MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
up vote 7 down vote accepted

No, this is not true. -- For example $p := 2^{2203}-1$ is a Mersenne prime (cf., 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 $13^2 = 169$.

share|cite|improve this answer
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:‎

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.