Can a non-rational polynomial be rational at all integers?

Question

Is there a polynomial $f \in \mathbb{R}[X]$ such that for every $x \in \mathbb Z,\>\> f(x)$ is rational but at least one of the coefficients of $f$ is irrational?

Answer 1

No, the values $f(x)$ are rational linear combinations of irrational numbers. Let's say you have a sequence $b_i=f(x_i)$ in rational points and let be $a_i$ the coefficients of the polynomial. Therefore you can solve for the coefficients if you have enough $b_i$ (1+polynomial order). This can be written in the form $$\vec{b}=A\vec{a}$$ where $A$ is an integer valued matrix (Vandermonde matrix, consisting of just powers of $x_i$). So all you need to do is $$\vec{a}=A^{-1}\vec{b}$$ Your assumption is that $\vec{b}$ is a rational vector. $A^{-1}$ is also rational (can be seen from any of the inverse computing algorithms, for instance, Cramer's rule). So following from your assumption of rational evaluation points $x_i$ and rational polynomial values $b_i$, it follows immediately, that $a_i$ is rational.

Answer 2

No.

If there were such a polynomial of degree $n$, choose $n+1$ distinct integers $x_0,\ldots x_n$. There is only one degree $n$ polynomial going through all the points $(x_i,f(x_i))$, and that is given by the Lagrange interpolation, and we have $$ f(x)=\sum_{i=0}^n \frac{(x_1-x)\cdots(x_{i-1}-x)(x_{i+1}-x)\cdots(x_n-x)}{(x_1-x_i)\cdots(x_{i-1}-x_i)(x_{i+1}-x_i)\cdots(x_n-x_i)}f(x_i) $$ and we note that if the function values and $x_i$ are all rational, then everything in that sum is rational, and we cannot have any irrational coefficients.