How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$?

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?

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 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 $(x,y,z)$