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A work of Cameron Stewart (the paper has appeared in Acta Mathematica), proving a conjecture of Erdos, Stewart shows that the largest prime factor of $2^n-1$ is at least $n \times \exp\Big( \frac{\log n}{104 \log \log n}\Big)$ , if $n$ is large enough.

Zsigmondy's theorem is on the same topic.

I would like to know about the largest prime factor of $2^n+1$. I have searched, but most of the time result of the largest prime factor of $2^n-1$ appears.

My questions is-

  1. Is there any result regarding the largest prime factor of $2^n+1$ ?

if you know anything related to the problem, please inform.

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    $\begingroup$ Hmm, because $2^n+1 = (2^{2n}-1)/(2^n-1)$ - isn't then the greatest prime factor of$2^n+1$ likely/in many cases that of $2^{2n}-1$ (because the greatest primefactor of the latter should-according to the formula- be larger than that of $2^n-1$) ? $\endgroup$
    – Gottfried Helms
    Commented Aug 13, 2016 at 17:36
  • $\begingroup$ @GottfriedHelms Why? The bound for the greatest prime factor of $2^{2n} - 1$ is smaller than $4n^2$, much smaller than $2^n - 1$. $\endgroup$
    – WhatsUp
    Commented Aug 14, 2016 at 3:24
  • $\begingroup$ At least when $n$ is odd, the result of Stewart still applies and gives the same bound. $\endgroup$
    – WhatsUp
    Commented Aug 14, 2016 at 3:33

1 Answer 1

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One has $2^n+1= \frac{2^{2n}-1}{2^n-1} = \prod_{d \mid 2n,~d\nmid n} \Phi_d(2,1)$. In particular, $\Phi_{2n}(2,1)$ divides $2^n+1$, and so the main theorem of Stewart's paper shows immediately that $P(2^n+1) > 2n \exp(\log(2n)/104\log\log(2n))$ for large $n$.

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