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Let $f(x) \in \mathbb{Q}[x]$, and suppose $f(n)$ is an integer for all large integer $n$. Prove that $f(n)$ is an integer for small positive integers $n$.

I read the answer from here is the hilbert polynomial integer-valued everywhere?, but I'm looking for a different way to prove it since I just start learning some basic algebra theories: groups, rings, fields...

I tried to look at $f(p)$, where $p$ is a prime less than some fixed integer $N$, and then consider $f(k.p)$, where $k$ is a positive integer. However, I couldn't conclude anything.

Thank you.

Sil
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dh16
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2 Answers2

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We can write $f(x)$ as $\frac{P(x)}{k}$, where $P(x)$ has integer coefficients, and $k$ is an integer.

We are told that $\frac{P(b)}{k}$ is an integer of all large enough $b$. That implies that $P(x)\equiv 0\pmod{k}$ whenever $x\equiv b\pmod{k}$, for all large enough $b$. Thus $P(x)\equiv 0\pmod{k}$ for all $x$, and therefore $k$ divides $P(x)$ for every integer $x$.

André Nicolas
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Hint $\ $ Write $\,f(x) = g(x)/d,\,\ d\in \Bbb Z,\,\ g\in \Bbb Z[x].\,$ Any nonintegral value $\,f(k) \not\in\Bbb Z\,$ implies that $\,f(x)\,$ takes nonintegral values for arbitrariy large arguments $\,x = k\!+\!nd\equiv k\pmod d\,$ since

$\quad f(k)\in\Bbb Z \!\iff\! g(k)\equiv 0\pmod d\!\iff\! g(k+\!n\!d)\equiv 0\pmod d\!\iff\! f(k\!+\!nd)\in\Bbb Z$

because, $\ {\rm mod}\ d\!:\,\ k\equiv k\!+\!nd\,\Rightarrow\, g(k)\equiv g(k\!+\!nd),\ $ by the Polynomial Congruence Rule.

Community
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Bill Dubuque
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