Irish Mathematical Olympiad

Let us first determine the smallest possible $N$ for which we find two solutions of the form $(2, y, z)$. If we let $M = N/3$, this amounts to finding the smallest possible positive integer $M$ for which the equation $(y^2 - 1)(z^2 - 1) = M$ has at least two solutions with $2 \le y \le z$. This is solved in problem 15, where it is shown that the smallest $M$ is equal to $360$. Therefore, we find two solutions $(2, 4, 5)$ and $(2, 2, 11)$ for $N = 3 \cdot 360 = 1080$.
The smallest value the expression $(x^2-1)(y^2-1)(z^2-1)$ can have if $3 \le x \le y \le z$ is $8^3 = 512$. But this happens only if $x = y = z = 3$. If $z \ge 4$, the least value will be $8^2 \cdot 15 = 960$, only occurring for $(x, y, z) = (3, 3, 4)$. If $z \ge 5$ or $y \ge 4$, we obtain at least $8^2 \cdot 24 = 1536 > 1080$ or $8 \cdot 15^2 = 1800 > 1080$, respectively.
Because $512$ is not divisible by $3$, there cannot exist a solution of the form $(2, y, z)$ for $N = 512$.
Finally, if $N = 960$ and $x = 2$, we have to solve $(y^2-1)(z^2-1) = 320 = 2^6 \cdot 5$. Because $3 \cdot 120 > 320$, we have $2 \le y \le z \le 10$. A look at the first table in the solution of problem 4 shows that the only numbers of the form $x^2 - 1$ for $x \le 10$, which have $2$ and $5$ as its only prime factors, are $8$ and $80$. But $8 \cdot 80 \ne 320$ and so for $N = 960$ we cannot find two different solutions. This proves that $N = 1080$ is the desired minimal value.