I this is a terrific question. Very interesting! I'll have to delay thinking about this till later, but meanwhile it suggests a few other related questions: Suppose the nth degree polynomial P(X) = Sum_{1 <= j <= n} a_j X^j has coefficients in some nice ring S containing Z. ((( Like S = Gaussian integers Z[i] (a_j = K + L*i), or S = Eisenstein integers Z[exp(2πi/3)] (a_j = K + L*exp(2πi/3), or S = Hurwitz quaternions Z[1, i, j, k, ±1/2 ± i/2 ± j/2 ± k/2] (a_j = integer linear combinations of those 24 quaternions), or even S = integral octonions (see http://math.ucr.edu/home/baez/octonions/integers/integers_1.html). The important thing is, they're power associative, so polynomials P(X) are well-defined when evaluated on octonions. (Question to algebraists: Are these "maximal orders" ?) ))) Suppose for any one of these S we have P(Z) = Z (what Allan calls Z —> Z). When are the a_j actually in Z ? When don't they need to be? Each of these rings S can be tensored with the rationals (S tensor Q just allows a rational linear combinations where you see an integer in those linear combinations above). Then: Same question for these rings S(x)Q (S tensor Q) —Dan ----- For a polynomial P with coefficients in Z, it's trivial that P(n) is integer if n is. The converse is not true. There are lots of polynomials that always give integer answers to integer questions, but whose coefficients are not integers. For example, n(n+1)/2 = (1/2)n + (1/2)n^2 always takes integer values for integer n, even though the coefficients aren't integers. Is there a way to quickly eyeball a polynomial in general to see if it is Z -> Z? If the coefficients are rational, one can find K = the LCM of the denominators, multiply through by K, and test it for all the integers from 0 to K-1 to see if the result is always divisible by K. But I am hoping there is a simpler way. If any of the coefficients are irrational, my intuition is that the polynomial is never Z->Z, but I haven't been able to think of an easy proof. -----