Consider the following order-7 magic square.

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The rows, columns, and diagonals all add up to the same sum: 175. Also, all the broken diagonals add up to the same sum, making this a pandiagonal magic square. The square is like a torus -- by shifting rows or columns to the other edge, any square can be put at the center.

Pick a number from the magic square, then shift about both squares so that your chosen number is at the center. Place a queen at the center of each square. Each queen will then cover one number, the same number, and attack 24 other numbers. Neither queen will attack the same number, and every number gets attacked on exactly one of the two squares.

In combinatorics, this is known as the 2-(49,7,1) design. With 49 elements, in blocks of 7, any 2 items can be found in exactly 1 of the blocks. Each of the two squares represents 7 rows, 7 columns, 7 left diagonals, 7 right diagonals, or 28 blocks in each square. So 56 blocks in total. This is also known as the Paley-49 graph. A similar trick can be done with 4x4 magic squares.

A Rook-3 representation of 2-(9,3,1) exists, Lo Shu + (123)(456)(789).
A restricted Queen-4 representation of 2-(16,4,1) exists where diagonal moves are only on main diagonals.
A restricted Queen-5 representation of 2-(25,5,1) exists where diagonal moves are only left diagonals.
A Queen-7 representation of 2-(49,7,1) is shown above.

In fairy chess, the amazon or superqueen has the combined moves of a queen and knight. At the center of a 9x9 board an amazon attacks exactly half of the other squares.

Does an Amazon-9 representation of 2-(81,9,1) exist?

Ideally, one of the two squares would be the numbers 1-81 in order.

One popular piece is the nightrider, a knight that can continue moving in a straight line. Here, give the nightrider the ability to move over an edge as if on a toroidal board, but on a $(2 n +1)(2 n +1)$ board limit the number of moves in a straight line to $n$.

The above board is also a Nightrider-7 design.


Does an Amazon-9 representation of 2-(81,9,1) exist?

No, because there are not enough residue classes modulo 9 that are coprime to 9.

Let there be vectors $\mathbf{u}=(a,b), \mathbf{v}=(c,d)$ with coordinates in $\mathbb{Z}_9$. The left-hand square would be a square $S_{ij}, i, j\in\mathbb{Z}_9$ with


where no non-zero multiple of any of $\mathbf{u}, \mathbf{v}, \mathbf{u\pm v}, \mathbf{u\pm}2\mathbf{v}$ or $2\mathbf{u\pm v}$ is a non-zero multiple of any of $(0,1), (1,0), (1,\pm1), (1,\pm2)$ or $(2,\pm1)$.

Besides, the amazon is so powerful that she attacks more than half the empty squares. On an $n\times n$ torus board, a rook attacks $2(n-1)$ squares; a bishop, $2(n-1)$; a queen, therefore, $4(n-1)$; a nightrider, $4(n-1)$; an amazon, therefore, $8(n-1)$. Thus the appropriate board for the queen is where

$$\begin{array}{rcl} 4(n-1) & = & \frac{n^2-1}2=\frac{n+1}2(n-1) \\ \Rightarrow n & = & 7 \end{array}$$

where, as you show, the desired arrangement exists. The appropriate board for the amazon is the $15\times 15$. This did not look hopeful, given that $15=3\cdot5$. There are $8$ coprime residues modulo $15$: $\pm(1, 2, 4, 7)$. Unfortunately, that is not enough. There are no suitable $a,b,c,d$ where $\mathbf{u},\mathbf{v}$ satisfy the above criteria.


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