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 $$x=\frac{\sqrt[5]{c}}{\sqrt[5]{d}}$$
 A: My way of fixing the problem in the accepted answer.

The unknown number $x$ is a zero of both $f(X)=X^5-q_1$ and $g(X)=X^2-(q_2/20)X+19/20$, where $q_1=c/d$ and $q_2=a/b$ are the rational numbers appearing in the question.
When we carry out polynomial division we get polynomials $q(X)$ and $r(X)$, both with rational coefficients, such that
$$
f(X)=q(X)g(X)+r(X).
$$
Furthermore, the remainder $r(X)$ is at most linear. The obvious point is that $r(x)=0$. It follows immediately from plugging in $x$ for $X$.
If $r(X)$ is a linear polynomial, then its only zero is a rational number that must be equal to $x$, and we are done. This is the essence of Krishnarjun's answer.
If $r(X)$ is a constant, that constant must be equal to zero for otherwise $r(x)\neq0$. This is a more problematic case. The key to handling it is that in this case $f(X)=q(X)g(X)$. Thus we can deduce that the zeros of $g(X)$ are also zeros of $f(X)$.

But the zeros of $g(X)$ are both real because one of them is. On the other hand $f(X)$ has only a single real zero, so this is a contradiction.

A remaining blemish is that the zeros of $g(X)$ might be equal. But $f(X)$ cannot have multiple zeros unless $q_1=0$, when also $x=0$. So we are done in that case also.
A: 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. 
A: The accepted answer has a fatal gap, so  I present an alternative simple argument. Note that
$\quad 20x+19/x = r\in \Bbb Q\!$ $\overset{\large\, \times\ x}\iff 20x^2+19 = rx\!$ $\overset{\large\, \div\ 20}\iff 0 = x^2-\frac{r}{20}\,x+\color{#c00}{\frac{19}{20}} =: \color{#0a0}{f(x)}$ 
So the claim is that any common root $\,x\,$ of $\,\color{#0a0}{f(X)}\,$ and $\,X^5-a/b,\,\ a/b\in \Bbb Q,\,$ is rational.
Suppose for contradiction $\,x\not\in\Bbb Q.\,$ Vieta $\,\Rightarrow\, \color{#c00}{x\bar x = 19/20}\,$ where $\,\bar x =$ conjugate of $\,x$.
By hypothesis $\, x^5 = a/b\in\Bbb Q\,$ so $\, {\bar x}^{\,5} = a/b\ $ by conjugating. Multiplying the two yields that $\, (\color{#c00}{19/20})^5 = (\color{#c00}{x\bar x})^5 = a^2/b^2\,$ $\Rightarrow\,19^5 b^2 = 20^5 a^2.\,$ But the prime $19$ has odd power in the prime factorization of $\,19^5 b^2$ vs. even power in $\,20^5 a^2,\,$ contra uniqueness of prime factorizations.
A: Given that $20x+\frac{19}{x}$ is rational. Therefore $x$ satisfies a quadratic polynomial with rational co-efficients. If we call that polynomial as $g$, we get $g(x)=0$. Now by the Euclidean algorithm $x^5=h(x)g(x)+f(x)$, where $f(x)$ is a linear polynomial on $x$ with rational co-efficients. Since $x^5\in\mathbb{Q}$ and $g(x)=0$, we get $f(x)\in\mathbb{Q}\Rightarrow x\in\mathbb{Q}$.
[Note by Bill D. $ $ The inference $f(x)\in\Bbb Q\,\Rightarrow\, x\in \Bbb Q\,$ fails if $\,\deg f < 1,\,$ so we must prove $\,\deg f = 1\,$ to complete the above argument. One way to remedy that is in Jyrki's answer, and another way is to explicitly compute the remainder $\,f(x) = r\, x + s\,$ then prove $\,r \neq 0\,$ (which is essentially equivalent to the method in the linked official solution - reproduced in Jack's answer)]
