2
$\begingroup$

Can you determine any number of magic squares, that when treated as matrices, can be applied mathematical operations to return a new magic-matrix?

You cannot use the same matrix twice!

The answer should be given as a mathematical equation using the matrices you find.

Magic squares follow the general magic square rules, they contain 1, .., $n \times n$ once, and each row, column, and diagonal sum to 1 number.

Allowed Operations: Add, Subtract, Multiply, Exponention, Logarithms

Improve this question
$\endgroup$
7
  • $\begingroup$ Must an n by n magic square contain each of the numbers 1 through $n^2$ exactly once? What are the allowed operations? $\endgroup$
    – Lopsy
    Dec 21, 2014 at 4:17
  • $\begingroup$ @Lopsy See my edit. $\endgroup$
    – warspyking
    Dec 21, 2014 at 4:26
  • $\begingroup$ Do you mean the operations to act on matrices or their elements? Precisely what operations are allowed? $\endgroup$
    – xnor
    Dec 21, 2014 at 6:59
  • $\begingroup$ Anybody have any idea who downvoted? I honestly don't see any reason... $\endgroup$
    – warspyking
    Dec 21, 2014 at 15:00
  • 1
    $\begingroup$ @warspyking "Are there any integer solutions to $2^x+\lfloor \sqrt{x} \rfloor = y^4$?" Scrap. "Are there any integer solutions to $x^n+y^n=z^n$?" 358 years of interest. The monk asks, "Why is one question more worthy than another?" His master answers, "Now you have asked a worthy question." $\endgroup$
    – Lopsy
    Dec 21, 2014 at 15:53

2 Answers 2

Reset to default
5
$\begingroup$

It is possible.

Fix $n$. Let $M$ denote the set of all $n$-by-$n$ magic squares. $M$ is really huge - for $n=6$, we have $|M| \sim 10^{19}$ (see OEIS).

Consider all $2^{|M|}$ sums of subsets of $M$. Each one is an $n$-by-$n$ matrix whose entries are positive integers which are at most $n^2|M|$. There are $(n^2|M|)^{n^2}$ matrices which fit this description.

If $|M|$ is large enough, then $2^{|M|} > (n^2|M|)^{n^2}$. Then by Pigeonhole, two of the sums of subsets are equal. Using this equality, we can express one of the magic squares as a sum/difference of others.

What is "large enough"? Taking logs on the inequality, we find that even $|M|=O(n^3)$ is large enough for sufficiently large $n$. For $n=4,5,6$ this works. I cannot find any references to back this up, but I bet there are enough magic squares that this works for all $n\geq4$.

Improve this answer
$\endgroup$
4
$\begingroup$

If we consider two magic squares that differ by a rotation distinct, then here's an example with $3\times 3$ squares. Let $$ M_1=\left[\begin{array}{ccc}4&9&2\\3&5&7\\8&1&6\end{array}\right],\,M_2=\left[\begin{array}{ccc}8&3&4\\1&5&9\\6&7&2\end{array}\right],\,M_3=\left[\begin{array}{ccc}6&1&8\\7&5&3\\2&9&4\end{array}\right],M_4=\left[\begin{array}{ccc}2&7&6\\9&5&1\\4&3&8\end{array}\right]. $$ Then we can compute $$ M_1+M_3=M_2+M_4=\left[\begin{array}{ccc}10&10&10\\10&10&10\\10&10&10\end{array}\right], $$ so that the equation $M_1+M_3-M_2=M_4$ gives a solution to the problem.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Your finished product doesn't include the numbers 1, ..., $n \times n$ not to mention, you didn't even add properly? 4 + 8 + 6 + 2 don't equal 10... $\endgroup$
    – warspyking
    Dec 21, 2014 at 15:03
  • $\begingroup$ @warspyking Read the answer more carefully. He is not just adding all four matrices. $\endgroup$
    – Lopsy
    Dec 21, 2014 at 15:15
  • $\begingroup$ Oh oops, I misread that = as a + (how on earth did I do that? Lol!) $\endgroup$
    – warspyking
    Dec 21, 2014 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.