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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: Solution | Comment 3 of 11 |
(In reply to Solution by Dej Mar)

Dej: Why does my (x,y) pair (109,1138) not satisfy the equation? 109 cubed is 1295029; 1138 squared is 1295044.

I tested by varying x from 1 to 999, getting x cubed, then adding 15, then taking the square root to get y. - all positive exact integers (not i).

I have tried extending to higher values of x, but I have only 17 or 18 significant digit precision, so I turn up several hundred "close" sets, but they all seem to miss integer y by the ninth of higher decimal digit; I don't want to test these all manually (or with the MS Calculator).  When I first posted, I was testing only x to three digits (to 998), and found only (1,4) so I posted that before trying other ranges.  Am I missing something?

  Posted by ed bottemiller on 2010-07-08 14:37:36
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