I was given:

$x + y^2 = y^3 ...(i) \\ y + x^2 = x^3...(ii)$

And was asked to find real $(x,y)$ solutions that satisfy the equation.

I substracted $(i)$ by $(ii)$:

$x^3 - y^3 + y^2 - x^2 + x - y = 0$

Then factored it out so I have:

$(x-y)(x^2 + xy + y^2 - x - y + 1) = 0$

Multiplying it by two, I get:

$(x-y)(2x^2 + 2xy + 2y^2 - 2x - 2y + 2) = 0 \\ (x-y)((x^2 - 2x + 1) + (y^2 - 2y + 1) + (x^2 + 2xy + y^2)) = 0 \\ (x-y)((x-1)^2 + (y-1)^2 + (x+y)^2) = 0$

I noticed that a solution exists only if $x=y$ because there are no real solutions for $x$ and $y$ that satisfies $(x-1)^2 + (y-1)^2 + (x+y)^2 = 0$.

Substituting $x=y$ into the first equation, I get: $y(y^2-y-1)=0$ where the roots are $y= 0, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}$. Hence, the real solutions of $(x,y)$ that satisfy are:

$(x,y) = (0,0), (\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}), (\frac{1-\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2})$.

What I would like to ask is: Is there a better way to solve the question? It's from a local university entrance test, where this kind of questions are aimed to be done in < 3 minutes. It took me a while to manipulate the algebraic stuffs above.

Someone in a local forum said something about symmetric systems which says that the solution does not exist for $x \neq y$. How do I know if the equation is a symmetric one? (Never heard of something before throughout high school here...) I would love to see a resource for this!

  • $\begingroup$ Suppose $(x_1,y_1)$ is a solution, is $(y_1,x_1)$ a solution to your system? You'll see that it is, because one equation is just the flip of the other. This is what symmetry means $\endgroup$ Jul 14, 2020 at 8:46
  • $\begingroup$ Do you have the mark scheme available? $\endgroup$ Jul 14, 2020 at 8:54

1 Answer 1


You can begin by noting that the $2$ functions are inverses of each other (and only involve odd non-zero exponents). Using the fact that inverse functions are reflections in the line $y=x$, we can now see that the intersection points must be along the line $y=x$. Substituting $y$ into $x$ or vice-versa, we obtain the equation you get and obtain the solutions you got. That would only take about 3 minutes. I hope that helps :)


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