Show that any rational zero of a polynomial $p\in\Bbb Z[X]$ with $\deg(p)\ge1$ belongs to $a_n^{-1}\Bbb Z$ (hint: consider $a_n^{n-1}p$.)
Can you check my proof please? I am "almost sure" that it is correct but still I need some confirmation or some comment about how to write it better. Thank you.
Let $q\in\Bbb Q$ a zero of $p$ and we will define $z:=a_n^{-1}q$, then we can rewrite the polynomial as
$$\sum_{k=0}^na_kq^k=0\iff z^n+\sum_{k=0}^{n-1}a_n^{n-1-k}a_kz^k=0\iff z^n=\sum_{k=0}^{n-1}c_kz^k\tag{1}$$
with $c_k=-a_ka_n^{n-1-k}$. If $z\in\Bbb Q\setminus\Bbb Z$ then we can write $z:=\frac{p}{q}$ where $p,q$ are coprime. Then rewriting $(1)$ we get
$$\frac{p^n}{q^n}=\sum c_k \frac{p^k}{q^k}\iff p^n=q^n\sum c_k\frac{p^k}{q^k}\tag{2}$$
where $p,q,c_k\in\Bbb Z$. Then $(2)$ to hold must be the case that $\frac{c_k}{q^k}\in\Bbb Z$ because $p^n,q^n\in\Bbb Z$ but if this is the case we have that $q$ and $p$ are not coprimes, what is a contradiction.
But if $z\in\Bbb Z$ we have that $z^n=\sum c_kz^k$ where for appropriate $c_k$ it is certainly possible, by example take the case
$$z=1\quad\text{and}\quad c_k=(-1)^{n-k}\binom{n}{k}\quad\text{and}\quad a_n=1$$
Then because $z=a_n^{-1}q$ we can conclude that if $q\in\Bbb Q$ is a zero of $p$ then $q\in a_n^{-1}\Bbb Z$.$\Box$
