Prove that $(2^n-1)(3^n-1)$ is not a perfect square.

I have tried this problem for a few days already and I feel I am really far from solving it. Most of my approaches have been analyzing how many times 2 divides the number, and how many times 3 divides it, as well as various mods. I am starting to think the proof is going to be factoring on a weird field or something like that instead.

We can see that if $n$ is odd then $3^n-1$ is divisible by $2$ exactly one time so the exponent of $2$ in the prime factorization of the number is $1$ and thus it is not a perfect square. Furthermore by lifting the exponent lemma we know that since $n$ is even the exponent of $2$ in the prime factorization of $3^n-1$ is $3-1+v_2(2) = 2+v_2(n)$ so we need $v_2(n)$ to be even. Therefore it is greater tan or equal to $2$ i.e $4$ divides $n$.

Similarly by lifting we can see that the exponent of $3$ in $2^n-1$ is $1+v_3(n)$ so we have $v_3(n)$ is odd i.e $3$ divides $n$.

Therefore if the expression is a perfect square we must have $12|n$.

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    $\begingroup$ You should state clearly what your problem is/what kind of help you want. $\endgroup$ – Henrik supports the community Jul 9 '16 at 18:15
  • $\begingroup$ There are several possible things we could do. Do you just want a complete proof, a hint a comment about whether what you've done is relevant or something I haven't imagined? $\endgroup$ – Henrik supports the community Jul 9 '16 at 18:47
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    $\begingroup$ Well, we have $$\Biggl(6^k-\frac{1}{2}\left(\frac{3}{2}\right)^k-\left(\frac{2}{3}\right)^k \Biggr)^2 <\left(2^{2k}-1\right)\left(3^{2k}-1\right)<\Biggl(6^k-\frac{1}{2}\left( \frac{3}{2}\right)^k-\frac{1}{2}\left(\frac{2}{3}\right)^k\Biggr)^2$$ for all $k=1,2,\ldots$. Not sure if one can show that there is no integer between $6^k-\frac{1}{2}\left(\frac{3}{2}\right)^k-\left(\frac{2}{3}\right)^k$ and $6^k-\frac{1}{2}\left(\frac{3}{2}\right)^k-\frac{1}{2}\left(\frac{2}{3}\right)^k$ for any $k=1,2,\ldots$. $\endgroup$ – Batominovski Jul 9 '16 at 19:15
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    $\begingroup$ @Peter The revision history could explain them. $\endgroup$ – Daniel Fischer Jul 9 '16 at 20:24
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    $\begingroup$ This result was proved by Szalay in the 1990s. A link to a generalization by Walsh is mysite.science.uottawa.ca/gwalsh/slov1.pdf. $\endgroup$ – Mike Bennett Jul 9 '16 at 20:33

That there are no solutions was proved by Szalay in 1997; a generalization to the equation $$ (2^n-1)(3^m-1) = z^2 $$ was given by Walsh in 2000 or so :


The proof follows from elementary arguments about (binary) recurrence sequences and local considerations at the primes $2$ and $3$.

  • $\begingroup$ Thanks for the reference. How come you had it at your fingertip ? $\endgroup$ – user230452 Jul 10 '16 at 12:45
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    $\begingroup$ Misspent youth, I suspect. $\endgroup$ – Mike Bennett Jul 10 '16 at 14:47
  • $\begingroup$ Are you a mathematician ? $\endgroup$ – user230452 Jul 10 '16 at 15:03
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    $\begingroup$ On occasion! It keeps me out of trouble. $\endgroup$ – Mike Bennett Jul 10 '16 at 15:23

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