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Subject says it all. I'm sure I've seen this somewhere, so if someone can link me to a wiki page or post that explains, that works for me.

Trevor
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    If $a$ has an odd divisor $d>1$ , you can write the number as $z^d+1$ , which is divisible by $z+1$ , being a non-trivial factor for $z\ge 2$ – Peter Feb 25 '20 at 16:33
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    The conclusion should be $a=2^k.$ – Thomas Andrews Feb 25 '20 at 16:36
  • @Peter Ah, well that would make more sense. – lulu Feb 25 '20 at 16:39
  • In particular , the primes of the form $\ 2^n+1\ $ , $\ n\ $ a positive integer , are Fermat primes. – Peter Feb 25 '20 at 16:39
  • In the list of the largest known primes, the Mersenne numbers play the major role, followed by the generalized Fermat numbers (numbers of the form $$a^{2^n}+1$$) The reason for such numbers to have a good chance to be prime is that every prime factor must be of the form $$k\cdot 2^{n+1}+1$$ assuming that $\ a\ $ is even. – Peter Feb 25 '20 at 16:45
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    Technically, if $x$ and $a$ are to be positive integers, the claim assumes $x\ge2$. Else we get $1+1=2$ for any $a$. When $x=1$ one of the identidied factors of $x^a+1$ for $a$ not a power of $2$ falls to $1$. – Oscar Lanzi Feb 25 '20 at 16:48
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    The title of my choice of duplicate target feels more specific than the current one, but its answer covers this question as well. See also this and this for more discussion and links. – Jyrki Lahtonen Feb 25 '20 at 18:11

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If $n$ is any odd integer $\ge 3$, then $X^n+1$ factorises as $$(X+1)(X^{n-1}-X^{n-2}+X^{n-3}-\cdots-X+1)$$ (Note that this factorisation fails if $n$ is even.)

So now if $a$ has any odd factor $n\ge 3$, we can put $a=kn$ for some $k$, and $X=x^k$, and this gives us a factorisation of $x^a+1$:

$$x^a+1=X^n+1=(X+1)(X^{n-1}-\cdots -X+1)$$

Therefore if $a$ has any odd factor $\ge 3$, then $x^a+1$ is not prime.

Now note that the numbers with no odd factors $\ge 3$ are precisely the powers of $2$.

TonyK
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