The centre of a 3x3 magic square is one third of the line sum

In a $3 \times 3$ magic square the centre entry is exactly one third of the magic constant.

Let $M$ be a $3 \times 3$ array of natural numbers whose three rows, three columns and two main diagonals all sum to the same number $s$. Then

where $M_{1,1}$ is the central entry (indices are Fin 3, so the centre is the index $1$).

The proof is the classical one: add the middle row, the middle column and the two diagonals. The centre is counted four times and every other cell exactly once, so the total is $3s + 3M_{1,1}$; but it is also $4s$, whence $s = 3M_{1,1}$.

Formalization Note Everything stays in $\mathbb{N}$, so no divisibility hypothesis is needed: the identity itself forces $3 \mid s$.