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QUESTION

Let $p=a^m+1$ be a prime, where $a\geq2$ and $m\geq1$ are
integers. Prove that $a$ must be even and $m=2^n$ for some
positive integer, $n$.



ANSWER

If $q$ is odd then we have the following identity between
polynomials with integral coefficients

$$t^q+1=(t+1)(t^{q-1}-t^{q-2}+\ldots-t+1).$$

For our purposes it would be sufficient to know that $t+1$ divides
$t^q+1$ in $Z[t]$. Now write $m=2^nq$ where $q$ is odd. Setting
$t=a^{2^n}$ yields

$$t^q+1=(a^{2^n})^q+1=a^{2^nq}+1=p.$$

Therefore $a^{2^n}+1$ divides $p$ in the integers. However, $2\leq
a^{2^n}+1$ for all $n$ and $a^{2^n}+1=p$ if and only if $q=1$.
Hence $p$ has proper divisors unless $n=2^n$. Even so, for any
$m,\ p$ would be even if $a$ were odd, so $a$ must be even.





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