Every order-three magic square of line sum 3e is the parametrized one

The parametrization is complete. Let $M$ be a $3\times3$ magic square with nonnegative integer entries and line sum $3e$. Then $M$ is exactly the parametrized array built from its two top corners:

where $a=M_{00}$ and $c=M_{02}$.

The eight line identities determine the remaining seven cells uniquely. The centre is $e$ (MacMahon's identity $3M_{11}=s$ with $s=3e$). The two diagonals give $M_{22}=2e-M_{00}$ and $M_{20}=2e-M_{02}$; row $0$ then gives $M_{01}=3e-M_{00}-M_{02}$; column $0$ and column $2$ give $M_{10}=e+M_{02}-M_{00}$ and $M_{12}=e+M_{00}-M_{02}$; and column $1$ gives $M_{21}=M_{00}+M_{02}-e$. Since $M$ is a genuine square over $\mathbb{N}$, none of these subtractions truncates.

Consequently a $3\times3$ magic square of line sum $3e$ is determined by its two top corners, and the pair $(M_{00},M_{02})$ satisfies precisely the admissibility inequalities — the count of such squares is therefore the count of admissible pairs, which is MacMahon's $2e^{2}+2e+1$.

Formalization Note The proof expands the eight line identities of IsMagic and closes each of the nine cell equalities by omega; no integrality hypothesis beyond working over $\mathbb{N}$ is needed.