12
$\begingroup$

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 $12|n$.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @Peter The revision history could explain them. $\endgroup$ – Daniel Fischer Jul 9 '16 at 20:24
  • 5
    $\begingroup$ 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
6
$\begingroup$

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 :

http://mysite.science.uottawa.ca/gwalsh/slov1.pdf

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

$\endgroup$
  • $\begingroup$ Thanks for the reference. How come you had it at your fingertip ? $\endgroup$ – user230452 Jul 10 '16 at 12:45
  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ On occasion! It keeps me out of trouble. $\endgroup$ – Mike Bennett Jul 10 '16 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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