<s>[INST] System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
  x+y^2 &= y^3\\
  y+x^2 &= x^3
\end{align*}
These are the solutions:
 \begin{align*}
  (0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2); 
\end{align*}
 [/INST] Here is a dynamical systems approach assuming that we only look for real solutions. Consider the map: $$f(x)=x^3-x^2=x^2(x-1)$$
Solving $f(x)=y$ and $f(y)=x$ is equivalent to solving $f(f(x))=x$. We want to show that a solution of the latter (a periodic orbit of period 2) is necessarily a fixed point, i.e. $f(x)=x$ which is trivially solvable.
So assume that $x=f(y), y=f(x)$. We may assume that $x\leq y$.
Now $f$ is strictly monotone increasing on $[1,+\infty)$ so if $1\leq x\leq y$ then $y=f(x)\leq x=f(y)$ implying $x=y$.
Similarly, as $f$ is strictly monotone on $(-\infty,0]$ then  $x\leq y\leq 0$ implies $x=y$.
Finally, $f$ maps $(-\infty,1]$ into $(-\infty,0]$ so $x\in (0,1]$ implies $x=f(f(x))\leq 0$ (contradiction). We conclude that $x=y$ as wanted.
Remark: Incidently, the same argument shows that any real orbit of period $p$ must also be a fixed point. So e.g. $x+y^2=y^3$, $y+z^2=z^3$, $z+x^2=x^3$ has the same (real) solutions.
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