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Subtract The Cubics, Get Product (Posted on 2007-07-22) Difficulty: 3 of 5
Determine all possible pairs of positive whole numbers (v, w) satisfying the following equation:

v3 w3 = vw + 61

See The Solution Submitted by K Sengupta    
Rating: 3.3333 (3 votes)

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Solution Comment 5 of 5 |

Clearly v>w>0, then let v = w+d with d>0.

The equation can then be rewritten as a quadratic in w:
(3d-1)w^2 + (3d^2-d)w + (d^3-61) = 0

If d>=4 then all the coefficients are positive, which means that there are no positive roots when d>=4.

If d=3 then the quadratic is 8w^2+24w-34=0 which has two irrational roots.  Similarily when d=2 the quadratic is 5w^2+10w-53=0 which also has two irrational roots.

If d=1 then the quadratic is 2w^2+2w-60=0 which has two integer roots, w=-6 and w=5.  Since w>0, there is only one integer solution, d=1, w=5, v=6


  Posted by Brian Smith on 2007-07-23 10:35:35
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