# 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.

• The answer is no if the $x_i$s are not all integers. Just take $P(x) = x^2$, $x_1 = 0, x_2 = \pi$: interpolation will give you the (linear) polynomial $\pi\cdot x$. – Marco Golla Jun 5 '14 at 8:14

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 $1/2$.
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.