Editing Pandiagonal magic square (section)

==3×3 pandiagonal magic squares==
It can be shown that [[Triviality (mathematics)|non-trivial]] pandiagonal magic squares of order 3 do not exist.  Suppose the square
:<math>\begin{array}{|c|c|c|}
\hline
\!\!\!\; a_{11} \!\!\! & \!\! a_{12}\!\!\!\!\; & \!\! a_{13} \!\!\\
\hline
\!\!\!\; a_{21} \!\!\! & \!\! a_{22}\!\!\!\!\; & \!\! a_{23} \!\!\\
\hline
\!\!\!\; a_{31} \!\!\! & \!\! a_{32}\!\!\!\!\; & \!\! a_{33} \!\!\\
\hline
\end{array}</math>
is pandiagonally magic with magic constant {{tmath|s}}. Adding sums {{tmath|a_{11} + a_{22} + a_{33},}} {{tmath|a_{12} + a_{22} + a_{32},}} and {{tmath|a_{13} + a_{22} + a_{31} }} results in {{tmath|3s}}.  Subtracting {{tmath|a_{11} + a_{12} + a_{13} }} and {{tmath|a_{31} + a_{32} + a_{33},}} we get {{tmath|1=3a_{22} = s}}
However, if we move the third column in front and perform the same argument, we obtain {{tmath|1=3a_{21} = s}}. In fact, using the [[Symmetry in mathematics|symmetries]] of 3&thinsp;×&thinsp;3 magic squares, all cells must equal {{tmath|\tfrac{1}{3}s}}. Therefore, all 3&thinsp;×&thinsp;3 pandiagonal magic squares must be trivial.

However, if the magic square concept is generalized to include geometric shapes instead of numbers – the [[geometric magic square]]s discovered by [[Lee Sallows]] – a 3&thinsp;×&thinsp;3 pandiagonal magic square does exist.