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Jan 3, 2017 at 21:41 comment added Noam D. Elkies Yes, that's what I'm guessing, based on the fact that for each $\epsilon>0$ random $x$ has all prime factors are less than $x^\epsilon$ with positive probability, but also has a prime factor greater than $x^{1-\epsilon}$ with positive probability. But the first probability decays rapidly enough ($\exp$ of something like $\log(\epsilon) / \epsilon$) that the first example with $2^p-1 \mid 2^{5p}-1$ might be quite large.
Jan 3, 2017 at 21:23 comment added Gerhard Paseman Fortunately you found an example involving a Mersenne prime and posted it just before I posted my comment. It would be nice to know if similar examples existed using other bases. Or even which N have N^4 + 5(N^3 + 2N^2 + 2N +1) break into 4 or more times as many prime factors as N has. Gerhard "Cell Phones Are Much Slower" Paseman, 2017.01.03.
Jan 3, 2017 at 21:20 comment added Noam D. Elkies ($p$ being a large prime.)
Jan 3, 2017 at 21:14 comment added Noam D. Elkies There aren't that many Mersenne primes. But there should be a positive probability that $2^p-1$ has a prime factor $> 2^{cp}$ for say $c = .8$, and also a positive and independent probability that $2^{3p}-1$ has no larger prime factor. So there should be a positive albeit small density of counterexamples among exponents $3p$. Likewise for $5p$, $7p$, etc. though rapidly decreasing.
Jan 3, 2017 at 21:14 vote accept Amir Baghban
Jan 3, 2017 at 21:10 comment added Gerhard Paseman Not necessarily. Of course a consequence is that 2n+1 divides 2m+1. It could happen that N is a Mersenne prime and that the larger number is N(N^2 +3N+3) and has N as its largest prime factor. A look through the Cunningham tables should find a small counterexample if there are any. Gerhard "Perhaps A Very Small Counterexample" Paseman, 2017.01.03.
Jan 3, 2017 at 21:03 answer added Noam D. Elkies timeline score: 3
Jan 3, 2017 at 20:50 history asked Amir Baghban CC BY-SA 3.0