Consider three positive integers x < y < z in arithmetic sequence, and determine all possible solutions of: x⁴ + y⁴ = z⁴ - 64

If x = y-k and z = y+k, then

  x^4 + y^4 = z^4 - 64 becomes

  y*(8*k^3 + 8*k*y^2 - y^3) = 64

Therefore, y = 1, 2, 4, 8, 16, 32, or 64.

Of these, only three are possible:

  y = 2    :    5 = k*(k^2 + 4),  k = 1 or 5

  y = 4    :   10 = k*(k^2 + 16), k = 1, 2, 5, or 10

  y = 8    :   65 = k*(k^2 + 64), k = 1, 5, 13, or 65

Of these, only

  y = 2  and  k = 1

  y = 8  and  k = 1

are valid. Therefore,

  x < y < z
 -----------

  1 < 2 < 3

  7 < 8 < 9

Posted by Bractals on 2006-06-27 21:01:09