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Well, the title actually describes what is the problem in question. I was just thinking a bit about magic squares and this question popped-up. It could be that it is not interesting but I do not see a reason why it should not be asked.

Suppose that we have two magic squares of the same order $n>2$.

We can view those two magic squares as matrices and we can multiply them as we multiply matrices in the usual way.

The question is:

Is it true that for every natural number $n>2$ there exist two magic squares $M_1$ and $M_2$ such that either their product $M_1M_2$ or the product $M_2M_1$ (viewed as the product of matrices) is a magic square?

I really do not know can the way in which we multiply matrices give birth to another magic square constructed by multiplying two magic squares as matrices, your ideas are welcomed.

Also, suppose that all entries in magic squares are different.

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  • $\begingroup$ What is your definition of magic square? Using the usual definition, wherein the entries are $1, \ldots, n$, the product of any two magic squares of size $n > 1$ (regarded as matrices) has some entry $> n$ and so is not a magic square. $\endgroup$ – Travis Jan 29 '16 at 17:23
  • $\begingroup$ @Travis thank you for your remark. I would like to allow any entries in the matrix, it is only crucial that the sum of rows and columns and of two main diagonals is the same number. $\endgroup$ – Farewell Feb 1 '16 at 17:35
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Two squares with the magic sum of 2 can be multiplied together to get 2 times a square with the magic sum of 2. The first two squares have consecutive entries. All squares have no internal repetition.

magic square multiplication

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