Existence if integer solutions

Question

How many integer solutions exist for:

\begin{cases} x+y=1-z \\ x^3+y^3=1-z^2 \end{cases}

How do I do this. I'm stuck

Answer 1

$$x^3+y^3+z^2=1$$ $$x^3+y^3 + \underbrace{ (1-x-y)^2}_\mathrm{= 1-(x+y)^2} =1$$ $$(x+y)(x^2-xy+y^2) + 1+ (x+y)^2 -2(x+y) =1$$ $$(x+y)(x^2-xy+y^2) + (x+y) (x+y-2)=0$$ $$(x+y)(x^2+y^2-xy+x+y-2)=0$$ Thus, if we have $x=k$ and $y=-k$ then we get $z=1$ and here we get infinite solutions $(x,y,z)=(k,-k,1)$

Answer 2

hint:$x^3+y^3 = (1-z)(1+z) = (x+y)(1+z) \to (x+y)(x^2-xy+y^2-1-z) = 0 \to x+y= 0$ or $x^2-xy+y^2=1+z = 1+(1-x-y)$