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Math Help - Steep Diagonals / Diabolics ! (Interesting)...

  1. #1
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    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}<br />
x & d & x & x & x & x & x\\ <br />
x & x & x & x & d & x & x\\ <br />
d & x & x & x & x & x & x\\ <br />
x & x & x & d & x & x & x\\ <br />
x & x & x & x & x & x & d\\ <br />
x & x & d & x & x & x & x\\ <br />
x & x & x & x & x & d & x<br />
\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?
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  2. #2
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    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!
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    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?
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    By the way, this also implies the determinant of such a matrix is 0 when n is odd.
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