We want to describe via a picture a set of subsets of a square which are something like diagonals, but are not quite the same. We’ll call them steep diagonals. One of them, labelled e, is illustrated in the square below; the other 6 are parallel to it.
Can someone give me a theorem and prove the theorem about under what conditions we can expect that the sums on the positive (or negative) steep diagonals are constant, when we’re dealing with a square full of consecutive integers starting at 0?
For a prime size square that is filled with the procedure we are taught as the easy way (start middle top, up and to the right number by number, down when you hit one that is already there) you can show that your steep diagonal will sum to the magic constant. The procedure given breaks the numbers from $1$ to $n^2$ into brackets $[1,n],[n+1,2n]\dots,[(n-1)n+1,n^2]$ and puts one number from each bracket into each row and column as well as one position in a bracket (the number added to the multiple of $n$) in each row and column. You can use the cyclic nature of the prime group to show your steep diagonal will do this, too. If $n$ is not prime, the same will work as long as the increments along your steep diagonal are coprime to $n$.
For general magic squares, I doubt there is anything you can say. There are lots of ways to fill in a magic square and your steep diagonal could hit generally small or large numbers.