Polynomials fixing Rationals

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#1 h Mar 1, 2025, 6:28 PM

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Suppose that $p(x) \in \mathbb{R}[x]$ be a polynomial with the property that there exists an interval $I$ and $J$ such that $p(x)$ is surjective on $\mathbb{Q} \cap I \to \mathbb{Q} \cap J$ . Prove that $p(x)$ is linear.

My approach :
One can say from the initial condition that coefficients of $p(x)$ must be rational. The rest is about proving that $p'(x)$ is constant. Although there might be other approaches.

This post has been edited 1 time. Last edited by Royrik123456, Mar 1, 2025, 6:28 PM

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#2 h Mar 3, 2025, 6:19 PM • 1 Y

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Indeed, from Lagrange interpolation we know that all coefficients of $p$ are rational. However, I am not sure if considering the derivative is helpful here; instead, one can use the number-theoretic properties of $\mathbb{Q}$ .
By scaling with a suitable rational number, we may assume that $P(x)=a_d x^d + a_{d-1} x^{d-1} + \cdots + a_1 x + a_0$ for integers $a_i$ with $a_d \ne 0$ . Suppose fsoc that $d \ge 2$ . We claim that any number of the form $\textstyle\frac{k}{p}$ where $p > |a_d|$ is prime and $(k, p) = 1$ cannot be in the image of $P$ . Since the interval $J$ certainly contains such a rational number for sufficiently large $p$ , this will be a contradiction.
Indeed, if $x=\textstyle\frac{b}{c}$ with $(b, c)=1$ satisfies $P(x) = \textstyle\frac{k}{p}$ then
$p \sum_{i=0}^d a_i b^ic^{d-i} = k c^d.$ In particular, $p \mid c$ . But then the RHS is divisible by $p^d$ while the LHS is not divisible by $p^2$ , since every term except $pa_db^d$ is divisible by $p^2$ and $(p, b)= (p, a_d)=1$ .

See also here.

This post has been edited 1 time. Last edited by Gryphos, Mar 4, 2025, 2:56 PM
Reason: fixed typo

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#3 h Tuesday at 10:55 PM

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Sorry for late reply. Thanks for the solution.

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