Magic Hexagon of Order 3 is Unique
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Theorem
Apart from the trivial order $1$ magic hexagon, there exists only one magic hexagon: the order $3$ magic hexagon:
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.
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Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $19$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $19$