Solving two simultaneous equations

Question

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 $z$.

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?

Answer 1

If $z = 1$, then $y = -x$, and all triples $(x,-x,1)$ are solutions. So let's suppose $z \neq 1$. Then we can divide

$$x^2 - xy + y^2 = \frac{x^3+y^3}{x+y} = \frac{1-z^2}{1-z} = 1+z. \tag{3}$$

Adding $(3)$ to $(1)$ yields

$$x^2 - xy + y^2 + x + y = 2.$$

Multiplying by $2$ and adding $2$ to both sides yields

$$\begin{align} 6 &= 2x^2 - 2xy + 2y^2 + 2x +2y + 2\\ &= (x^2 -2xy+y^2) + (x^2+2x+1) + (y^2+2y+1)\\ &= (x-y)^2 + (x+1)^2 + (y+1)^2. \end{align}$$

There aren't many combinations of $x$ and $y$ left to check.

Answer 2

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.