Estonian Mathematical Olympiad

Notice that for $n = 360$, the pairs $x = 2, y = 11$ and $x = 4, y = 5$ satisfy the condition.

We will show that for smaller numbers $n$, there do not exist two distinct suitable pairs $(x, y)$. If $x = 1$, then $(x^2 - 1)(y^2 - 1)$ is not positive. Thus, we can assume that $1 < x \le y$. Now, as $x$ or $y$ increases, $(x^2 - 1)(y^2 - 1)$ also increases. Let's examine the cases.

* If $x = 2$, then $y = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \dots$ give us respectively $(x^2 - 1)(y^2 - 1) = 9, 24, 45, 72, 105, 144, 189, 240, 297, 360, \dots$.
* If $x = 3$, then $y = 3, 4, 5, 6, 7, \dots$ give similarly $(x^2 - 1)(y^2 - 1) = 64, 120, 192, 280, 384, \dots$.
* If $x = 4$, then $y = 4, 5, \dots$ give $(x^2 - 1)(y^2 - 1) = 225, 360, \dots$.

No positive number smaller than 360 appeared repeatedly. If we continue the inspection for $x = 5, 6, \dots$, even the first case $y = x$ would give $(x^2 - 1)(y^2 - 1) > 360$, since $x = 4$ gives 360 or a larger number for the same $y$. In conclusion, 360 is the smallest number with the required property.