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$x \in Z$. Prove, that all prime divisors of $x^{16}+1$, which are not equal to 2, are equal $1 \pmod{32}$.

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    I upvoted to reverse anonymous downvote. I am uncomfortable with immediate (i.e. $<$ 7 minutes) downvote for new user as opposed to leaving a comment such as "what is your background, what have you tried?" and giving them a chance to respond. New users have a negative emotional reaction to downvotes. – user2661923 Oct 04 '20 at 11:10
  • What have you tried? It is helpful to give a bit more context. – Dietrich Burde Oct 04 '20 at 11:21
  • Proof is the same as in the case $x=2$ in the dupe. – Bill Dubuque Oct 04 '20 at 18:19

2 Answers2

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Hint: If $x^{16}+1 \equiv 0 \bmod p$, then $x$ has multiplicative order $32$ mod $p$.

lhf
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Indeed, downvotes have a negative impact. Next time, show your thinking, show that at least you tried and people will happily help. Here is a proper solution:

Suppose $p|x^{16}+1$. Then $x^{16}\equiv-1\pmod{p}$, leding to $x^{32}\equiv1\pmod{p}$, if we square.

Denote with $ord_x(p)$ the order of $2 \pmod{p}$. It is known that $ord_x(p)|k$ if $x^k\equiv1\pmod{p}$

Thus, $ord_x(p)|32$, so let $ord_x(p)=2^a$, $1\leq a\leq 5$. If $a<5$, then $ord_x(p)|16$ which would imply that $x^{16}\equiv 1\pmod{p}$, contradiction.

So $ord_x(p)=32$. As I said before, $ord_x(p)|k$ if $x^k\equiv1\pmod{p}$. From Fermat's little theorem, because $x^{\phi(p)}=x^{p-1}\equiv1\pmod{p}$ (note that $gcd(p;n)=1$) we have $32|p-1$, so $p\equiv 1\pmod{32}$.