Solve the system of equations

Question

I found system of equations on internet, I want to practice math and to solve it.

All $x$,$y$ and $z$ are integers.

$$\begin{cases} x+y+z=3 \\ x^3+y^3+z^3=3 \end{cases} $$

Should I start by watching all combinations that sum up to 3, please give me some hint to start.

Answer 1

Consider the identity $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx). $$ Then since $x+y+z=3=x^3+y^3+z^3 $, thus we need to solve $$1-xyz=x^2+y^2+z^2-xy-yz-zx .$$

Answer 2

If you are looking for solutions with $x,y,z$ real, then you can just substitue $z=3-x-y$ in the second equation to find $$ y=\frac{-9+6x-x^2\pm\sqrt{x^4-18x^2+32x-15}}{2x-6} \\ z = 3-x-y \\ x\leq -5 \mbox{ or }x=1\mbox{ or }x\geq 3 $$ (The quartic inside the square root is easy to factor because $x=1$ is a double root.)

So that is a set of three curves, one each in quadrants I, II, and IV.

If you are looking for solutions with $x,y,z\in \Bbb Z$ this immediately gives $$ \{x=1,y=1,z=1\}, \{x=-5,y=4,z=4\} $$ and by symmetry you get two others that lie along the curves in the first and second quadrants $$\{x=4,y=-5,z=4\},\{x=4,y=4,z=-5\}$$

So now we need to wonder whether there are any other integer solutions.

For that to happen, a necessary (but perhaps not sufficient) condition is that $$ x^4-18x^2+32x-15=k^2$$ for some integer $k\not\in \{0,3,4\}$.

We can show that this cannot happen, as follows:

For $x\geq 5$, $$ (x^2-1)^2 < x^4-18x^2+32x-15 + 16(x^2-1)^2 <x^4-18x^2+32x-15 < x^4=(x^2)^2 $$ so x$^4-18x^2+32x-15$ lies striclty between two consecutive integers.

Answer 3

For a complete solution see this duplicate.So let me just add a comment. There is a stronger conjecture, namely that the Diophantine equation $$ x^3+y^3+z^3=3 $$ only has $(1,1,1)$ and $(4,4,-5)$ and its permutations as integral solutions, see this MO-question. Cassels showed that any integral solution $(x,y,z)$ has to satisfy $$ x\equiv y\equiv z \bmod 9. $$ Furthermore, Oppenheim studied the Diophantine equation $$ x^3+y^3+z^3=x+y+z. $$