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x^3 = y^2 15 (Posted on 2010-07-08) Difficulty: 3 of 5
Determine all possible pair(s) (x, y) of positive integers that satisfy the equation: x3 = y2 15

No Solution Yet Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

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re: Fascinating | Comment 6 of 11 |
(In reply to Fascinating by broll)

I was somewhat surprised to find an isolate solution beyond the obvious (1,4), but less so because of the " - 15 " part of the equation.  Without that, x-cubed and y-squared would both need to have the same unique complete factorization, but would not appear "independent".  It is not obvious what 109 and 1138 have in common (I did not find these by any "analytic" approach, but just "brute force" testing.) . Perhaps instead of -15 we could try with similar subtractions.  Maybe KS will let us know his take on the question.

EDIT: For curiosity I tried x**3 = y**2 - 17 instead.  The lowest and only solution I found (in range examined) was x=5234 and y=378661 (the cube is 143384152904).  No solutions in range for subtracting 11, 13, 19 ...  .   We know that if x is odd, y must be even, and v.v. if we are subtracting an odd number.

Edited on July 9, 2010, 12:54 pm

Edited on July 9, 2010, 1:36 pm
  Posted by ed bottemiller on 2010-07-09 12:36:35

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