Prove that if $x,y$ are coprime numbers then every odd factor of $x^{2^n}+y^{2^n}$ where $n$ is a positive integer , is of the form $2^{n+1}m+1$.

Asked Feb 07 '23 at 16:47

Active Feb 23 '23 at 22:15

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Prove that if $x$ and $y$ have no common factor then every odd factor of $x^{2^n}+y^{2^n}$ where $n$ is a positive integer , is of the form $2^{n+1}m+1$.

I have tried using order. Let $k$ be a odd factor of $$x^{2^n}+y^{2^n}.$$ I assumed the order of $x^{2^n}$ w.r.t. $k$ be $m$. So $x^{(2^n)^m} \equiv 1 \pmod{k}$. Now how do I proceed with this idea? I would like to get some hints.

edited Feb 23 '23 at 22:15

Bill Dubuque

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asked Feb 07 '23 at 16:47

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Does this answer your question? Number Theory Problem Using Group Theory - found using an Approach0 search. Also, an alternate, with some more detail, explanation is in the AoPS thread order problem. ... – John Omielan Feb 07 '23 at 20:02
(cont.) In addition, these two proof methods of the problem are similar to that outlined about Édouard Lucas' theorem that "Any prime divisor $p$ of $F_n=2^{2^n}+1$ is of the form $k2^{n+2}+1$ whenever $n\gt 1$". Finally, FYI, this other question's answer has a link to Fermat numbers and Wieferich primes, with this having some results & references which might be interesting to you regarding your question. – John Omielan Feb 07 '23 at 20:12

Prove that if $x,y$ are coprime numbers then every odd factor of $x^{2^n}+y^{2^n}$ where $n$ is a positive integer , is of the form $2^{n+1}m+1$.

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