Definition:
$n$-th root free:
Let $\dfrac{p}{q}$ be in lowest terms and if $b=\left(\dfrac{p}{q}\right)^{\!1/n}$ for some$\ n\in\mathbb{N}$, $b$ is called $n$-th root free if $b$ cannot be expressed as $b=\dfrac{p'}{q'}\left(\dfrac{p''}{q''}\right)^{\!1/n}$ for relatively prime pairs $p', q'$ and $p'',q''.$
Question:
Prove or disprove that the following polynomial $f(x)$ does not have a integer root:
$$f(x)=\sum_{k=0}^na_kx^k\in \mathbb{R}[x],$$
where
$(i) \; a_i\in\{\mathbb{R}\setminus\mathbb{Q}\}\cup\{0\}$$\ \forall \ i=1,2,...,n-1$.
$(ii)\;a_n=1$; $a_0\in\mathbb{R}\setminus\mathbb{Q}.$
$(iii)\; a_i$ and $a_j$ are pairwise distinct for all $i\neq j$.
$(iv)\; $ Let $\dfrac{p}{q}$ be in lowest terms and if $a_i=\left(\dfrac{p}{q}\right)^{\!1/n}$ for some$\ n\in\mathbb{N}$; then $a_i$ is $n$-th root free.
My Comments:
$1.$ It seems to be true.
$2.$ I have proved it for polynomials up to $\deg(f)=4$ since their roots could be easily seen through the radical method.
$3.$ It seems quite a difficult task for $\deg(f)\geq 5$.
Can anyone help me with this?
It's false. Examine the function $f(x)=(x-\pi)(x-2)=x^2-(2+\pi)x+2\pi.$ Now neither $2\pi$ nor $-(2+\pi)$ are expressible as the $n$th root of rationals, so the definition of being $n$th root free is vacuously true. This polynomial therefore satisfies all criteria, but certainly has an integer root.