<s>[INST] proving that $ab$ is a perfect square. let $a,b \ge 2$ be integers, such that for every positive integer $n$, the expression :
$(a^n-1)(b^n-1)$ is a perfect square.
How to prove that $ab$ is a perfect square.
 [/INST] It is obvious that for $a=b$ the product $(a^n-1)(b^n-1)$ is square for all $n$. We  prove that this is the only possibility.
Let $b^n=a^n+h_n$; one has $(a^n-1)(a^n-1+h_n)=x_n^2$ so
$$(a^n-1)^2+h_n(a^n-1)-x_n^2=0\Rightarrow 2(a^n-1)=-h_n\pm\sqrt{h_n^2+4x_n^2}$$ so we have a Pythagorean triple
$$(h_n, 2x_n,z_n)=(r_n^2-s_n^2,\space 2r_ns_n,\space r_n^2+s_n^2)$$ and
$$2(a^n-1)=-(r_n^2-s_n^2)\pm(r_n^2+s_n^2)$$ thus for all $n$ $$a^n=s_n^2+1\\b^n=r_n^2+1$$  For $n=2$ we have
$$a^2-s_2^2=1\\b^2-r_2^2=1$$ We are done.
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