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 $$x=\frac{\sqrt[5]{c}}{\sqrt[5]{d}}$$

Solution 1:

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.

Solution 2:

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)]