<s>[INST] Solving two simultaneous equations Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero)  such that we get the following relationships simultaneously


*

*$x + y = 1 - z$ and   

*$x^3 + y^3= 1 - z^2$


Find all such $x$, $y$ and $z$.
I have tried to determine the values of $x$ and $y$ by fixing some values of $z$ but it is obvious that this method will not work. I also tried to determine the bounds of $x$ and $y$ and the related terms like $xy$ and ($x$$2$ +$y$$2$) but this this method failed as well. What then is the way out?
 [/INST] To find all integer solutions, we begin by substituting $z=1-x-y$, into $z^2= 1-x^3-y^3$ to get
$$(1-x-y)^2=1-x^3-y^3.$$
Since  $y=-x$ is already a clear solution, we will force the factor $(y+x)$ in what follows.
\begin{align*}
& (1-x-y)^2=1-x^3-y^3 \\
\Rightarrow & 1-2x+x^2-2y+2xy+y^2=1-x^3-y^3 \\
\Rightarrow & y^3+y^2+2xy-2y+x^3+x^2-2x=0 \\
\Rightarrow & y^3+y^2+y(2x-2)+(x^3+x^2-2x)=0.
\end{align*}
Divide the left by $(y+x)$ to see that
\begin{align*}
&\frac{y^3+y^2+y(2x-2)+(x^3+x^2-2x)}{y+x} \\
=&y^2+y(1-x)+(x^2+x-2) \\
=&y^2-xy+y+x^2+x-2,
\end{align*}
with a zero remainder. Hence
\begin{align*}
& y^3+y^2+y(2x-2)+(x^3+x^2-2x)=0 \\
\Rightarrow & (y+x)\left( y^2-xy+y+x^2+x-2 \right)=0 \\
\Rightarrow & (y=-x) \, \operatorname{or}\, \left(y^2-xy+y+x^2+x-2 =0 \right).
\end{align*}
Treating the latter case as a quadratic in $y$, and applying the quadratic formula, we get
\begin{align*}
& y^2-xy+y+x^2+x-2 =0 \\
\Rightarrow& y^2 +(1-x)y+\left( x^2+x-2 \right)=0 \\
\Rightarrow&y=\frac{-(1-x) \pm \sqrt{(1-x)^2-4(x^2+x-2)}}{2} \\
&\,\,\,\,=\frac{(x-1)\pm\sqrt{3(1-x)(x+3)}}{2}.
\end{align*}
Now 
$$\sqrt{3(1-x)(x+3)}\geq 0 \Rightarrow -3 \leq x \leq 1.$$
Of these 5 integer test values for $x$, $x \in \{ -3,-2,0,1 \}$ produce integers $y$, and it produces them for both the $\pm$ on the solution $y$. This leaves the set of all integer solutions,
$$(x,-x,1),$$
from the $+$ branch on the solution $y$,
$$\{(-3, -2, 6), (-2, 0, 3), (0, 1, 0), (1, 0, 0)\},$$
and from the $-$ branch on the solution $y$, two more solutions arise not in that prior set,
$$\{(-2, -3, 6), (0, -2, 3)\}.$$
This constitutes all integer solutions.
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