Find all integers $a$, $b$, and $c$ such that $a + b =1 - c $, $a^3 + b^3 =1 - c^2$

Question

Find all integers a,b, and c such that: \begin{align*}a + b &= 1 - c, \\a^3 + b^3 &= 1 - c^2. \end{align*} My approach is $(a + b)(a^2-ab+b^2) = (1 - c)(1+c)$, so one solution would be $(a,b,c)=(k,-k,1)$ where $k \in \mathbb{Z}$, but if we then assume $a+b \neq 0$ then we get $a^2-ab+b^2 = 1+c$, and I am not really sure what to do from there.

Answer 1

We have $c = 1-a-b$ by the first equation, which we can plug into the second equation. So we just have one equation, namely $$a^3 + b^3 + (1-a-b)^2 - 1 = 0$$ or equivalently $$(a+b)(a^2+b^2+a+b-ab-2)=0$$ So there are two cases, $b=-a$ (hence $c=1$) and $$a^2+b^2+a+b-ab-2 = 0$$ The latter describes an ellipse, which you can see here.

It is located in the interval $[-3,1] \times [-3,1]$, so there are not so many integers to check. The solutions are also clearly symmetric since the equation is symmetric in $a,b$. The solutions up to this symmetry are $(-3,-2)$, $(-2,0)$, $(0,1)$.