2
$\begingroup$

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

$\endgroup$

3 Answers 3

8
$\begingroup$

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 $13^2 = 169$.

$\endgroup$
3
  • 1
    $\begingroup$ Is $5$ here the only prime divisor occuring with multiplicity more than $1$ ? $\endgroup$ May 2, 2013 at 11:37
  • 3
    $\begingroup$ 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. $\endgroup$ May 2, 2013 at 15:20
  • 4
    $\begingroup$ 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 $4$). $\endgroup$ May 2, 2013 at 17:44
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – BHZ
    May 2, 2013 at 10:14
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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