# Are all the solutions of the system $\begin{cases} f(x,y)=0 \\ f(y,x)=0 \end{cases}$ (if there are any) pairs of equal numbers?

I've had to solve this system of two equations with two unknowns over the reals:

$$\begin{cases} x^3-x-y=0 \\ y^3-y-x=0 \end{cases}$$

The simple substitution $y=x^3-x$ works, of course. The solutions are: $(0,0)$, $(\sqrt{2},\sqrt{2})$, $(-\sqrt{2},-\sqrt{2})$. It seems rather interesting to me that all the solutions are pairs of equal numbers, and I've tried to figure that out in a different way.

Our system of equations has the form $$\begin{cases} f(x,y)=0 \\ f(y,x)=0 \end{cases}$$

Simply put, one equation is obtained from another by replacing $x$ with $y$ and vice versa. If we plot the graphs of the equations $f(x,y)=0$ and $f(y,x)=0$ on the same coordinate system, the obtained geometric figures will be reflections of one another in the line $y=x$. The reason is that $x$ and $y$ exchange their roles when we pass from one equation to another.

Thus, if our system of equations has solutions, they do have the form of pairs of equal numbers. Otherwise, the symmetry of the graphs would be violated.

My question is whether this approach is mathematically valid.

• Consider a function that is already symmetric, such as $f(x,y)=x^2+y^2-1$. – Harald Hanche-Olsen Mar 11 '17 at 13:10
• Thank you for the counterexample. Anyway, can my argument be modified to become a correct one? For example, by making $f(x,y)$ a non-symmetric function of $x$ and $y$. – Alex Konrad Mar 11 '17 at 13:36
• @HaraldHanche-Olsen, as far as I can understand, the requirement for $f(x,y)$ being non-symmetric makes the graph of $f(x,y)$ non-symmetric with respect to the line $y=x$. Then the intersections of this graph (if there are any) with $f(y,x)=0$ lie on the line $y=x$, don't they? – Alex Konrad Mar 11 '17 at 13:58

This is not logically sound — no, they don't have to be pairs of equal numbers. Yes, the symmetry of the graph implies the symmetry of solutions, as in symmetric relations: if $(x,y)$ is a solution, then $(y,x)$ also is a solution. So the correct general conclusion is that solutions with nonequal numbers have to come in pairs.
Here's a quick nonsymmetric example: let $f(x,y)=x^2-1$. Then the system of equations is \left\{\begin{align} x^2-1&=0 \\ y^2-1&=0 \end{align}\right. whose solutions are $(\pm1,\pm1)$, half of which have nonequal $x$ and $y$.
Consider the system: $\begin{cases} x+y=0 \\ y+x=0 \end{cases}$ We have $f(x,y)=x + y$ and, let's say $x=1,y=-1$ is solution for the system.
• @martycohen The point is $x \ne y$ – user261263 Mar 11 '17 at 16:47