Classification of symmetric order-three magic squares

Symmetry pins down the square.

Let $M=(M_{ij})$ be a $3\times3$ array of nonnegative integers which is symmetric ($M_{ij}=M_{ji}$) and magic of line sum $3e$. Then $M$ is completely determined by its top-left corner:

Proof. Symmetry identifies $M_{01}=M_{10}$, $M_{02}=M_{20}$ and $M_{12}=M_{21}$, so the eight line identities collapse to five: the three rows, the main diagonal $M_{00}+M_{11}+M_{22}=3e$, and the anti-diagonal $M_{02}+M_{11}+M_{20}=2M_{02}+M_{11}=3e$. The anti-diagonal forces $M_{02}=M_{11}=e$ (subtracting it from the main diagonal gives $M_{00}+M_{22}=M_{02}+M_{20}$, and the middle row then pins $M_{11}$). The rows and the remaining diagonal identities then express every other cell in terms of $a=M_{00}$ and $e$:

This is exactly the shape packaged as symmMagic3 e a in the companion definition MagicSquaresSpecial3.

Context. The statement is the order-three instance of the classical observation that symmetric magic squares are highly constrained: the symmetry removes three of the eight line conditions, and the remaining ones leave a one-parameter family rather than a two-parameter one. Iterating it gives the count $S_{3}(3e)=2e+1$. Unlike the panmagic case, the surviving family is genuinely non-trivial — the parameter $a$ ranges over $0,\dots,2e$ — which is why the symmetric and panmagic counts of order three differ.