System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$

Question

Solve over reals:

$$ \begin{cases} x^3+y=y^2\\ y^3+z=z^2\\ z^3+x=x^2\\ \end{cases} $$

I think the only solution is $x=y=z=0$. If the variables are non-zero multiplying the equations:

$$(x-1)(y-1)(z-1)=x^2y^2z^2 > 0$$

Here, there are two cases: all three variables are greater than $1$ or only one.

If $x,y,z>1$, it's easy to prove there are no solutions because $y^2>x^3$ a.s.o. gives $xyz<1$.

How can I prove there are no solutions when only variable, say $x$, is greater than $1$?

Answer 1

I will prove by contradiction that there exists no solutions for $x > 1$ and $y,z < 1$. Suppose $(x,y,z)$ is such a solution.

Iteration #1: First, note that: $$ z^3 = x^2 - x = x(x - 1) > 0 \implies z > 0 $$ Thus, $0 < z < 1$. We see that: $$ y^3 = z^2 - z = z(z - 1) $$ For $0 < z < 1$, $-\frac{1}{4} < z(z - 1) < 0$, so $-2^{-\frac{2}{3}} < y < 0$. For this range of values of $y$, since $y(y - 1)$ is decreasing for $y < 0$, we have that: $$ x^3 = y^2 - y < 2^{-\frac{4}{3}} + 2^{-\frac{2}{3}} = 1.0268108 \implies x < 1.00885823 $$ Iteration #2: Applying this inequality to the third equation again, we have: $$ z^3 = x(x - 1) < 1.00885823(1.00885823 - 1) = 0.0089367 \implies z < 0.20752 $$ For $0 < z < 0.20752$, we have that: $$ 0.20752(0.20752 - 1) < y^3 < 0 \implies -0.54787 < y < 0 $$ Once again, we have: $$ x^3 = y(y - 1) < -0.54787 \left(-0.54787 - 1\right) < 1 \implies x < 1 $$ This contradicts that $x > 1$. Therefore, no non-trivial solutions exist.

Answer 2

Since only $x>1$ Then $x> y$

Using the first equation,

$x^3=y(y-1)$

For the case $0\leq y <1$, the RHS is negative,

So $y<0$

Taking the third equation,

$z^3=x^2-x$

The RHS is always positive hence $0<z<1$

Taking the second equation,

$y^3=z^2-z$

Differentiate RHS to find stationary point,

$2z-1=0$

$z=1/2$,

$-1/4\leq y^3 <0$

Using the first equation again,

$x^3=y(y-1)$

Differentiating RHS to find min/max for $x$

$y=-1/4$,

$x^3=5/16$, which contradicts that $x>1$

Answer 3

If $xyz=0$ so $(x,y,z)=(0,0,0).$

Let $xyz=r\neq0.$

Thus, by the WE Tutorial School's work we obtain: $$(x+1)(y+1)(z+1)=1.$$ or $$x+y+z+xy+xz+yz=-r.$$ In another hand, by your work we obtain: $$(x-1)(y-1)(z-1)=x^2y^2z^2,$$ which gives $$x+y+z-xy-xz-yz=r^2-r+1,$$ which gives $$x+y+z=\frac{(r-1)^2}{2}$$ and $$xy+xz+yz=-\frac{r^2+1}{2}.$$ But after summing of our equations we obtain: $$\sum_{cyc}(x^3-x^2+x)=0$$ and since $$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz$$ and $$x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz),$$ we have: $$\frac{(r-1)^6}{8}+\frac{3(r-1)^2(r^2+1)}{4}+3r-\frac{(r-1)^4}{4}-r^2-1+\frac{(r-1)^2}{2}=0$$ or $$r^6-6r^5+19r^4-24r^3+11r^2+6r+1=0$$ or $$(r^3-3r^2+3r+1)^2+4r^2(r^2-2r+2)=0,$$ which is impossible for real value of $r$.

Id est, $(0,0,0)$ is an unique solution of the system.