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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.

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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 $x = y = 0$.

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

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