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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?

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3 Answers 3

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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 $x\ne y$.

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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 $x=y$.

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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.

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