Magic Constant of Magic Square

Theorem
The magic sum of a magic square of order $n$ is given by:


 * $S_n = \dfrac {n \left({n^2 + 1}\right)} 2$

Proof
Let $M_n$ denote a magic square of order $n$.

By Sum of Terms of Magic Square, the total of all the entries in a magic square of order $n$ is given by:


 * $T_n = \dfrac {n^2 \left({n^2 + 1}\right)} 2$

There are $n$ rows in $M_n$, each one with the same magic sum.

Thus the magic sum $S_n$ of the magic square $M_n$ is given by: