Magic Constant of Order 3 Magic Square/Proof 1

Proof
Let $M_3$ denote the order $3$ magic square

By Sum of Terms of Magic Square, the total of all the entries in the $M_3$ is given by:
 * $T_3 = \dfrac {3^2 \left({3^2 + 1}\right)} 2 = \dfrac {9 \times 10} 2 = 45$

As there are $3$ rows of $M_3$, the magic sum of $M_3$ is given by:
 * $S_3 = \dfrac {45} 3 = 15$