Solving system of nonlinear equations by substitution I encountered this system of nonlinear equations:
$$\begin{cases}
x+xy^4=y+x^4y\\
x+xy^2=y+x^2y
\end{cases} $$
My ultimate goal is to show that this has only solutions when $x=y$. I didn't find any straight forward method to solving this. But then I came up with the following solution.

First, if $x=0$, then clearly $y=0$ and for the solution we need to
have $x=y$.
Then, assume that $x\ne 0$. Therefore there exists a real number $t$
s.t. $y=t x$. By substituting this to the equations, we find (by
comparing the coefficients) that $t=1$ and therefore $x=y$.
Therefore the system has only solutions of form $x=y$, and every pair
(x, y=x) is a solution.

So is this kind of method OK? If I checked the case $x=0$ separately?
 A: The approach is fine, but since you did not show us your computations, I cannot tell you whether or not the full solution is correct.
Here's how I would do it. Note that\begin{align}x+xy^2=y+yx^2&\iff x-y=yx^2-xy^2\\&\iff x-y=xy(x-y)\end{align}and so if $x\ne y$, $xy=1$. But (still assuming that $x\ne y$)\begin{align}x+xy^4=y+yx^4&\iff x-y=xy(x^3-y^3)=xy(x-y)(x^2+xy+y^2)\\&\iff1=x^2+1+y^2\text{ (since $xy=1$ and $x-y\ne0$)}\\&\iff x^2+y^2=0\\&\iff x=y=0.\end{align}But we were assuming that $x\ne y$. So, there is no solution with $x\ne y$.
A: Proof by contradiction:
Take the difference of the two equations and divide out common factors to get $y^3-y=x^3-x$.  This is a cubic in either variable in terms of the other, giving  three solutions in each case, possible duplicates (x=y will appear in both sets).  Use synthetic division by $x-y$ to get quadratics in both cases to get remaining solutions.
Remaining solutions: $x=\frac{-y\pm \sqrt{4-3y^2}}{2}$ and $y=\frac{-x\pm \sqrt{4-3x^2}}{2}$
However these possible solutions do not in general satisfy the original equations, leaving $x=y$ as the only possible.  An example: $y=1$ leads to $x=0$ and $x=-1$, which fail.
A: Square the equation $x(1+y^2)=y(1+x^2)$ and we have (Note that $2x^2y^2$ will cancel)
\begin{eqnarray*}
x^2(1+y^4)=y^2(1+x^4).
\end{eqnarray*}
Now divide by the first equation and we have $x=y$.
