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



Is it true that $p^2+1$ is square free if $p>7$ is a Mersenne prime
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 $13^2 = 169$.


