Diophantine Equation : $x+y+z=3$ and $x^3+y^3+z^3=3$

Question

Solve, in integers, the system of equations.

$x+y+z=3$

$x^3+y^3+z^3=3$

I'm not sure how to approach this question, as I have only dealt with linear diophantine equations.

Answer 1

Eliminating $z$ from $$x+y+z=3,\quad x^3+y^3+z^3=3$$ gives $$\begin{align}&x^3+y^3+(3-x-y)^3=3\\&\Rightarrow x^3+y^3+27-27x+9x^2-x^3-27y+18xy-3x^2y+9y^2-3xy^2-y^3-3=0\\&\Rightarrow 24-27x+9x^2-27y+18xy-3x^2y+9y^2-3xy^2=0\\&\Rightarrow 8-9x+3x^2-9y+6xy-x^2y+3y^2-xy^2=0\\&\Rightarrow (3-y)x^2-x(3-y)^2-3y(3-y)=-8\\&\Rightarrow (3-y)(x^2-x(3-y)-3y)=-8\\&\Rightarrow (y-3)(x^2-x(3-y)-3y)=8\\&\Rightarrow (y-3)(x+y)(x-3)=8\end{align}$$

So, we have $$\small(y-3,(x+y)(x-3))=(1,8),(2,4),(4,2),(8,1),(-1,-8),(-2,-4),(-4,-2),(-8,-1).$$

Now checking if each of these is sufficient gives that the followings are the only solutions : $$(x,y,z)=(1,1,1),(-5,4,4),(4,-5,4),(4,4,-5).$$

Answer 2

(This is only a partial answer to the question.)

First, we have: $$x^3 + y^3 + z^3 = x + y + z = 3.$$

Assume that $x$, $y$, and $z$ are all positive.

Note that, by the Arithmetic Mean-Geometric Mean Inequality: $$3 = x + y + z = x^3 + y^3 + z^3 \geq 3\sqrt[3]{{x^3}{y^3}{z^3}} = 3xyz,$$ so that $xyz \leq 1$.

Since $x$, $y$, and $z$ are (assumed to be positive) integers, the only solution that works is $x = y = z = 1$.