3$ is a Mersenne prime?

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Apr 6, 2012 at 21:53	comment	added	Zack Wolske		Just to clarify, such primes are called Wieferich, and only two are known: 1093, and 3511.
Apr 6, 2012 at 20:52	comment	added	Philip van Reeuwijk		Yes, the little Fermat part I had figured out already; the part about $p^2$ not being a factor is beyond me, I fear. Ribenboim's book looks interesting, I'll check it out. Thanks for the tip!
Apr 6, 2012 at 20:35	comment	added	Zack Wolske		If you could that would be something interesting. Showing that $p$ divides is an application of little Fermat. Conjecture $W'_2$ on $p. 343$ of Ribenboim's "New Book of Prime Numbers" is "There exist only finitely many primes with $2^{p-1} \equiv 1 \mod p^2$", and he goes on to show that this would imply solutions to other well-known open problems.
Apr 6, 2012 at 19:32	comment	added	Philip van Reeuwijk		I have checked a few cases, and if $m = 2^p-1$, it actually seems the case that $p$ is a factor of $n = \frac{m^2-1}{2}$ and $p^2$ isn't. This even seems to work when $m$ isn't a prime but just the $p$-th Mersenne number with $p$ prime. I can't prove it, though...
Apr 6, 2012 at 19:13	history	answered	Zack Wolske	CC BY-SA 3.0