If $1+a^n$ is prime for some $a\geq 2$ and $n\geq 1$, show that $n$ must be a power of $2$.
I made this; $a^n + 1$ is prime. Then $a^n$ must be even. Then $a^n=2k$. That's it :)
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Clearly, $a^n$ must be even $\iff a$ must be even
Now if $n$ has an odd factor $f(>1),n= d\cdot f$ (say)
$\displaystyle a^n+1=a^{df}+1=(a^d)^f-(-1)^f$ will be divisible by $a^d-(-1)=a^d+1$ which is definitely $>1$
References :
lab bhattacharjee
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What The fact that it's divisible tells us here? Why is a even? – AYARcom Jun 11 '14 at 16:24
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@AYARcom, If $a$ is odd, so must be $a^n$. Hence $a^n+1$ will be even and $>2$ for $a\ge2,$ right? Hence, composite – lab bhattacharjee Jun 11 '14 at 16:26