(In reply to

Possible solution; not a proof by Larry)

The equation a^3 + ab - ab^2 - b^2 = 0 can be factored.

Group in to (a^3-ab^2) + (ab-b^2) = 0. Then factor each group:

a*(a-b)*(a+b) + b*(a-b) = 0

(a-b)*[a*(a+b)+b] = 0

Then either a-b = 0 or a*(a+b)+b = 0. The first equation has solutions not permitted by the problem's conditions, therefore a*(a+b)+b = 0.

Now solve for b:

a*(a+b)+b = 0

a^2 + (a+1)*b = 0

b = -a^2/(a+1)

b = (1-a^2)/(a+1) - 1/(a+1)

b = 1 - a - 1/(a+1)

To be integers, a+1 must be a factor of -1. Therefore a+1=1 or a+1=-1. Then a=0 or -2. a=0 can be discarded which leaves a=-2.

Then b = 1 - (-2) - 1/(-2+1) = 1 + 2 + 1/1 = 4

Then (-2)^3 + (-2)*4 + c = 0, which makes c = -(-8 + -8) = 16.

Then the solution to the problem is **(a,b,c) = (-2, 4, 16)**.