AIME 2015 Problem 5

In an $n \times n$ grid, there are $(n-2)^{2}$ squares such that there are $4$ squares adjacent to the square you chose, there are $4n-8$ squares such that there are $3$ squares adjacent to the square you chose, and lastly, the $4$ corner squares such that there are $2$ possible adjacent squares. This means we want the smallest integer $n$ such that

$\frac{(n-2)^{2}}{n^{2}} \times \frac{4}{n^{2}-1}+\frac{4n-8}{n^{2}} \times \frac{3}{n^{2}-1}+\frac{4}{n^{2}} \times \frac{2}{n^{2}-1}<\frac{1}{2015}$

We can simplify the inequality to

$\frac{4n^{2}-4n}{n^{4}-n^{2}}<\frac{1}{2015}$ $\frac{(n)(4)(n-1)}{n^{2}(n+1)(n-1)}<\frac{1}{2015}$ $n(n+1)>8060$

And so we can deduce that the smallest $n$ is $n=\boxed{90}$