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Let be $p$ and $q$ two arbitrary Mersenne numbers.

Is there a simple proof that $p\cdot q-1$ can never been a square?

$p\cdot q-1$ can instead be a 3rd power for:

$p=3,q=3$

$p=7,q=31$

$p=63,q=127$

In these cases it is interesting to see that $p\cdot q+1$ is an even semi-prime, as in the case $3\cdot 3 +1=10$ or $7\cdot 31+1=218$ or $63\cdot 127 +1=4001\cdot 2$

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  • $\begingroup$ By "power of 3" I think you mean "3rd power". $\endgroup$
    – Gerry Myerson
    Commented May 4, 2019 at 11:17
  • $\begingroup$ yes surely it is so 3rd power $\endgroup$
    – Enzo Creti
    Commented May 4, 2019 at 11:22
  • $\begingroup$ but it is hopeless to find another 3rd power $\endgroup$
    – Enzo Creti
    Commented May 4, 2019 at 11:23
  • $\begingroup$ @Garry Myerson anything to do the 3rd powers with a sequence oeis.org/A002897 $\endgroup$
    – Enzo Creti
    Commented May 4, 2019 at 11:44
  • $\begingroup$ If it's hopeless to find another 3rd power, then it can't have much to do with that infinite sequence, can it? $\endgroup$
    – Gerry Myerson
    Commented May 4, 2019 at 12:39

2 Answers 2

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Write $p=2^m-1$ and $q=2^n-1$ where $m>n$ without loss of generality. If $pq-1=(2^m-2^{m-n}-1)2^n$ is a square, then so is $2^m-2^{m-n}-1$. But then $m=n+1$, as $m\ge n+2$ implies $2^m-2^{m-n}-1\equiv 3\pmod 4$. Now, $2^m-2^{m-n}-1=2^m-3$, which clearly is not a square if $m$ is even, and which is not a square if $m$ is odd either as in this case $2^m-3\equiv (-1)^m\equiv -1\pmod 3$.

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  • $\begingroup$ Aren't you missing a factor of $2$? $n$ is going to be an odd number, so it's $2(2^m+2^{m-n}-1)$ that's a square. $\endgroup$
    – Gareth McCaughan
    Commented May 4, 2019 at 12:48
  • $\begingroup$ Not quite necessarily odd -- we could on the face of it have n=2. But in every other case n is odd. $\endgroup$
    – Gareth McCaughan
    Commented May 4, 2019 at 12:55
  • $\begingroup$ @GarethMcCaughan: No, I am not missing anything. The numbers $2^m-2^{m-n}-1$ and $2^n$ are coprime; if the product of two co-prime numbers is a square, then so is each of them. $\endgroup$
    – Seva
    Commented May 4, 2019 at 13:21
  • $\begingroup$ D'oh! Of course they are. You're quite right. Sorry. $\endgroup$
    – Gareth McCaughan
    Commented May 4, 2019 at 14:51
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    $\begingroup$ By the way, you don't need the case $m$ even. $2^n$ must also be a square, forcing $n$ to be even, yielding that $m=n+1$ is odd, and thus, $2^m-3\equiv -1\pmod{3}$, contradiction. $\endgroup$
    – hookah
    Commented May 6, 2019 at 16:34
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Suppose $(2^m-1)(2^n-1)-1=2^{m+n}-2^m-2^n$ is a square. Obviously we can't have $m=n$, so the largest power of 2 dividing this is $2^{\min(m,n)}$. If neither of our Mersenne primes is 3, then that exponent is an odd number, which can't possibly be the largest power of 2 dividing a square.

So one of our primes, say $2^m-1$, is 3 and we have $2^{n+2}-2^n-4$ where $n$ is an odd prime. This is 4 times $2^n-2^{n-2}-1$. Unless $n=3$ this is $-1$ mod 4 and therefore can't be a square.

So the only remaining case to check is $m=2,n=3$ yielding $3\cdot7-1=20$ which happens not to be a square.

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  • $\begingroup$ You're doing Mersenne primes, where OP only says Mersenne numbers. $\endgroup$
    – Gerry Myerson
    Commented Aug 16, 2021 at 12:59
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    $\begingroup$ Yikes! I confess I didn't even notice that OP didn't just mean Mersenne primes, but since they use 63 as an example it's clear that they didn't. (I don't think I've heard "Mersenne number" used to mean "power of 2 minus 1" before.) $\endgroup$
    – Gareth McCaughan
    Commented Aug 16, 2021 at 15:55

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