Let n be an odd positive number. Prove that n | $(2^{n!}-1)$.

Question

Let n be an odd positive number. Prove that n | $(2^{n!}-1)$. I don't know how to start this. If n is prime I might try Fermat's Little Theorem or something but as n is merely odd I don't know what to do.

There's a generalization of FLT, that uses the Euler's function. It may prove useful in this case — Exodd, May 16 '20 at 17:29
Thank you! @rtybase I believe these two questions are related. However I cannot seem to develop a proof for the case n as the link shows (n+1)(n-1) — RnHdw, May 16 '20 at 17:37
@RnHdw you will need to go through all the details, 6 minutes may not be enough. But the main takeaway is $\varphi(m)<m$. — rtybase, May 16 '20 at 17:41

score 2 · Answer 1 · answered May 16 '20 at 17:34

2

Hint Let $\phi(n)$ be the Euler totient function. Since $\phi(n) \leq n$ we have $\phi(n)|n!$.

answered May 16 '20 at 17:34

N. S.

132,525

Let n be an odd positive number. Prove that n | $(2^{n!}-1)$.

1 Answers1

Linked