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Let $g\in \mathbb{Q}[x]$ be a polynomial of degree 2022. Prove that there exist infinitely many rational $q\in (0,1)$ such that $\sqrt[5]{g(q)}\notin \mathbb{Q}$.

I encountered the above problem in the problem solving seminar. I have seen the following somewhat similar problem;

Let $f(x)\in \mathbb{Z}[x]$ be a polynomial of degree 3. Prove that there exist infinitely many integer $n$ such that $f(n)$ is not a perfect square.

However, the methods used in the latter seem to be not applicable to the former. For the second one, we consider $f(x)=ax^3+bx^2+cx+d$ with $a\neq 0$. Then, if $a<0$, it's trivial. Finally, consider the case when $a>0$ which required results from several other problems we proved in the seminar.

At this point, I don't know other approaches. Since the problem is dealing with a polynomial of degree 2022 this time, I bet there should be a clever way. In addition, the instructor told us this one is much simpler.

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    It might be enough to show that any interval will contain a rational number $q$ such that $\sqrt[5]{q} \notin \mathbb{Q}$, then argue using the intermediate value theorem. – Satvik Saha Dec 01 '22 at 03:28
  • Thank you, Satvik. Can you elaborate your idea, please? –  Dec 01 '22 at 06:13
  • Sorry, this doesn't quite work. I thought that since $g$ maps intervals to intervals, we'd find $q = g(x)$ in the image of an interval with $\sqrt[5]{q} \notin \mathbb{Q}$; but there's no way to guarantee that $x \in \mathbb{Q}$. – Satvik Saha Dec 01 '22 at 07:34
  • Oh ok, at least now I know one way that doesn't work, thank you for followup! –  Dec 01 '22 at 07:45
  • Just an idea: The polynomial $p(x)=\sqrt 2x+1$ has the property that $p(a)$ is irrational for every rational $a\neq 0$. So it is necessary to use the condition that the polynomial is in $\Bbb Q[x]$. – ajotatxe Dec 01 '22 at 16:12

2 Answers2

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Hint:

For $g \in \mathbb{Z}[x]$, consider the denominator of $g(1/m)$, where $m$ is any integer relatively prime to the leading coefficient of $g(x)$.

For $g \in \mathbb{Q}[x]$, multiply by an appropriate $5$th power to reduce to the previous case.

Benjamin Wright
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    I came up with the same argument, but g has rational coefficients so to be safe I took m to be a prime not appearing in any of the coeffs of g. Then the m-adic valuation of the leading term is definitely different from any other term – usr0192 Dec 02 '22 at 00:14
  • I believe this argument works for any pair $N, m$ in my setup (different $m$ from the $m$ in your answer) with $m \not\mid N$. – Dustan Levenstein Dec 02 '22 at 00:42
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    @usr0192 You're right, I was thinking of $g \in \mathbb{Z}[x]$. Fixed. – Benjamin Wright Dec 02 '22 at 04:23
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Just some observations which won't fit in a comment.

For fixed $N, m \in \mathbb N$ ($N=2022, m=5$ in this question), the following are equivalent:

  1. For every $g \in \mathbb Q[x]$ of degree $N$, there are infinitely many $q \in (0, 1) \cap \mathbb Q$ such that $\sqrt[m]{g(q)} \notin \mathbb Q$.
  2. For every $g \in \mathbb Q[x]$ of degree $N$, there is at least one $q \in (0, 1) \cap \mathbb Q$ such that $\sqrt[m]{g(q)} \notin \mathbb Q$.
  3. For every $g \in \mathbb Q[x]$ of degree $N$ and $a< b$ both rational numbers, there is at least one $q \in (a, b) \cap \mathbb Q$ such that $\sqrt[m]{g(q)} \notin \mathbb Q$.
  4. For every $g \in \mathbb Q[x]$ of degree $N$, the set of $q \in \mathbb Q$ with $\sqrt[m]{g(q)} \notin\mathbb Q$ is dense in $\mathbb Q$.

This is easy to see because affine transformation of the domain of a polynomial preserves the degree $N$ condition (i.e., for $c, d \in \mathbb Q$ with $c \neq 0$ and $f(x) = g(cx+d)$, we have $\deg f = \deg g$).

My other observation is this question evokes the proof of the rational root theorem somehow; we're trying to find $q$ such that $g(q) = s^m$ has no solution in $\mathbb Q$. Let $D$ be the LCM of the denominators in $g$, and let $D \mid r^m$ for some $r \in \mathbb Z$. Then $r^m q \in \mathbb Z[x]$. Suppose that $a/b \in \mathbb Q$ with $a, b \in \mathbb Z$ such that $$b^{Km}r^mg(a/b) = s^m$$ for sufficiently large $K$ so that the denominators are cleared in $r^mg(a/b)$; $Km \ge N$ does the trick. A rational solution for $s$ is then necessarily an integer by the rational root theorem.

Let's write $$r^m g(x) = \sum_{n=0}^N C_n x^n,$$ where $C_n \in \mathbb Z$ by construction. Then $$b^{Km} r^m g(a/b) = b^{Km-N} \sum_{n=0}^N C_n a^nb^{N-n}.$$ Taking $a, b$ to be relatively prime, we can conclude that $a \mid s^m-C_0b^N$. Since $m \not\mid N$ holds in the original question (and is a necessary condition for the above equivalent conditions to hold), we can also conclude that $Km-N>0$ so $b \mid s^m$. This isn't as clean as the standard rational root theorem, but maybe this setup is usable somehow?

Dustan Levenstein
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