Linear Factor of Polynomial with Integer Coefficients

Question

I am trying to prove the following: if $f(x) \in \mathbb{Z}[x]$, and $m/t \in \mathbb{Q}$ is a root of $f(x)$, where $gcd(m,t)=1$, then $m-ct \mid f(c)$, for all $c \in \mathbb{Z}$.

What I have tried so far: let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. If $m/t$ is a root of $f(x)$, clearing denominators we obtain $$a_nm^n+a_{n-1}m^{n-1}t+\cdots+a_1mt^{n-1}+a_0t^n=0 $$ Also, $$a_n(c^nt^n)+a_{n-1}t(c^{n-1}t^{n-1})+\cdots+a_1t^{n-1}ct+a_0t^n=f(c)t^n$$ Subtracting the two, we obtain

$$f(c)t^n=a_n((ct)^n-m^n)+a_{n-1}t((ct)^{n-1}-m^{n-1})+a_1t^{n-1}(ct-m) $$

Of course $ct-m \mid (ct)^i-m^i$, for all $1 \leq i \leq n$. Thus, $ct-m \mid f(c)t^n$. But from there I can't conclude. I also tried to use the Rational Root Theorem, but I have not been able to use it in any meaningful way. Any hints?

Answer 1

Since $t$ and $m$ are coprime, $ct-m$ and $t^n$ have no common factors (if $p$ is a common prime factor, then it divides $t$ from $t^n$ and hence $m$ from $ct-m$, contradiction). Thus $ct-m$ divides $f(c)$, as you wanted.