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 $0,1,\ldots,n^2-1$.



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> 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?
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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.



Increasing sequence of normal magic squares
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.


