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Let $f_n(x):=x^n-a\in\mathbb{Q}[x]$ be a polynomial such that $a\notin\{-1,0,1\}$.

For fixed $a$, how to show there is only finitely many $n\in\mathbb{Z}_{\geq 1}$ such that for each $n$, there exists $x\in\mathbb{Q}$ and $f_n(x)=0$.

Edit. For $a=p_1^{n_1}...p_k^{n_k}\in \mathbb{Z}$, $f_n(x)$ has rational solution if $n\mid n_i \;(1\leq i\leq k)$, see this, which implies number $n$ should be finite. For $a\in \mathbb{Q}$ I guess we may do the same argument, however not so clear to me.

Dual
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Nov 25 '24 at 12:44
  • @Guruprasad. I am asking whether number of $n$ is finite, not for fix $n$ whether $x$ is finite :) – Dual Nov 25 '24 at 13:24
  • What do you mean by "$\ f_n(x)\ $ admits a rational solution"? Do you mean that $\ x\ $ is a rational root of $\ f_n \ $ (i.e. that $\ f_n(x)=0\ $)? – lonza leggiera Nov 25 '24 at 13:39
  • @lonzaleggiera Yes you are correct, let me edit it – Dual Nov 25 '24 at 13:45
  • If $(s/t)^n=u/v$, with $(s,t)=(u,v)=1$, choose any prime $p$ dividing $u$ or $v$, and look at its powers on the lhs and rhs. – TonyK Nov 25 '24 at 14:04
  • Yes $n/n_i$ is good – EDX Nov 25 '24 at 15:16

1 Answers1

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You can extend the argument to $a\in \mathbb{Q}, \ a=p/q$ with $\gcd(p,q)=1 $ and $(p,q) \in \mathbb{Z}\times \mathbb{Z}^*$

So writting :

$$ p= p_{i_1}^{\alpha_{i_1}}...p_{i_k}^{\alpha_{i_k}} $$ $$ q= p_{j_1}^{\alpha_{j_1}}...p_{j_m}^{\alpha_{j_m}} $$

You want $n | \alpha_{i_l}$ with $l \in [1,k] \ $ and $n | \alpha_{j_l}$ with $l \in [1,m]$.

Otherwise you will have a root in nominator or denominator say $p_o^{\alpha_v}$ such that $\sqrt[n]{p_o^{\alpha_v}}$ is irrational, which is excluded according to what we seek.

EDX
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