# Suppose that x , y and z are three integers (positive,negative or zero) such that w

Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero) such that we get the following relationships simultaneously
1. $x+y=1-z$
and
2. ${x}^{3}+{y}^{3}=1-{z}^{2}$
Find all such $x$, $y$ and $z$.
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Norah Baxter
If $z=1$, then $y=-x$ and all triples $\left(x,-x,1\right)$ are solutions. So let's suppose $z\ne 1$. Then we can divide
$\begin{array}{}\text{(3)}& {x}^{2}-xy+{y}^{2}=\frac{{x}^{3}+{y}^{3}}{x+y}=\frac{1-{z}^{2}}{1-z}=1+z.\end{array}$
Adding ($\left(3\right)$ to $\left(1\right)$ yields
${x}^{2}-xy+{y}^{2}+x+y=2.$
Multiplying by $2$ and adding $2$ to both sides yields
$\begin{array}{rl}6& =2{x}^{2}-2xy+2{y}^{2}+2x+2y+2\\ & =\left({x}^{2}-2xy+{y}^{2}\right)+\left({x}^{2}+2x+1\right)+\left({y}^{2}+2y+1\right)\\ & =\left(x-y{\right)}^{2}+\left(x+1{\right)}^{2}+\left(y+1{\right)}^{2}.\end{array}$
There aren't many combinations of $x$ and $y$ left to check.