There is a supposed solution to the same question here but that I frankly don't understand what's going on there. I would appreciate feedback as to whether my method is correct.
Exercise:
Show that any rational zero of a polynomial $p = \sum_{k = 0}^n a_k X^k \in \mathbb Z [X]$ of degree $n \geq 1$ is in $a_n^{- 1} \mathbb Z$. (Hint: Consider $a_n^{n - 1}p$.)
My attempt:
Taking the hint, we have
\begin{align*} a_n^{n - 1}p &= a_0 a_n^{n - 1} + a_1 a_n^{n - 2}(a_n X) + a_2 a_n^{n - 3} (a_n X)^2 + \dots + a_{n - 1}(a_n X)^{n - 1} + a_n a_n^{n - 1}X^n\\ &= \sum_{k = 0}^{n - 1} a_k a_n^{n - 1 - k} (a_nX)^k + (a_nX)^n. \end{align*}
If we let $Y = a_n X$ then we have a polynomial whose leading coefficient is $1$, and it is known that any rational zero of such a polynomial must be an integer; call it $z \in \mathbb Z$. We therefore have
\begin{align*} 0 &= \sum_{k = 0}^{n - 1} a_k a_n^{n - 1 - k} z^k + z^n. \end{align*}
Since $z = a_n x, x \in \mathbb Z$, this implies that $x = a_n^{-1}z$.
Assuming that this is correct, and I'm not quite sure it is, then we have found a zero of $a_n^{n - 1}p$ in terms of $z$. What I need is a zero of $p$ itself in terms of $z$.
Is this method salvageable? I appreciate any help.
(Analysis I by Amann and Escher, page 90)
