What is the smallest integer $$n$$ for which the number $$\left(2^{2}-1\right)\cdot \left(3^{2}-1\right)\cdot \left(4^{2}-1\right)\cdots \left(n^{2}-1\right)$$ is the square of an integer? 
Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Questions involving Square Numbers
Using the difference of two squares, we can get:  $$\left(2^{2}-1\right)\cdot \left(3^{2}-1\right)\cdot \left(4^{2}-1\right)\cdots \left(n^{2}-1\right) = 1\times 3\times 2\times 4\times 3\times 5\times \cdots \times (n-1) \times (n+1)$$  you will notice that this is equal to:  $$1\times 2\times 3^{2} \times 4^{2} \times \cdots \times (n-1)^{2} \times n \times (n+1)$$  So this will only be a square if $2n(n+1)$ is a square, which happens when $n=8$ as $2n(n+1) = 144=12^{2}$.  Answer: $B$ 