TITLE: Suppose that $x^5$ and $20x+\frac {19}x$ are rational numbers. Then $x$ is also rational
QUESTION [11 upvotes]: 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 $$x=\frac{\sqrt[5]{c}}{\sqrt[5]{d}}$$

REPLY [7 votes]: 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.