Magic Constant of Magic Square

Theorem

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

$S_n = \dfrac {n \paren {n^2 + 1} } 2$

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:

$T_n = \dfrac {n^2 \paren {n^2 + 1} } 2$

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

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

\(\ds S_n\)

\(=\)

\(\ds \dfrac {T_n} n\)

\(\ds \)

\(=\)

\(\ds \dfrac {n^2 \paren {n^2 + 1} } {2 n}\)

Sum of Terms of Magic Square

\(\ds \)

\(=\)

\(\ds \dfrac {n \paren {n^2 + 1} } 2\)

$\blacksquare$

Sequence

The sequence of the magic constants of magic squares of order $n$ begins:

$1, (5,) \, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, \ldots$

However, note that while $5 = \dfrac {2 \paren {2^2 + 1} } 2$, a magic square of order $2$ does not actually exist.