Setting $x=y$, we get that $W(x)$ is rational for all rational $x$. If it has degree $n$, it's uniquely determined by $n+1$ points and equals it's Lagrange interpolation polynomial [*]. Since we can choose these points to have both coordinates rational, the Lagrange interpolation polynomial and thus $W$ itself is rational. [*]: For given points $(a_i,b-_i)$, $i=1,2,3,...n$, it is defined via $\sum_{i=1}^n b_i \prod_{j=1, i \neq j}^n \frac{x-a_j}{a_i-a_j}$. It's an easy check that it runs through all those points $(a_i,b_i)$ and has rational coefficients only if the $a_i,b_i$ are rational.

The previous post gives a necessary, but not sufficient condition. The answer is that the only such polynomials $ W$ are of the form $ x\to ax+b$ where $ a$ and $ b$ are rational. Indeed these are the only continuous functions $ W$ from the reals to the reals such that if $ x+y$ is rational then $ W(x)+W(y)$ is rational. Proof: Fix a rational number $ q$. Then $ x\to W(x)+W(q-x)$ takes only rational values (since $ x+(q-x)=q$ is rational), and is continuous. Therefore it is constant. Thus, if $ x+y$ is rational, then $ W(x)+W(y)=W(x+y)+W(0)$. If we let $ Z(x)=W(x)-W(0)$, this says that if $ x+y$ is rational, then $ Z(x)+Z(y)=Z(x+y)$. In particular, $ Z$ is additive on the rationals. It follows by an easy induction that $ Z(nx)=nZ(x)$ for any integer $ n$ and rational number $ x$. Thus, $ nZ(\frac{m}{n})=Z(m)=mZ(1)$, so $ Z(q)=qZ(1)$ for all rational $ q$. Thus $ W(x)=W(0)+x(W(1)-W(0))$ for all $ x$ in the set of all rationals. Since the rationals are dense, this is true for all $ x$. QED Luke See my puzzle blog at http://bozzball.blogspot.com[/url]

Lagrange interpolition