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Further Arithmetic Integers (Posted on 2006-06-27) Difficulty: 3 of 5
Consider three positive integers x < y < z in arithmetic sequence, and determine all possible solutions of: x4 + y4 = z4 - 64

See The Solution Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

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Solution Solution | Comment 3 of 10 |

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
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