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I am currently studying magic squares and ran into a bit of trouble. The concept I am learning about is a regular square. Below are the conditions of a regular square.

We can say that an $n$-by-$n$ square is regular provided that:

  1. Each of the integers from $0$ to $n^2 − 1$ appears in exactly one cell, and each cell contains only one integer (so that the square is filled), and

  2. If we express the entries in base-$n$ form, each base-$n$ digit occurs exactly once in the units’ position, and exactly once in the $n$’s position.

Example of a Regular Square:

$\begin{pmatrix} 00 & 23 & 32 & 11 \\ 21 & 02 & 13 & 30 \\ 12 & 31 & 20 & 03 \\ 33 & 10 & 01 & 22 \end{pmatrix} = \begin{pmatrix} 0 & 11 & 14 & 5 \\ 9 & 2 & 7 & 12 \\ 6 & 13 & 8 & 3 \\ 15 & 4 & 1 & 10 \end{pmatrix}$

Is every regular square necessarily magic? If so why is it true, or why isn't it true? Can someone please provide me an insightful proof as to why this is or isn't true?

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    $\begingroup$ This question seems to be completely identical to yesterday's question why every regular square is necessarily magic $\endgroup$ – MY USER NAME IS A LIE Feb 21 '15 at 21:06
  • $\begingroup$ @MYUSERNAMEISALIE Do you know how to answer it? $\endgroup$ – Sophia Feb 21 '15 at 21:22
  • $\begingroup$ Maybe your condition (2) needs to be adjusted a little? Was "in each row and column" supposed to be in it? $\endgroup$ – Ken Feb 21 '15 at 23:11
  • $\begingroup$ Also posted to (but closed on) MO, mathoverflow.net/questions/198081/regular-magic-squares $\endgroup$ – Gerry Myerson Feb 21 '15 at 23:13
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    $\begingroup$ Note that these squares are NOT magic by the usual definition, which requires the diagonals to work. $\endgroup$ – Gerry Myerson Feb 21 '15 at 23:16

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