x+y=1-zx+y=1−z
x^3+y^3=1-z^2x3+y3=1−z2
Dividing term to term the second equation by the first we have
(x^3+y^3)/(x+y) = ((1-z)(1+z))/(1-z)x3+y3x+y=(1−z)(1+z)1−z or
x^2-xy+y^2=1+zx2−xy+y2=1+z
Adding this equation with the first we have
x^2-x y+y^2+x+y=2x2−xy+y2+x+y=2. Solving for xx we obtain
x = 1/2 (-1 + y pm sqrt[3] sqrt[3 - 2 y - y^2])x=12(−1+y±√3√3−2y−y2)
Here
3 - 2 y - y^2 ge 03−2y−y2≥0 so
-3 le y le 1−3≤y≤1 but y in NN so y in {-3,-2,-1,0,1}
Checking we have
{y=-3,x=-2,z=6}
{y=-2,x=-3,z=6}
{y=-2,x=0,z=3}
{y=0,x=-2,z=3}
{y=0,x=1,z=0}
{y=1,x=0,z=0}
for y = -1 the solutions, are not integer solutions.
