Integral interpolation by polynomial  This question arises from a discussion with my friends on a commonly encountered IQ test questions: "What's the next number in this series 2,6,12,20,...". Here a "number" usually means an integer. I was wondering whether there is a systematical way to solve such problems.Let us call a point on a plane integer point if all the components of it are integers. I want to know the following:
Give a finite set of integer points, Can we always find a corresponding polynomial that passes all these points and maps integers to integers? 
 A: Let the points be $(x_j, y_j), j=1\ldots n$.  If the $x_j$ are consecutive, the Lagrange interpolating polynomial will take integers to integers: easy proof by induction, using the difference operator $\Delta(p)(x) = p(x+1) - p(x)$.  If not, choose arbitrary integers for the $y$ values to fill in the gaps.   
A: Just to make a few comments:
1) As noted, If we have a list of values $a_0,a_1,\cdots, a_n$ of integers then there is a (unique) polynomial $f(t)$ of degree no more than $n$ with integer coefficients which maps integers to integers and such that $f(k)=k$ for $0 \le k \le n$.
2) There is a method involving the differences and differences of the differences etc. which reveals $f(n)$ as an integer linear combination of the polynomials $\binom{t}{j}$ for $0 \le j \le n$. Furthermore, the polynomials of this form are exactly the polynomials sending integers to integers. These (specialized) Newton Polynomials are very similar to the Taylor series which uses the basis $\frac{t^k}{k!}$ 
3) If you just want the next term (as predicted by this polynomial) then you don't need to explicitly find the polynomial, just extend the differences. Many test takers realize this.
$$\begin{matrix}2&\ &6&\ &12&\ &20&\ &\mathbf{30}\\\ &4&&6&&8&&\mathbf{10}&\\\ &&2&&2&&\mathbf{2}&&\end{matrix}$$ corresponds to $f(n)=2+4n+2\binom{n}{2}=n^2+3n+2$
4) There is also a polynomial of degree 3 that gives $2,6,12,20,\mathbf{2011}$ so there is no unique extension.
5) If the given sequence is $1,2,4,8,?$ then the polynomial interpolation gives $15$ next from $\binom n0+\binom n1+\binom n2+\binom n3$ although most tests would favor another continuation.
A: It's a topic I liked to cover when I was still teaching junior-level Algebra, even if it didn't fit in well with the other topics. You start with a function defined on the set of integers from $0$ to $n$ inclusive, and end with a polynomial of degree $\le n$ that agrees at those $n+1$ points, and if the values you started with are integers, your function always sends integers to integers. You take successive differences, as indicated by Robert Israel above, and then you list $f(0)$, $\Delta f(0)$, $\Delta^2f(0)$, up to $\Delta^n$, and use these as coefficients, which you multiply to $C_0(x)=1$, $C_1(x)=x$, $C_2(x)=x(x-1)/2$, etc., the binomial polynomials. Your assignment is to try it out for a few examples, and then prove that the method works.
