Take the 2-minute tour ×

The questions below are motivated by pure curiosity. I heard of the first question from my former advisor. I have no idea how difficult they are, since I have no experience with magic squares.

By a normal magic square of order $n$ I mean a $n\times n$ magic square whose terms are all of the numbers $0,1,\ldots,n^2-1$.

  1. Is it possible to construct an infinite sequence $M_n$ of normal magic squares such that $M_{n}$ is a block submatrix of $M_{n+1}$ lying in the centre of $M_{n+1}$ (i.e. to obtain $M_{n}$ we remove from $M_{n+1}$ the $k$ top rows, $k$ bottom rows, $k$ columns from the left and $k$ columns from the right)?

  2. Can one construct a normal magic square of odd order (greater than $1$) with $0$ as the central element of the square?

Note that a positive answer to the second question gives a positive answer to the first one. Let $A=[a_{ij}]_{0\leq i,j\leq n}$ be a magic square as in question 2. For a number $c$ we shall use the notation $A+c=[a_{ij}+c]_{0\leq i,j\leq 2n}$ Then: $$A'=\left[\begin{matrix}A+(n+1)^2a_{00} & \dots & A+(n+1)^2a_{0n} \\ \vdots & \ddots & \vdots\\ A+(n+1)^2a_{n0} & \dots & A+(n+1)^2a_{nn}\end{matrix}\right]$$ is a normal magic square with $A$ in the centre.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Here you are: $$ \begin{array}{|c|c|c|c|c|} \hline 8& 15& 21& 2& 14\cr \hline 20& 7& 13& 19& 1\cr \hline 12& 24& 0& 6& 18\cr \hline 4& 11& 17& 23& 5\cr \hline 16& 3& 9& 10& 22 \cr \hline \end{array} $$

In fact, you may start with any weakly magic square (that is --- with no conditions on the diagonals); then you just need to permute rows and/or columns to satisfy the conditions on the diagonals.

I took a standard algorithm of constructing the magic squares of odd order; amazingly, I needed only to permute the columns. It seems that a similar method should work for higher orders as well.

share|improve this answer
Thank you for your answer. –  Michał Kukieła Apr 5 '13 at 20:27

Your Answer


