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Let $p=2^a-1$ be a Mersenne prime and so $a$ is an odd prime if $p>7$. We know that if $p=7$, then $(p^2+1)/2$ is equal to $5^2$. Can we prove that if $p>7$ , then $(p^2+1)/2$ is not equal to the powers of a prime number $t$, i.e. there exists no prime number $t$ such that $(p^2+1)/2=t^\beta$ for some $\beta>0$?

One more question:

Can we prove that if there exists a prime number $q$ such that $q$ divides $(p^2+1)/2$, then $q^2$ does not didivde $(p^2+1)/2$?

Thanks for your help BHZ

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    $\begingroup$ Cases where $\beta$ is even would have to produce solutions of Pell's equation $p^2-2q^2=-1$. This would make $2p$ very close to a power of $2$ and to a power of $3+2\sqrt{2}$. $\endgroup$
    – Douglas Zare
    Commented May 1, 2013 at 10:24
  • $\begingroup$ Thanks for your answer, Do you mean that $2p=2^r(3+2\sqrt(2))^s$, for some $r$ and $s$, where $\beta$ is even? of course by a result of Crescenzo we know that the only possibility for $\beta$ is $1$ or $2$, and I interest to prove that there exists at least two distinct prime divisors for $(p^2+1)/2$. $\endgroup$
    – BHZ
    Commented May 1, 2013 at 10:32
  • $\begingroup$ Actually, I had the coefficient for $p^2-2q^2=1$ rather than $p^2-2q^2=-1$. The solutions $p$ for the correct $-1$ are the nearest integers to $(\sqrt{2}-1)/2 (3+2\sqrt{2})^n: 1, 7, 41, 239...$. Perhaps you can eliminate the possibility that a power of $2$ is so close to $(\sqrt{2}-1)/2$ times a power of $(3+2\sqrt{2})$, except for the initial cases of $1$ and $7$. $\endgroup$
    – Douglas Zare
    Commented May 1, 2013 at 13:02
  • $\begingroup$ Dropping the assumption that $n=2^a-1$ be prime leads to the sequence 1,5,25,113,481,... -- see oeis.org/A092440 -- for which, as far as I could easily check, the numbers (except for 25) are all square-free. Did I not check far enough? $\endgroup$
    – Barry Cipra
    Commented May 1, 2013 at 14:42
  • $\begingroup$ Thanks for your taking time to answer my question, but I need a proof for this fact in order to use it for proving some result in Group Theory. Could you prove that $(p^2+1)/2$ is square free in general by the assumption that $p$ is a Mersenne prime. $\endgroup$
    – BHZ
    Commented May 2, 2013 at 5:09

1 Answer 1

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If $p=2^a-1$ and $(p^2+1)/2=t^b$, then $2^{2a-1}-2^a+1=t^b$. If $a \equiv 1 \pmod 4$ and $b$ is even then we get an integral point on the curve $y^2=2x^8-2x^4+1$, so there are only finitely many such $p$. Likewise, we can deal with $a \equiv 3 \pmod 4$, using another curve. For $b>1$ odd, you get a curve that depends on $t$, so it's not immediate, but something similar might work. Note that I didn't use that $p$ was prime, except to assume that $a$ is odd.

Now, I don't see an obstruction right away in the case $b=1$, in particular there should be infinitely many primes $t=2^{2a-1}-2^a+1$. With the additional requirement that $p$ is prime, things get a little muddier.

Finally, for the last question, it should follow from the ABC conjecture.

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  • $\begingroup$ Thanks for your answer. As I checked the ABC conjecture does not proved yet? Also if it is proved how we can get this result from this conjecture. $\endgroup$
    – BHZ
    Commented May 2, 2013 at 5:09

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