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 $12|n$.
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 $3$.