# Steep Diagonals / Diabolics ! (Interesting)...

• Jun 4th 2010, 04:17 AM
Samson
Steep Diagonals / Diabolics ! (Interesting)...
Hello All, I have an interesting problem I would like to propose to the number theory masterminds here. (For reference, this question came from the "For Thought" section of my text book, that is why it may seem a little odd)

Using a picture, we can describe a set of subsets within a square that resemble diagonals but aren't quite the same. We shall call these "steep diagonals". Notice one of them labelled 'd' in the square below, and there are 6 others parallel to it.

$\begin{pmatrix}
x & d & x & x & x & x & x\\
x & x & x & x & d & x & x\\
d & x & x & x & x & x & x\\
x & x & x & d & x & x & x\\
x & x & x & x & x & x & d\\
x & x & d & x & x & x & x\\
x & x & x & x & x & d & x
\end{pmatrix}$

It can be proven under certain conditions that the sums of the positive (or negative) steep diagonals are constant providing that we are dealing with a square full of consecutive integers starting at 0.

My question to you all is: How can this be proven? What are the certain conditions that it is referencing?
• Jun 6th 2010, 09:52 AM
Samson
Does anyone have any idea about this? This has been up here for a few days and I haven't had a single response.... I figured this might be difficult, but somebody should get this here! Any help is appreciated!
• Jun 6th 2010, 12:08 PM
Bruno J.
The sum is constant if and only if the "slope" of the diagonal (in your example, the slope is $2$) is relatively prime to the dimension of the matrix. In this case, we obtain exactly one number in each row or column. In fact, if you pick $n$ integers from the matrix, such that no row or column contains two of the integers, then the sum will always be the same.

Can you prove it?
• Jun 6th 2010, 12:18 PM
Bruno J.
By the way, this also implies the determinant of such a matrix is $0$ when $n$ is odd.