N is an integer ≥ 2 and each of X, Y and Z is a positive integer with X ≤ Y ≤ Z

Determine all possible solutions to this equation :

X^N + Y^N + Z^N = 2008

Extra Challenge: Solve this puzzle without using a computer program.

DEFDBL A-Z
FOR n = 2 TO 99999
  FOR x = 1 TO 2008 ^ (1 / n) / 3
  FOR y = x TO 2008 ^ (1 / n) / 2
    zn = 2008 - INT(x ^ n + .5) - INT(y ^ n + .5)
    z = INT(zn ^ (1 / n) + .5)
    IF INT(z ^ n + .5) = zn THEN
       PRINT n, x; y; z, x ^ n; y ^ n; z ^ n
    END IF
  NEXT
  NEXT
NEXT

finds

 N             X  Y   Z      X^N  Y^N  Z^N
 2             6  6  44      36  36  1936
 2             10  12  42    100  144  1764
 3             4  6  12      64  216  1728

Posted by Charlie on 2013-11-08 13:34:46