In an order-three magic square, opposite cells sum to twice the centre

Opposite cells in an order-three magic square. Let $A$ be a $3\times3$ magic square with line sum $S$. Then for every pair of centrally opposite cells,

Equivalently, an order-three magic square is automatically associative with complement constant $2A_{11}$. Since MacMahon's identity gives $S = 3A_{11}$, each such pair sums to $\tfrac23 S$, and in particular the square is determined by its centre: every opposite pair is pinned to twice it.

The four pairs are $(1,1)$–$(3,3)$ and $(1,3)$–$(3,1)$ (the two diagonals) and $(1,2)$–$(3,2)$, $(2,1)$–$(2,3)$ (the middle column and middle row); the fifth "pair" is the centre with itself, which is trivial. This is the structural fact behind the classical parametrisation of $3\times3$ magic squares by two corner entries: once $A_{11}$ and one corner are chosen, all remaining cells follow.

Formalization Note Cells are indexed by Fin 3 and Fin.rev is the reversal $i\mapsto 2-i$, so $(i,j)$ and $(\mathrm{rev}\,i,\mathrm{rev}\,j)$ are the centrally opposite pair. The proof expands the nine line identities and finishes by linear arithmetic; no integrality hypothesis beyond $\mathbb{N}$ is needed.