Allan, The theorem is as follows: for a positive integer n define f_n(x) = x(x-1)...(x-n+1)/n!, and f_0(x) = 1. These are clearly is integer valued at all integers. Every polynomial with rational coefficients which is integer valued at all integers is an linear combination of the f_n(x) with integer coefficients. The forward direction is obvious. To get the converse, use the difference operater: (Delta g)(x) = g(x+1) - g(x). We have (Delta f_n)(x) = ((x+1) x (x-1) ...(x-n+2) - x(x-1) ...(x-n+1))/n! = (x+1 - (x-n+1)) x (x-1) ... (x-n+2))/n! = f_{n-1}(x). So (Delta^j f_n)(0) = 1 if j=n and 0 otherwise. If g(x) is an integer valued polynomial, then so is (Delta g)(x) Since every polynomial with rational coefficients can be written as sum_j a[j] * f_j(x), applying Delta^r and evaluating at 0, yields a[r]. On Thu, Nov 15, 2018 at 6:08 PM Allan Wechsler <acwacw@gmail.com> wrote:
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. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun