# What is $n>1$ so that there is an $n\times n$ board with a perfect square number of squares?

There is an $n\times n$ board ($n>1$) such that the total number of different squares it contains is a square of an integer. For example, a $2\times 2$ board contains $5$ different squares. Find $n$.

• Welcome to Mathematics Stack Exchange! I would suggest you to explain a little bit how you tried to solve it, so other people could help you better. Good luck! – iadvd May 29 '15 at 9:07
• I thought there wouldn't be a single answer. Would induction be involved? – Greg May 29 '15 at 9:08
• The overall number of squares is $\sum\limits_{k=1}^{n} k^2 = \frac{1}{6}n(n+1)(2n+1)$ if it helps. – Alexey Burdin May 29 '15 at 9:20

The number of different squares contained in a $n\times n$-board is $$\sum_{j=1}^n j^2=\frac{n(n+1)(2n+1)}{6}$$, which is a square for
$$n=0,1,24$$
So, the smallest (and probably only) non-trivial example is $n=24$.
There is no other $n\le 10^8$, for which the sum is a square.