, 2020

We assert that $m = 13$ works.
There are infinitely many integers $b$ satisfying $4b^2 - 3 = 13k^2$ for some $k$, because $(b, k) = (2, 1)$ is a solution, and $18^2 - 13 \cdot 5^2 = -1$, so one can generate arbitrarily many solutions using Pell equations. Specifically, we take $(2 + \sqrt{13})(18 - 13\sqrt{5})^{2m}$ to get a norm of $3 \cdot (-1)^{2m} = 3$.
Then, let $a = b(4b^2 - 3)$. Now,
$$a - 1 = (2b + 1)^2(b - 1)$$
$$a + 1 = (2b - 1)^2(b + 1).$$
Therefore,
$$(a^3 - a)(b^3 - b) = [(b-1)(b)(b+1)(2b-1)(2b+1)]^2 \cdot (4b^2 - 3)$$
which is a square.