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

The zero case for panmagic squares.

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

It complements the evaluation $P_{3}(3e)=1$: together they give the complete count $P_{3}(t)=1$ for $3\mid t$ and $P_{3}(t)=0$ otherwise.

Proof. A panmagic square is in particular a magic square, because the broken diagonals of offset $0$ are exactly the two main diagonals. For a magic square of order three and line sum $t$, the classical centre identity center_of_order_three gives $3c=t$ where $c$ is the centre entry. Hence $3\mid t$, and the square cannot exist when $t$ is not divisible by three.

Context. This is the standard divisibility obstruction for order-three magic squares: the centre is forced to be one third of the line sum, so line sums indivisible by three are impossible for any class that contains the magic squares. The same argument already gave the vanishing of the plain magic count magic_count_three_otherwise and of the semi-magic count; what the present statement adds is only the observation that panmagicity is stronger than magicity, so the obstruction is inherited.