Thank you, Victor, Dan, and Jeffrey. That is a satisfying answer to a question that I think has been stewing in my brain since I first learned the general formula for triangular numbers. On Thu, Nov 15, 2018 at 7:24 PM Jeffrey Shallit <shallit@uwaterloo.ca> wrote:
A very old theorem is that a polynomial has the property you want iff it is an integer linear combination of the binomial coefficient polynomials (x choose n). That property can be tested easily by solving a linear system.
On 11/15/18 6:06 PM, Allan Wechsler 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