Magic Constant of Order 3 Magic Square

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Theorem

The magic constant of the order $3$ magic square is $15$.


Proof 1

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

By Sum of Terms of Magic Square, the total of all the entries in $M_3$ 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 constant of $M_3$ is given by:

$S_3 = \dfrac {45} 3 = 15$

$\blacksquare$


Proof 2

Let $M_n$ denote the magic square of order $n$.

By Magic Constant of Magic Square, the magic constant of $M_n$ is given by:

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

Setting $n = 3$:

$S_3 = \dfrac {3 \times 10} 2 = 15$

$\blacksquare$


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