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Here's the conjecture:

For $n\in\mathbb{N}^*$, $(2^n+1)$ always has a prime factor with form $(2nk+1)$ where $k\in\mathbb{N}^*$, with an exception when $n=3$.

For example, when $n=5$, we have $2^5+1=33=3\times 11$, here the factor $11$ can be written as $2\times 5\times 1+1$, which means $k=1$ satisfies the conjecture.

I've verified this conjecture for $n\le 100$ but can't come up with an idea to prove or disprove it. I noticed that Zsigmondy's Theorem has a similar form but I can't find a good way to apply it in this problem. Any help would be appreciated!

Edit: Note that $2^n+1$ is not necessarily a Fermat number. Therefore, avoid conflating this query with those about prime divisors of Fermat numbers.

Mivik
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  • this is what you nedd. – Peter Aug 23 '21 at 13:23
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    With $a=2$ and $b=1$ you have the desired statement. "primitive" means that the order is maximal ($2n$) for some prime factor and hence this prime factor must be congruent to $1$ modulo $2n$. – Peter Aug 23 '21 at 13:38
  • Fermat primes appear to be an exception to your conjecture. – ordptt Aug 23 '21 at 14:18
  • @ordptt 65537 itself is the desired prime factor since 65537=2162048+1 – Mivik Aug 23 '21 at 14:20
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    I see. Well, then at least it is easy to check that the conjecture holds for every fermat prime. – ordptt Aug 23 '21 at 14:44
  • @Peter Why do you say "'primitive' means that the order is maximal (2n) for some prime factor"? I can't quite see the relation here, though the later part is ok for me. Could give some detailed explanation on this? Thank you so much! – Mivik Aug 23 '21 at 14:51
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    I meant : There is a primitive prime factor (which means that it has maximum order). Maybe , this was formulated confusing. – Peter Aug 23 '21 at 15:58
  • I have found the following OEIS entry that is related to this question: https://oeis.org/search?q=3%2C5%2C17%2C11%2C13%2C43%2C257%2C19%2C41&sort=&language=english&go=Search

    It does not provide a proof; however, this is a strong evidence that the conjecture is true.

     – Josh Aug 23 '21 at 20:48

1 Answers1

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By Zsigmondy's theorem there is an odd prime $p$ which divides $2^n+1$ but does not divide any number of the form $2^k+1$ for $k< n$. We have $2^n\equiv -1\pmod{p}$, so $2^{2n}\equiv 1\pmod{p}$. If you manage to prove that $2n$ is exactly the order of $2$ in $\mathbb{Z}/p\mathbb{Z}^*$ you are done, since the order of an element has to divide the order of the group, and $2n\mid (p-1)$ is equivalent to $p=2nk+1$. Essentially you just have to prove that $2^d\not\equiv 1\pmod{p}$ for any divisor $d$ of $2n$.

Bill Dubuque
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Jack D'Aurizio
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  • Actually you don't have to proof that the other is exactly $2n$ since we already have $2^{2n}=1\pmod p$ and if we prove that $2n<p$ then it's enough to say that $2n$ is a factor of $\varphi(p)=p−1$. – Mivik Aug 24 '21 at 11:01
  • The argument using "order" can be done w/o any knowledge of group theory using the Order Test, e.g. see here and here. – Bill Dubuque May 14 '24 at 16:15