Let $p\neq 3$ be a Mersenne prime. Is it true that $(p^21)/2$ has a prime divisor $r$ such that $r^{2}$ does not divide $n$?

1$\begingroup$ The condition "for every prime $q$ dividing $n$, $q^2$ also divides $n$" is called being powerful in the literature. So after dividing out the factors of $2$, you'd like to know if $2^{p1}  1$ is ever powerful when $2^p  1$ is prime. Ribenboim has some conjectures related to similar questions in "The New Book Of Prime Number Records", in the section on Wieferich primes. $\endgroup$ – Zack Wolske Apr 6 '12 at 18:25
Well, maybe I'm missing the point, but the first Mersenne prime is $3=2^21$, and $\frac{3^21}{2} = 4 = 2^2$, so... no, I guess.
Considering there are only 47 known Mersenne primes, and finding the factors of $2^p1$ is a difficult task, I'm not sure this question is fully tractable. But we can certainly show that all of the known Mersenne primes satisfy your question. First, take $m=2^p1$ to be a Mersenne number, and rewrite $n= \frac{m^2  1}{2} = 2^{2p1}  2^p$. This is powerful when $p=2$, which gives Philip van Reeuwijk's counterexample. For larger $p$, we know $4$ will always divide $n$, so we can ignore powers of $2$ and just consider the odd part $n'= 2^{p1}  1$.
We know $p$ is prime, and odd since it is not $2$, so we are looking at $n' = 4^k  1$, with $k=\frac{p1}{2}$. Clearly $3$ divides $n'$, and $9$ divides iff $3$ divides $k$. So we require that $p \equiv 1 \mod6$. This is not enough; plenty of known Mersenne primes have this property.
Since $3$ divides $k$, we can write $n'=64^{k'}1$. Then $7$ divides $n'$, and $7^2$ divides $n'$ iff $7$ divides $k'$. So now we require that $p \equiv 1 \mod42$. Sadly, again this is not enough.
One step further, we see that when $7$ divides $k'$, $43$ divides $64^{k'}1$, and $43^2$ divides iff $43$ divides $k'$. Now we're happy (for the time being), because no known Mersenne prime has $p\equiv 1 \mod 1806 = 43*7*6$. But it seems there's no reason they can't have this property, so you may have to continue your search once such a Mersenne prime is found. Expect one by the 504th instance: $504 = \phi(1806)$.

$\begingroup$ I have checked a few cases, and if $m = 2^p1$, it actually seems the case that $p$ is a factor of $n = \frac{m^21}{2}$ 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... $\endgroup$ – Philip van Reeuwijk Apr 6 '12 at 19:32

$\begingroup$ 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 $2^{p1} \equiv 1 \mod p^2$", and he goes on to show that this would imply solutions to other wellknown open problems. $\endgroup$ – Zack Wolske Apr 6 '12 at 20:35

$\begingroup$ 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! $\endgroup$ – Philip van Reeuwijk Apr 6 '12 at 20:52

$\begingroup$ Just to clarify, such primes are called Wieferich, and only two are known: 1093, and 3511. $\endgroup$ – Zack Wolske Apr 6 '12 at 21:53