# Integral Polynomial Interpolation

## Question

It's well known that given some sequence points of \( n \) points \( (x_i, y_i) \) with \( x_i \) distinct, that there exists a unique polynomial \( P \) of degree \( d < n \) such that for all \( i \), \( P(x_i) = y_i \), the **Lagrange Interpolating Polynomial**.

If all \( x_i, y_i \) are integers, we can glean from the Lagrange Interpolation Formula that it is guaranteed that \( P(X) \in \mathbb{Q}[X] \). But does there exist a polynomial \( F(X) \) of higher degree satisfying the same property (\( F(x_i) = y_i \)) such that its coefficients are all integers?

## Solution

Strangely enough there exists no such polynomial if \( P \) has some non integer coefficient. Proof is thanks to this mathoverflow answer:

First define \( D(X) \in \mathbb{Z}[X] \) as \( D(X) = \prod (X - x_i) \), ie the monic polynomial of degree \( n \) whose roots are the \( x \) coordinates of our points.

Next, suppose \( F(X) \) has coefficients all integers. Since \( D(X) \) is monic we can write:

\[ F(X) = D(X)Q(X) + R(X) \]

Where \( Q(X), R(X) \in \mathbb{Z}[X] \), and also \( deg(R(X)) < deg(D(X)) \).

Now we must have \( F(x_i) = R(x_i) \), but this implies that \( R(X) \) is the Lagrange interpolating polynomial since it has degree less than \( n \) and satisfies \( R(x_i) = P(x_i) \) for all \( i \). Thus we can re-write:

\[ P(X) = F(X) - D(X)Q(X) \]

Which implies that \( P(X) \in \mathbb{Z}[X] \) since that latter set is closed under addition and multiplication.