If $f(x)$ is a polynomial and $f(n)$ is an integer for all $n\in \mathbb{Z}$, then $f(x)$ must have integer coefficients.

Question

If $f(x)$ is a polynomial and $f(n)$ is an integer for all $n\in \mathbb{Z}$, then $f(x)$ must have integer coefficients.

I need to disprove the statement via a counter-example. But I can't seem to think of one. Could $3x^2+2x^{-1}+1$ be a possible example?

Answer 1

Another family of simpe counter-examples: By Fermat's little theorem, $x^p\equiv x\pmod p$ for all $x\in\Bbb Z$ if $p$ is prime. Therefore $$\frac1px^p-\frac1px $$ maps integers to integers.

Answer 2

Are you after a polynomial $p(x)$ such that not all of its coefficients are integer but such that nevertheless you have $p(n)\in\mathbb Z$ for each integer $n$? You can take $p(x)=\frac12x^2+\frac12x$, for instance.

Answer 3

If you want something more than a quadratic try $$\frac 16(x^3-x)$$

Answer 4

It is well known that the real polynomials with the given property are exactly the integer linear combinations of polynomials of the form $\binom{x}{n}$ (that is a binomial coefficient). José Carlos Santos' answer is a very good example which is almost exactly the simplest nontrivial one with non-integer coefficients, as $\frac{1}{2}x^2+\frac{1}{2}x = \binom{x}{2}+x$.

Try to prove the above by induction on the degree, or directly, it is not so hard. Then you have many counterexamples.