There is exactly one panmagic square of order three

The panmagic squares of order three.

A $3\times3$ array of nonnegative integers is panmagic (or pandiagonal) of line sum $s$ when all three rows, all three columns, and all six broken diagonals — the three descending and the three ascending ones, read modulo three — sum to $s$. Writing $P_{n}(t)$ for the number of panmagic squares of order $n$ and line sum $t$, the theorem states

and the single square in question is the constant array, all of whose nine entries equal $e$. In particular $P_{3}(t)=0$ whenever $3\nmid t$, since the twelve line sums of a panmagic square are all equal.

Context. Order three is the degenerate case of the panmagic problem. For $n\ge5$ odd (and $n=4$) panmagic squares are abundant and their enumeration is a genuine problem, so it is the shape of the order-three answer that is informative: the two-dimensional family of magic squares of order three (MacMahon's $2e^{2}+2e+1$ squares) collapses to a single point once the six broken diagonals are imposed. This is the counterpart, for symmetric requirements, of the Lo Shu classification: there, normality pins the square to eight $D_{4}$-images; here, panmagicity pins it to the constant square.

Proof. Let $M$ be panmagic of line sum $3e$ with entries $a,b,c;d,m,f;g,h,i$. The twelve line equations are linear in the entries, and the four broken diagonals

combine with the rows and columns to force $a=b=\cdots=i=e$: from the two broken diagonals through the centre one gets $d=m=f$, then $m=e$ from the middle row, and the remaining equations give $a=b=c$ and $g=h=i$, whence $3a=3e$. Since the nonnegative integers are an integral domain, $a=e$. The constant square $\texttt{constSquare3}\ e$ is panmagic for every $e$, so $P_{3}(3e)=1$.