<s>[INST] Suppose that $x^5$ and $20x+\frac {19}x$ are rational numbers. Then $x$ is also rational Let $x\neq0$ be a real number such that $x^5$ and $20x+\frac {19}x$ are rational. How can we prove that $x$ is also rational? (This was a question from the RMO 2019 in India.)
My attempt: Let $a,b,c,d$ be integers such that $20x+\frac {19}{x}=\frac ab$ and $x^5 = \frac{c}{d}$. 
Then we have $$x=\frac{a\pm\sqrt{a^2-1520 b^2}}{40b}$$ so $x$ is rational iff $\sqrt{a^2-1520 b^2}$ is rational. 
However, I don't know how to prove that $\sqrt{a^2-1520 b^2}$ is rational using that $$x=\frac{\sqrt[5]{c}}{\sqrt[5]{d}}$$
 [/INST] It seems that one answer has been posted online. The argument is elementary and straightforward. I will repeat it here. 
Let $r=20x+\frac{19}{x}\in\mathbb{Q}$. It follows that
$$
20x^2-rx+19=0
$$
and the quadratic formula implies that
$$
x=r_1\pm\sqrt{r_2},\quad 
r_1=\frac{r}{40},\quad 
r_2={\frac{r^2-4\cdot 20\cdot 19}{40^2}}\geq 0.
$$
The binomial theorem implies that
$$
x^5=(r_1\pm\sqrt{r_2})^5=r_1^5+r_2^2\sqrt{r_2}
+5r_1^4\sqrt{r_2}+5r_1r_2^2
+10r_1^3r_2+10r_1^2r_2\sqrt{r_2}\\
=q\pm(r_2^2+5r_1^4+10r_1^2r_2)\sqrt{r_2},\quad q\in\mathbb{Q}.
$$
But $x^5\in\mathbb{Q}$, and $r_2^2+5r_1^4+10r_1^2r_2\neq 0$ (since $r_2\geq 0$ and $x\neq 0$). It follows that $\sqrt{r_2}\in\Bbb{Q}$. We are done. 
</s>