AP in Chessboard The natural numbers $1,2,...,n^2$ are arranged in a $nXn$ chessboard. In how many ways can we arrange the numbers such that the numbers on every row and every column are in arithmetic progression?
I know that $1$ has to be put in one of the corners. I have also observed that the given criteria is maintained if the numbers are arranged one after another like $1^{st}$ row- $1,2,...,n$. $2^{nd}$ row- $n+1....2n$ and so on. In this way I get $4$ different ways(by rotating the board). Trying to prove that these are the only possible ways. Am I on the right path? Please explain.

If $n \ge 2$ the $8$ are the only possibilities. First prove that the condition of arithmetic progressions means that if cell $(0,0)$ is the top-left, which must contain $1$, and cell $(1,0)$ contains $1+a$ and cell $(0,1)$ contains $1+b$ then cell $(x,y)$ must contain $1+ax+by$ for any positive integers $x,y < n$. Clearly $1 \le a,b \le n$ and $a \ne b$. It suffices to consider the case where $a < b$, since the other case is just a reflection across the diagonal. If $b < n$, cell $(0,a)$ and cell $(b,0)$ both contain $1+ab$, which is not allowed. Therefore $b = n$. Thus $a = 1$ otherwise no cell contains $2$. [Prove it!]
