Magic Constant of Order 5 Magic Square

Theorem

The magic constant of the order $5$ magic square is $65$.


Proof 1

Let $M_5$ denote an order $5$ magic square

By Sum of Terms of Magic Square, the total of all the entries in $M_5$ is given by:

$T_5 = \dfrac {5^2 \paren {5^2 + 1} } 2 = \dfrac {25 \times 26} 2 = 325$

As there are $5$ rows of $M_5$, the magic constant of $M_5$ is given by:

$S_5 = \dfrac {325} 5 = 65$

$\blacksquare$


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 $n = 5$:

$S_5 = \dfrac {5 \times 26} 2 = 65$

$\blacksquare$


Sources

  • 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $65$
  • 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $65$
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