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I had this question because I had to prove for a problem that $2^n+1 \not \mid (n-1)!$, so I thought it would be nice for there to be a prime factor of $2^n+1$ that is greater than $n-1$. I checked this using code and it's true up to $n=89$, but $2^n$ grows pretty fast. I also tried using assuming all prime divisors $p$ of $2^n+1$ are less than $n$ for contradiction, but I couldn't get anything from it aside from the fact that if $2^n+1$ is semiprime then it has a prime factor that is greater than $n$. However there exist some of these numbers that aren't at most semiprime ($2^9+1=513=3^3\cdot19$).

Is there something very simple I'm missing? Is there a really big counterexample I don't know about?

Bill Dubuque
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Semicolumn
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  • Idea: have you looked at the gcd of $2^n+1$ and $(n-1)!$? – Johannes Kloos May 14 '24 at 07:59
  • Doesn't that require information about the divisors of $2^n+1$? – Semicolumn May 14 '24 at 08:08
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    If $p$ is a prime divisor of $2^n+1$ that is not a divisor of any of $2^{n/q}+1$, where $q$ is a prime factor of $n$, then the order of $2$ modulo $p$ is equal to $2n$, and consequently we can deduce that $p\equiv1\pmod{2n}$. Your example $2^9+1=3^2\cdot19$ is a case in point: $19\equiv1\pmod{2\cdot9}$. – Jyrki Lahtonen May 14 '24 at 08:54
  • A "famous" case of that mechanism in action is that of Fermat numbers, when $n$ is a power of two. Check out this oldie and the questions linked to it. – Jyrki Lahtonen May 14 '24 at 08:59
  • Thank you! I wonder though, is there a guarantee that $2^n+1$ will introduce a new prime factor in relation to $2^{n/q}+1$ for any prime divisor $q$? I remember seeing a theorem like this once but I forgot what it was called. – Semicolumn May 14 '24 at 09:05
  • The line in your question "...so there should be a prime factor of $2n+1$ that is greater than $n−1$" is incorrect. $a\nmid b$ does not imply that $a$ has a prime factor that doesn't divide $b$; for instance, $4\nmid 6$ but all the prime factors of $4$ divide $6$. Obviously, if you can prove that the largest prime factor of $2^n+1$ is greater than $n-1$ you get the result, but it may not be the only way. (For reference, OEIS has the sequence of largest prime factors of $2^n+1$ here: https://oeis.org/A002587 ) – Chris Lewis May 14 '24 at 09:09
  • "So there should be a prime factor" is less of a formal statement and more of a heuristic. $2^n+1$, at the time, felt like it should have a prime factor greater than $n-1$. – Semicolumn May 14 '24 at 09:12
  • Edited the post to explain the logic of looking for such a prime factor better. – Semicolumn May 14 '24 at 09:21
  • For future reference, it's Zsigmondy's theorem. – Semicolumn May 14 '24 at 09:35
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    This time Google worked better than ApproachZero. – Jyrki Lahtonen May 14 '24 at 10:13

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