# $a,b$ are integers , both greater than $1$ , such that $(a^n-1)(b^n-1)$ is a perfect square for every positive integer $n$ , then $a=b$ ?

If $a,b$ are integers , both greater than $1$ , such that $(a^n-1)(b^n-1)$ is a perfect square for every positive integer $n$ , then is it true that $a=b$ ?

Hint: is it sufficient to consider $p$-adic heights, given by $$\nu_p(n)=\max\{m\in\mathbb{N}:p^m\mid n\}.$$ If you manage to prove that, assuming $a\neq b$, for some prime $p$ and some integer $n>0$ $$\nu_p\left((a^n-1)(b^n-1)\right) = \nu_p(a^n-1)+\nu_p(b^n-1)$$ is odd, then you're done, since that implies that $(a^n-1)(b^n-1)$ cannot be a square.