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.
 A: The case $n=1$ is trivial.
The case $n=2$ is special; in this case, all $4!=24$ arrangements fulfil the conditions.
Now assume $n\ge3$. You already know that $1$ has to be in one of the corners (because otherwise there would have to be non-positive numbers on one side of it). Since we can rotate and reflect any solution, we can assume without loss of generality that $1$ is in the corner $(0,0)$.
For similar reasons, $2$ must be either next to $1$ or also in a corner. It clearly cannot be in an adjacent corner; but it also cannot be in the opposite corner, since that would require $(n-1)d=1$, where $d$ is the sum of the common differences of the left column and bottom row, which is impossible for $n\ge3$. Thus $2$ is next to $1$, and without loss of generality we can assume that it is to the right of $1$, at $(1,0)$. That fixes the first row.
Again for similar reasons, $n+1$ has to be in one of the corners of the remaining rectangle. The only one that works out is $(0,1)$. That fixes the first column.
Again for similar reasons, $n^2$ must be in a corner. There's only one corner left, $(n-1,n-1)$, so that's where it must go. That fixes the last row and column, and then everything else is determined by the arithmetic progression conditions.
Thus, for $n\ge3$, there are indeed only the $8$ arrangements obtained from the standard arrangement by rotations and reflections.
A: 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!]
