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Let $f(x)\in \mathbb{Z}[x]$ and let $\frac{r}{s} \in \mathbb{Q}$ be a root of $f(x)$ with $(r,s) =1$. Show that $r-ms$ is a divisor of $f(m)$ for any integer $m$.

The book’s suggestion is: Write $f(x)$ in the form $\sum a_r(x-m)^r$. Show that every $a_r$ is an integer. Then, evaluate $f(x)$ in $r/s$.

I don’t see how $f(x)$ could be written in that form so without the book’s suggestion I have $$f(x)=(x-r/s)(g(x))$$ for some $g(x)\in\mathbb{Z}$ which implies $g(x)=(-1/s)(h(x))$ if $(r-ms)\mid f(m)$. I also tried by dividing $\frac{f(x)}{(x-r/s)}$ but still get no proof of this assumption. Thanks in advance.

Note: $f(x)=a_nx^n+\cdots a_1x + a_0$, $s\mid a_n$ and $r\mid a_0$

Octavio Berlanga
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2 Answers2

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The fact that $f(x)$ can be written in the form $\sum_{k=0}^d a_k (x-m)^k$ is due to long division; see here for instance.

Then in this form, we obtain $0=f(r/s) = \sum_{k=0}^d a_k (r/s-m)^k$ which implies $0 = \sum_{k=0}^d a_k s^{d-k} (r - sm)^k$ and thus $$-\sum_{k=1}^d a_k s^{d-k} (r-sm)^k = s^d a_0 = s^d f(m).$$

angryavian
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Easier: $ $ we can pull out the factor $\ sx-r\ $ using a slight generalization of polynomial division.

By nonmonic division: $\ s^kf(x) = (sx-r)g(x)\ $ for $\,g \in \Bbb Z[x],\ k\in \Bbb N$

Evaluating at $\,x\! =\! m\!:\ s^k f(m) = (sm-r)g(m)$

So $\, (s^k,sm\!-\!r)= 1,\ sm\!-\!r\mid s^k f(m)\,\Rightarrow\, sm\!-\!r\mid f(m)\,$ by Euclid's Lemma,

via $\ (s,sm\!-\!r) = (s,r) = 1$

Bill Dubuque
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