Suppose that there is a polynomial $P$ with integer coefficients such that $P(x\_i)=y\_i$ for $i=1,\ldots,n$. Is it true that the result of Lagrange interpolation through the data $(x\_i,y\_i)$ is a polynomial with integer coefficients?
Suppose there are $2$ cases :


$\bullet$ $x\_i$ are integers


$\bullet$ $x\_i$ are not all integers


What are the conclusions? Note $\deg P$ is not specified. 



Lagrange Interpolation and integer polynomials
The answer is yes if the $x\_i$ are integers. Write
$$
B(X):=\prod\_{i=1}^n (X-x\_i),
$$
so that $B(X)$ is a monic polynomial in $\mathbf{Z}[X]$. For any $P(X)\in\mathbf{Z}[X]$, we can write
$$
P(X) = B(X) Q(X) + R(X)
$$
where $Q,R\in\mathbf{Z}[X]$ and $\deg(R)<n$. Here $R(x\_i)=P(x\_i)=y\_i$ for every $i$. Since there is a unique polynomial $f(X)\in\mathbf{Q}[X]$ of degree less than $n$ which satisfies $f(x\_i)=y\_i$ for $i=1,2,\dots,n$, and both $R(X)$ and the Lagrange interpolation polynomial have the properties required of $f(X)$, it follows that the Lagrange interpolation polynomial equals $R(X)$ and hence has integer coefficients.


The answer is no if the $x\_i$ are not integers. For instance put $x\_1=y\_1=1/2$ (and $n=1$), then $P(X):=X$ satisfies $P(x\_1)=y\_1$, but the Lagrange interpolation polynomial is $1/2$.


