How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.

Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}

My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)=z(z-3) \implies y=\frac{z(z-3)}{3x}$ and $x=\frac{z(z-3)}{3y} $.
Note that since $z$,$x$,$y$ is an integer, we must have $3\mid z$.
So, let $z=3k$.
So we have $y=\frac{3k(3k-3)}{3x}=\frac{k(3k-3)}{x}$ and $x=\frac{z(z-3)}{3y}=\frac{k(3k-3)}{y}$ .
Then I am not able to proceed.
Hope one can give me some hints and guide me.
Thanks in advance.
 A: We have $(1-z)(x^2-xy+y^2)=1-z^2.$
If $z=1$, so $x+y=0$ and we obtain $(t,-t,1)$, where $t$ is an integer.
Let $z\neq1$.
Thus, $$x^2-xy+y^2=z+1$$ and $$x+y=1-z,$$ which gives $$(1-z)^2-3xy=z+1$$ or
$$3xy=z^2-3z.$$
Thus, $z$ is divisible by $3$ and $$(1-z)^2-\frac{4}{3}(z^2-3z)\geq0$$ or
$$z^2-6z-3\leq0$$ or
$$3-\sqrt{12}\leq z\leq 3+\sqrt{12},$$
which gives $$0\leq z\leq 6$$
Can you end it now?
A: Guide:
Case $1$: If $z=1$. Check what happens here.
Case $2$: If $z \ne 1$, then $x+y \ne 0$,
$$x^2-xy+y^2=1+z=1+(1-(x+y))$$
$$x^2+y^2-xy+x+y = 2$$
$$(x^2-xy+x)+(y^2+y)=2$$
$$(x^2-x(y-1))+(y^2+y)=2$$
$$\left(x - \frac{y-1}2\right)^2-\left(\frac{y-1}2 \right)^2 + (y^2+y)=2$$
$$(2x-y+1)^2 -(y^2-2y+1) + 4y^2+4y=8$$
$$(2x-y+1)^2 + 3y^2+6y-1=8$$
$$(2x-y+1)^2+3(y^2+2y+1)=12$$
$$(2x-y+1)^2+3(y+1)^2=12$$
Hence we have $|y+1| \in \{1,2\}$.

*

*If $|y+1|=1$, then $|2x-y+1|=3.$

*If $|y+1|=2$, then $|2x-y+1|=0.$
I will leave the rest as an exercise.
A: A general way:
By elimination of $z$ we get
$$(x+y)(x^2+y^2-xy+x+y-2)=0$$
Case 1: So two branches one: $x+y=0 \implies z=1,x=n, y=-n$, where $n\in I$
Case 2: The other set of solutions are given by
$$(x^2+y^2+xy+x+y-2)=0$$, write this as a quadratic of $x$ and treat $y$ as constant then
$$x=\frac{-(1-y)\pm \sqrt{(1-y)^2-4(y^2+y-2)}}{2}$$
$$x=\frac{y-1\pm\sqrt{-3[(y-3)(y+1)}}{2}~~~~(1)$$
The reality demands that $-3(y-3)(y+1)\ge 0 \implies (y-3)(y+1)\le 0$
\implies that $-1\le y\le 3$. So the possible integral values of $y$ are: $-1, 0, 1,2,3$ out of these only $y=1$ gives $x$ as irrational.
$$y=-1 \implies  x=-1, y=0 \implies x=1,-2; y=2 \implies x=-1,2; y=3 \implies x=1$$
We get six integral pairs of $(x,y)$, for them $z=1-x-y$ will yield six triplets of $(x,y,z)$
