How to prove that $p-mq \mid f(m)$ where $m$ is any integer,
$$f(x) = A_0 + A_1 x + A_2 x^2 + ... + A_{n-1} x^{n-1} + A_n x^n$$
$$f(x)∈ ℤ[x]$$
$p/q$ is a zero for $f(x)$ and $p$ and $q$ are coprime $⇒ \, ∃u,w ∈ ℤ \mid 1 = up + wq$.
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suppose $f(x)=a+bx$ can you do??? – Feb 27 '15 at 14:18
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Maybe, if I do induction for n. That would be the base case. – Ronald Becerra Feb 27 '15 at 14:21
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do that first... then let me know if you get some idea for other $n$ – Feb 27 '15 at 14:22
2 Answers
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By nonmonic polynomial division algorithm $\, q^n f = (qx\!-\!p) g + k\,$ for $\,g\in\Bbb Z[x],\ n\in\Bbb N,k\in\Bbb Z.$ Evaluating at $\:x=p/q\:$ yields $\:k=0.\ $ Evaluating at $\:x =m\:$ yields $\:\color{#c00}{qm\!-\!p\mid q^n}\color{#0a0}{f(m)}.\,$ But $\,(qm\!-\!p,q)=(p,q)\!=\!1,\,$ so by Euclid $\,(\color{#c00}{qm\!-\!p,q^n})=1,\,$ therefore $\,qm\!-\!p\mid \color{#0a0}{f(m)}.$
Bill Dubuque
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Note first that for any $p,q$, $q^nf\bigl(\frac pq\bigr)$ is an integer. Now if $\frac pq$ is a root of $f$, \begin{align*} q^nf(m)=q^nf(m)-q^nf\Bigl(\frac pq\Bigr)&=\sum_{k=0}^n A_k q^{n-k}\Bigl(m^kq^k - p^k\Bigr)\\ &=\sum_{k=0}^n A_k q^{n-k}(mq-p)\Bigl((mq)^{k-1}+(mq)^{k-1}p+\dots+p^{k-1}\Bigr) \end{align*} which proves $q^nf(m)$ is divisible by $p-mq$. As $p$ and $q$ are coprime, $p-mq$ and $q^n$ are coprime too, so that, by Gauß's lemma, $p-mq$ divides $f(m)$.
Bernard
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Instead of [ Akq^(n-k)(m^k*q^k-p^k) ], I get [ Ak(m^k-q^(n-k)p^k)] – Ronald Becerra Feb 27 '15 at 15:30
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That comes from $f(x)=\sum_{k=0}^n A_kx^k$, replacing $x$ with $\dfrac pq$ and cancelling the denominators with (part of) $q^n$. – Bernard Feb 27 '15 at 15:30
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You should get $A^k(m^k q^n-q^{n-k}p^k)$, then you can factor out $q^{n-k}$. – Bernard Feb 27 '15 at 15:35
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Oops! I wondered what was the problem, and forgot to check that side of the formula! Thanks for pointing it! Proof completed. – Bernard Feb 27 '15 at 23:23
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@Ronald I added a proof of the nonmonic polynomial division algorithm to help out with problems like this - see the link in my answer. – Bill Dubuque Mar 02 '15 at 22:46
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