Is it true that $p^2+1$ is square free if $p>7$ is a Mersenne prime For a problem in group Theory I need some information about the Mersenne primes:
Let $p=2^a-1>7$ be a Mersenne prime and so $a$ is an odd prime. Is it true that $p^2+1$ is square free. i.e.
 if there exists a prime number $q$ such that $q$ divides $(p^2+1)/2$, then $q^2$ does not didivde $(p^2+1)/2$?
Also is there any result about the number of distinct prime divisors of $p^2+1$ by these assumptions?
Many thanks for your help 
BHZ
 A: No, this is not true. -- For example $p := 2^{2203}-1$ is a Mersenne prime
(cf. http://en.wikipedia.org/wiki/Mersenne_prime), but $p^2+1$ is divisible by $5^2 = 25$.
Edit: To answer D. Burde's question: $p := 2^{11213}-1$ is a Mersenne prime as well, and
$p^2+1$ is divisible by $13^2 = 169$.
A: The question, how many integers $n$ are there, say with $n\le x$, such that $n^2+1$ is squarefree,
has been studied a lot. For references see the article of Heath-Brown: arxiv.org/pdf/1010.6217‎
It is easy to construct intervals $(x, x + c \log x]$ with a small positive constant
$c$, such that $n^2 + 1$ has a non-trivial square factor for every $n$ in the interval.
As the example $n=239$ shows, $n^2+1=57122=2\cdot 13^4$ is not squarefree.
In the question here, $n=2^a-1$ is of a special form. Then $n^2+1$  is "very often" squarefree, for smaller $a$,
not depending on whether $n$ is a Mersenne prime or not. On the other hand, this should not hold in general.
Edit: I just saw that there is a counterexample also for Mersenne primes: $p=2^{2203}-1$, given by Stefan Kohl.
It may be difficult to give an answer in general for such questions, though - see  Square free sum of two squares.
A: The folowing result was proved by Crescenzo about this problem:
With the exceptions of the relations $(239)^2-2(13)^4=-1$ and
$3^5-2(11)^2=1$ every solution of the equation $$p^m-2q^n=\pm 1;\ p,\ q\  prime;\  m,n>1$$ has exponents $m=n=2$; i.e. it comes from a unit $p-q\cdot 2^{1/2}$ of the quadratic field
$Q(2^{1/2}$) for which the coefficients $p$ and $q$ are primes.
