Magic Hexagon of Order 3 is Unique

From ProofWiki
Jump to navigation Jump to search

Theorem

Apart from the trivial order $1$ magic hexagon, there exists only one magic hexagon: the order $3$ magic hexagon:


MagicHexagon.png


Proof

We first prove that only order $3$ magic hexagons exist apart from the order $1$ magic hexagon, by calculating the magic constant for each order.


If we ignore the center tile, the rest of the hexagon can be divided into $3$ parallelogram-like parts:

For example, the above can be divided into $\set {3, 17, 18, 19, 7, 1}$, $\set {11, 6, 8, 9, 14, 15}$ and $\set {2, 4, 13, 16, 12, 10}$.

Each part has $n \paren {n - 1}$ tiles.

Thus the order $n$ hexagon has $3 n \paren {n - 1} + 1$ tiles.


Since we require each row to have the same sum, and there are $2 n - 1$ rows:

The sum of numbers in the hexagon $= \paren {2 n - 1} \times$ magic constant

By Sum of Arithmetic Sequence, the sum of numbers in the tiles is:

$\dfrac {\paren {3 n \paren {n - 1} + 1} \paren {3 n \paren {n - 1} + 2}} 2$

Hence the magic constant is:

\(\ds \) \(\) \(\ds \frac {\paren {3 n \paren {n - 1} + 1} \paren {3 n \paren {n - 1} + 2} } {2 \paren {2 n - 1} }\)
\(\ds \) \(=\) \(\ds \frac {9 n^4 - 18 n^3 + 18 n^2 - 9 n + 2} {2 \paren {2 n - 1} }\)


Since the magic constant is a sum of integers, it must be an integer.

Then:

$32 \paren {\dfrac {9 n^4 - 18 n^3 + 18 n^2 - 9 n + 2} {2 \paren {2 n - 1}}} = \dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1}$ is also an integer.


Since:

$\dfrac {144 n^4 - 288 n^3 + 288 n^2 - 144 n + 32} {2 n - 1} = 72 n^3 - 108 n^2 + 90 n - 27 + \dfrac 5 {2 n - 1}$

We must require $\dfrac 5 {2 n - 1}$ to be an integer.

This happens only if $n = 1$ or $n = 3$.


Thus there are no magic hexagons that are not of order $1$ or $3$.


It remains to show that the order $3$ magic hexagon is unique.




Sources