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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


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3 Answers 3

up vote 7 down vote accepted

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 $13^2 = 169$.

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Thank you very much for your answer. –  BHZ May 2 '13 at 9:28
Is $5$ here the only prime divisor occuring with multiplicity more than $1$ ? –  Dietrich Burde May 2 '13 at 11:37
Just to flesh out Stefan's answer a bit, 25 is a divisor any time the exponent is congruent to 3 mod 20 (2 is a primitive root of unity mod 25), which is the case for 2203 and several other Mersenne primes. –  Barry Cipra May 2 '13 at 15:20
Indeed, given any $q\equiv1$ (mod $4$), the condition that $q^2 \mid (p^2+1)$ means $p^2\equiv-1$ (mod $q^2$), which is equivalent to $p$ lying in one of two residue classes modulo $q^2$. For example, when $q=5$, those residue classes are $7$ and $18$ (mod $25$). Solving $2^a-1\equiv7$ or $18$ (mod $25$) gives $a\equiv3$ or $18$ (mod $20$). Therefore $(2^a-1)^2+1$ is divisible by $5^2$ if and only if $a\equiv3$ or $18$ (mod $20$). While Mersenne primes are rare, nothing seems to keep them out of the residue class $3$ (mod $20$). Similar calculations hold for any $q\equiv1$ (mod $4$). –  Greg Martin May 2 '13 at 17:44

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.

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Thanks for your help and your answer. Hae you any idea about the special case of this question, i.e. By the above assumptions is there any prime $q$ such that $q$ divides $p^2+1$ and $q>p$, if $p$ is a Mersenne prime? –  BHZ May 2 '13 at 10:14

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.

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