No symmetric magic squares of order three when the line sum is not divisible by three

The zero case for symmetric magic squares.

Writing $S_{n}(t)$ for the number of symmetric magic squares of order $n$ and line sum $t$, the theorem is

the companion of the evaluation $S_{3}(3e)=2e+1$.

Proof. A symmetric magic square is in particular magic, and for a magic square of order three and line sum $t$ the centre entry $c$ satisfies $3c=t$ by center_of_order_three. Hence $3\mid t$ is necessary for existence, and the filtered finset symmetricMagicSquares 3 t is empty otherwise.

Context. Symmetry does not produce new line sums beyond those already admitted by the magic squares — it only cuts each magic fibre down. So the same divisibility obstruction applies, and the symmetric count, like the panmagic and the plain magic counts, vanishes off the multiples of three.