Lagrange Interpolation and integer polynomials 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. 
 A: 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 $1/2$.
A: Seems to me that it might be wrong even with integers; consider the following conditions for instance:
$$P(1)=1$$
$$P(2)=1$$
$$P(3)=2$$
Then the Lagrange interpolation through the data $(1,1), (2,1), (3,2)$ gives:
$$P(X)=\frac{1}{2}X^2 - \frac{3}{2}X + 2$$
Even though $P(X)$ is an integer for every integer $X$, its coefficients are not integers.
