$

If $k>0$ and $a>1$ then this is a factorisation (if $a=1$ then $a^n+1$ is prime); but if $k=0$, i.e. if $n$ is a power of 2, then it leaves you where you started so it is still possible that $a^n+1$ could be prime.