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 3nČ-3n+1.
Prove that the normal magic hexagon exists only for n=1(trivial) and n=3.
For an order n hexagon there are 2n-1 sums in each direction so the sum of the numbers must be divisible by this.
Since there are 3n^3-3n+1 numbers, their sum is
(3n^3-3n+1)(3n^3-3n+2)/2 = (9n^4 - 18n^3 +18n^2 - 9n +2)/2
divide this by 2n-1 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/(2n-1) is an integer.
So 2n-1 must equal ±1,±2,±5,±10 with n an whole number.
Only n=1 and n=3 work.
Posted by Jer
on 2010-11-12 16:04:01