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Some Powers Sum 2008 (Posted on 2013-11-08) Difficulty: 3 of 5
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 :

XN + YN + ZN = 2008

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

No Solution Yet Submitted by K Sengupta    
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Solution re: more results Comment 3 of 3 |
(In reply to more results by Benny)

Apparently I underestimated how far I should check.

Revised program:

DEFDBL A-Z
CLS
FOR n = 2 TO 99999
  FOR x = 1 TO 2008 ^ (1 / n)
  FOR y = x TO 2008 ^ (1 / n)
    zn = 2008 - INT(x ^ n + .5) - INT(y ^ n + .5)
    IF zn > 0 THEN
      z = INT(zn ^ (1 / n) + .5)
      IF INT(z ^ n + .5) = zn AND z >= y THEN
         PRINT n, x; y; z, x ^ n; y ^ n; z ^ n
      END IF
    END IF
  NEXT
  NEXT
NEXT

giving all of Benny's results:

N             X   Y   Z     X^N Y^N Z^N
2             6  6  44      36  36  1936
2             6  26  36     36  676  1296
2             10  12  42    100  144  1764
2             18  28  30    324  784  900
3             2  10  10     8  1000  1000
3             4  6  12      64  216  1728

  Posted by Charlie on 2013-11-08 17:39:58
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