# Thread: prove f is a rational polynomial

1. ## prove f is a rational polynomial

$\displaystyle f(x)$ is polynomial with complex coefficients. $\displaystyle \forall n\in Z$, $\displaystyle f(n)$is integer, prove: coefficients of $\displaystyle f(x)$ are rational numbers, and give some examples about rational case.

### Prove:

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* consider coefficients are integers, of course $\displaystyle f(n)$ are integers.

* consider coefficients are rationals, we have $\displaystyle f(x)=\frac{1}{2}x(x+1)$, two consecutive integer can be divided by $\displaystyle 2$, there must be one even number.

* real coeffs

* complex coeffs

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And can you give me some more examples?

2. ## Re: prove f is a rational polynomial

The following ideas are based on this article (PDF). It turns out that every integer-valued polynomial p(x) has the form

$\displaystyle p(x) = a_0+a_1\binom{x}{1}+ \dots+ a_n\binom{x}{n}$ (*)

Indeed, the coefficients (omitting the constant term, which is zero) of polynomials $\displaystyle \binom{x}{1}$, $\displaystyle \binom{x}{2}$, ..., $\displaystyle \binom{x}{n}$ form a triangular matrix, so these polynomials form a basis, i.e., every polynomial of degree ≤ n with no constant term can be expressed as a linear combination of these polynomials, perhaps with complex coefficients. So, every polynomials from ℂ[x] of degree ≤ n can be written in the form (*) with complex $\displaystyle a_k$. Next prove by induction on k that $\displaystyle a_k\in\mathbb{Z}$.