Determine all possible pairs of positive whole numbers (v, w) satisfying the following equation:

v^{3} – w^{3} = vw + 61

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**