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

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    $\begingroup$ Helpful allen.ac.in/pncf/rmo/pdf/Paper-With-Solution-RMO.pdf $\endgroup$ – Fallen_Prince Nov 19 '19 at 11:21
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    $\begingroup$ Take the expression you have and take it to the power of 5 using binomial theorem, it will be some fractions involving $a$ and $b$ and the surd you gave. Since you know that is also rational, you can conclude that the surd is rational too. $\endgroup$ – David Nov 19 '19 at 11:33
  • $\begingroup$ The answer you accepted has a fatal gap, so I posted another simple method. $\endgroup$ – Bill Dubuque Nov 19 '19 at 23:30
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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)]

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  • $\begingroup$ That's a really nice solution, better than the provided one! Welcome to MSE! $\endgroup$ – YiFan Nov 19 '19 at 11:41
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    $\begingroup$ +1. Small point. $\mathbb{Z}$ has a Euclidean algorithm. $\mathbb{Z}[x]$ doesn't. $\endgroup$ – Ethan Bolker Nov 19 '19 at 11:49
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    $\begingroup$ @YiFan The argument is incorrect (or incomplete) since it doesn't consider the case that the remainder $\,f(x)\,$ is a constant. in which case the argument break down. $\endgroup$ – Bill Dubuque Nov 19 '19 at 19:38
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    $\begingroup$ @BillDubuque Yes, the proof does need a patch. I don't see one right away. I hope there is one since the argument is very pretty. $\endgroup$ – Ethan Bolker Nov 19 '19 at 22:04
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    $\begingroup$ An obvious way to try to patch it is to actually do the long division and show that the linear term of the remainder must be nonzero. In fact, it is clear that the linear coefficient of the remainder will be some polynomial in the rational value of $20x+\frac{19}{x}$ and then you just have to check that this polynomial has no rational roots. The calculation looks rather annoying to do by hand though. $\endgroup$ – Eric Wofsey Nov 19 '19 at 23:12
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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.

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  • $\begingroup$ This is nice. It may not be immediately obvious to the contestants what conjugation means here, but the worthy ones can prove $x'$ is also a solution of $x'^5=a/b$ with the aid of the quadratic formula and the binomial formula. $\endgroup$ – Jyrki Lahtonen Nov 20 '19 at 10:47
  • $\begingroup$ Note that the "official" solution simply applies the binomial formula for the solution of the quadratic. That may be unavoidable. Either for their argument or Bill's. $\endgroup$ – Jyrki Lahtonen Nov 20 '19 at 10:57
  • $\begingroup$ My approach makes the need for conjugation to go away. Not much to it. $\endgroup$ – Jyrki Lahtonen Nov 20 '19 at 11:25
  • $\begingroup$ @Jyrki Yes, I considered that way too but number theory won the day for me (not surprisingly). Both are instructive. $\endgroup$ – Bill Dubuque Nov 20 '19 at 13:45
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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.

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  • $\begingroup$ Nice, this looks right to me. (+1) $\endgroup$ – YiFan Nov 20 '19 at 12:46
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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.

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