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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(3): Fascinating | Comment 9 of 11 |
(In reply to re(2): Fascinating by broll)

I agree we have not proven any limitation on possible solutions to the original or variants.  I have tested all the odd subtrahends up thru 51.  We've already reviewed (x=109,y=1138) for the original, and I noted (x=5234,y=378661) for "- 17".

The only other case I found (other than obvious single-digit solutions for subtractions of 1, 3, 37, and 41) of interest was for (x=243,Y=3788) if subtracting 37.  All of my testing was limited by the number of significant digits I could work with, testing each equation for 0 < x < 10000.  (1259712**2 = 11664**3 without subtracting anything on the right side).  The only way to disprove a conjecture would seem to be some construction related to factorization.  We at least know that higher solutions would be very sparse, if existing at all.  An interesting problem -- hope KS has a solution for us. 

  Posted by ed bottemiller on 2010-07-09 15:00:27
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