# Finding the roots of a 2 variable polynomial

Consider the polynomial $$x^{n+1}+y^{n+1}+yx^n$$ where $$n\in\mathbb{N}$$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $$(x,y)=(0,0)$$. So far, I've been able to show that if we have a root different than $$(0,0)$$ then either $$x>0,y<0$$ or $$x<0,y>0$$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.

## 2 Answers

Using the well-known inequality $$ab \le \frac{a^p}{p} + \frac{b^q}{q}$$ for $$a, \, b \ge 0$$, where $$p^{-1} + q^{-1} = 1$$, one can estimate $$yx^n \ge - |y||x|^n \ge - \frac{|y|^{n+1}}{n+1} - \frac{n|x|^{n+1}}{n+1}$$ and therefore, since $$n+1$$ is even, $$y^{n+1} +y^{n+1} + yx^n \ge \frac{n}{n+1}y^{n+1} + \frac{1}{n+1}x^{n+1} \, .$$ This is positive except if $$x = y = 0$$.

• Wow, did not expect to take this path, thank you! – Ray Bern Mar 9 at 17:19

As the polynomial is homogeneous, you solve

$$t^{n+1}+t+1=0$$ and all $$(x,tx)$$ are solutions.

But it doesn't take long to see that the polynomial has no real roots.