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.