A magic hexagon of order n is an arrangement of numbers in a centered hexagonal pattern with n cells on each edge, in such a way that the numbers in each row, in all three directions, sum to the same magic constant.
A normal magic hexagon contains the consecutive integers from 1 to 3nČ3n+1.
Prove that the normal magic hexagon exists only for n=1(trivial) and n=3.
http://mathworld.wolfram.com/MagicHexagon.html
For an order n hexagon there are 2n1 sums in each direction so the sum of the numbers must be divisible by this.
Since there are 3n^33n+1 numbers, their sum is
(3n^33n+1)(3n^33n+2)/2 = (9n^4  18n^3 +18n^2  9n +2)/2
divide this by 2n1 and divide the coefficients gives
[(9/4)n^4(18/4)n^3+(18/4)n^2(9/4)n+(1/2)]/[n  1/2]
synthetic division gives
(9/4)n^3(27/8)n^2+(45/16)n(27/32)+(5/64)/(n  1/2)
This magic constant cannot be whole number unless 5/(n  1/2)= 10/(2n1) is an integer.
So 2n1 must equal ±1,±2,±5,±10 with n an whole number.
Only n=1 and n=3 work.

Posted by Jer
on 20101112 16:04:01 