
2. Solve for integers $x, y, z$ :

$$
x+y=1-z, \quad x^{3}+y^{3}=1-z^{2}
$$


Sol. : Eliminating $z$ from the given set of equations, we get

$$
x^{3}+y^{3}+\{1-(x+y)\}^{2}=1
$$

This factors to

$$
(x+y)\left(x^{2}-x y+y^{2}+x+y-2\right)=0
$$

Case 1. Suppose $x+y=0$. Then $z=1$ and $(x, y, z)=(m,-m, 1)$, where $m$ is an integer give one family of solutions.

Case 2. Suppose $x+y \neq 0$. Then we must have

$$
x^{2}-x y+y^{2}+x+y-2=0 \text {. }
$$

This can be written in the form

$$
(2 x-y+1)^{2}+3(y+1)^{2}=12
$$

Here there are two possibilities:

$$
2 x-y+1=0, y+1= \pm 2 ; \quad 2 x-y+1= \pm 3, y+1= \pm 1
$$

Analysing all these cases we get

$$
(x, y, z)=(0,1,0),(-2,-3,6),(1,0,0),(0,-2,3),(-2,0,3),(-3,-2,6) .
$$
