Solution to a system of equations

Question

Find all triples $(x,y,z)$ of real numbers satisfying the following system: $$ \begin{cases} (x+1)(x^2+1) = y^3+1\\ (y+1)(y^2+1) = z^3+1\\ (z+1)(z^2+1) = x^3+1 \end{cases} $$ I tried adding and multiplying the three equations and I still can't see how to proceed further!

Answer 1

• If $x>0$ then $$ x^2+1> x^2-x+1$$ and $y,z>0$ so $$ y^2+1> y^2-y+1$$ and $$ z^2+1> z^2-z+1$$ and thus $$ (x^2+1)(y^2+1)(z^2+1)> (x^2-x+1)(y^2-y+1)(z^2-z+1)$$ which is a contradiction since (notice that $x,y,z\ne -1)$ $$ (x^2+1)(y^2+1)(z^2+1)= (x^2-x+1)(y^2-y+1)(z^2-z+1)\;\;\;\;(*)$$ We get $(*)$ if we multiply all given equations and cancel equation we get with $x+1$, $y+1$ and $z+1$.
• If $x=0$ then $y=z=0$.
• If $x<0$ then $y^3 = x(x^2+x+1)<0$ so $y<0$ and $z<0$ Now if $x=-1$ then $y=z=-1$. So suppose $x\ne -1$ then $y\ne-1$ and $z\ne -1$ so we have again $(*)$. But now we have also: $$ x^2+1< x^2-x+1$$ and $$ y^2+1< y^2-y+1$$ and $$ z^2+1<z^2-z+1$$ and thus $$ (x^2+1)(y^2+1)(z^2+1)< (x^2-x+1)(y^2-y+1)(z^2-z+1)$$ which is a contradiction with $(*)$ again.

So $x=y=z=0$ or $x=y=z=-1$ are the only solutions.

Answer 2

Calling

$$ y = f(x)\\ z = f(y)\\ x = f(z) $$

we have

$$ y = f^3(y)\\ z = f^3(z)\\ x = f^3(x) $$

Then $x,y,z$ are fixed points of $f^3$ but the only fixed points are $\{-1,0\}$

hence $x=y=z=-1$ and $x = y = z = 0$

NOTE

$$ f(x) = \sqrt[3]{(x+1)(x^2+1)-1} $$