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I've been struggling with this problem for a while, I really don't know where to start:

Let $f(x) \in \mathbb{Z}[X]$ be a polynomial such that for every value of $a \in \mathbb{Z}$, $f(a)$ is always a multiple of $101$ or $107$. Prove that $f(a)$ is always divisible by $101$ for all values of $a$, or that $f(a)$ is divisible by 107 for all values of $a$.

Zev Chonoles
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aram
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  • I'm not sure I understand your question. Are you trying to find such a polynomial? If so are there any restrictions? – john Jun 02 '13 at 07:08
  • @john: no, OP wants to show that any $f$ that satisfies the given hypotheses also satisfies the desired conclusion. Can you be specific about what you don't get? – anon Jun 02 '13 at 07:10
  • ahh, just noticed the difference in wording – john Jun 02 '13 at 07:13

3 Answers3

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If neither of the statements "$f(x)$ is always divisible by $101$" or "$f(x)$ is always divisible by $107$" is true, then there exist $a,b\in{\bf Z}$ so that $107\nmid f(a)$ and $101\nmid f(b)$. It follows from hypotheses that

$$\begin{cases} f(a)\equiv 0\bmod 101 \\ f(a)\not\equiv0\bmod 107\end{cases}\qquad \begin{cases}f(b)\not\equiv 0\bmod 101 \\ f(b)\equiv 0\bmod 107\end{cases}$$

Let $c\in{\bf Z}$ be $\equiv a\bmod 107$ and $\equiv b\bmod 101$. Is $f(c)$ divisible by $101$ or $107$?

anon
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  • $f(c)$ is divisible by neither, contradiction to the hypothesis, then its always divisible by 101 or 107 ? – aram Jun 02 '13 at 08:27
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    Contradiction to the hypothesis, meaning at least one of the two statements in quotes in my answer must hold true. – anon Jun 02 '13 at 08:31
  • sorry i meant its awlways divisible by 101 or its always divisible by 107 – aram Jun 02 '13 at 08:32
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Hint: Divisiblility of $f(x)$ by $101$ depends only on the residue class of $x \mod 101$.

Robert Israel
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Let's abstract away the details by replacing $101, 107$ with any two ideals $\mathfrak{a}$, $\mathfrak{b}$ of some ring $R$ such that $\mathfrak{a}+\mathfrak{b}=R$. The condition is: $f(R)\subset \mathfrak{a} \cup \mathfrak{b}$, and we would like to show that either $f(R)\subset \mathfrak{a}$ or $f(R)\subset \mathfrak{b}$.

Suppose for contradiction that this wasn't the case. Then pick $a,b\in R$ such that $f(a)\not\subset\mathfrak{a}$ and $f(b)\not\subset\mathfrak{b}$. But now choose any $y\in (a+\mathfrak{a})\cap(b+\mathfrak{b})$ (the intersection is non-empty by the hypothesis on $\mathfrak{a}, \mathfrak{b}$) to obtain $f(y)\not\subset \mathfrak{a}\cup\mathfrak{b}$, contradiction. $\square$

pre-kidney
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