Irish Mathematical Olympiad

Because $N > 0$, we cannot have $x_i = 1$ and so $2 \le x_1$. Define $f(x) = (x^2 - 1)/3$ and for a given $N < 40 \cdot 3^k$ we let $M = N/3^k$. The equation $(x_1^2 - 1)(x_2^2 - 1) \cdots (x_k^2 - 1) = N$ is equivalent to $f(x_1)f(x_2) \cdots f(x_k) = M$. If $x \ge 2$ is an integer, $f(x) \ge 1$ and so $f(x_k) \le M < 40$. As $f(11) = 40$, we have $x_k \le 10$ for any solution. Here is a table of the relevant values of $f$:

We need to prove that there is no positive rational number $M < 40$ which is a product in two different ways of numbers from the second row of this table. If one of the values $f(x_i)$ is greater than $\frac{8}{3}$, it is at least 5, hence the product of the other factors is below 8. This shows that $\frac{8}{3}$ and 5 are the only possible factors greater than 1 if at least two factors are not equal to 1. The expressions of the form $(\frac{8}{3})^r 5^s$ below 40 are: $1, \frac{8}{3}, \frac{64}{9}, \frac{512}{27}, \frac{40}{3}, \frac{320}{9}, 5$ and 25. Each of them is obtained from a unique pair of integers $(r, s)$ and, except $1, \frac{8}{3}$ and 5, none of them appears as a value $f(x)$ in the table. This shows that there is no positive rational number $M < 40$ which is a product in two different ways of numbers of the form $f(x)$ with $2 \le x \le 10$ which proves that, up to permutation, there can be at most one solution in positive integers to $(x_1^2-1)(x_2^2-1)\cdots(x_k^2-1) = N$ for any positive integer $N < 40 \cdot 3^k$.

Remark: For $(2, \dots, 2, 4, 5)$ and $(2, \dots, 2, 2, 11)$ we obtain $N = 40 \cdot 3^k$.