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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(2): Fascinating | Comment 8 of 11 |
(In reply to re: Fascinating by ed bottemiller)

The way I saw the equation was x^3-1 = (y+4)(y-4); the differences between successive terms on the RHS are just the odd numbers, so one might instinctively (but obviously quite wrongly!) think that there should be an infinity of solutions - e.g. 1259712^2-15 = 11664^3; but as you mentioned earlier, there always seems to be a problem in the sixth place of decimals. But I would like to see some proof that there are in fact no such solutions, before writing this one off!

  Posted by broll on 2010-07-09 14:20:51
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