Determine all possible pair(s) (x, y) of positive integers that satisfy the equation:
x^{3} = y^{2} – 15
(In reply to
re: Fascinating by ed bottemiller)
The way I saw the equation was x^31 = (y+4)(y4); 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. 1259712^215 = 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 20100709 14:20:51 