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 February 2nd, 2010, 03:15 PM #1 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Integer-valued polynomials Let $P:\mathbb{Z}\to\mathbb{Z}$ be a non-constant polynomial with integer coefficients and a leading coefficient of 1, so that $P(n)=n^r+\sum_{j=0}^{r-1}a_jn^j$ for some $a_0,\ldots,a_{r-1}\in\mathbb{Z},$ with $r\,>\,0.$ Moreover, we will say that an integer $m$ divides $P$ (as usual, denoted $m|P$) if for all $n\in\mathbb{Z},$ we have $m|P(n).$ For $m\in\mathbb{N}$, let $r_m$ denote the smallest integer $r\in\mathbb{N}$ such that there exists a polynomial $P$ as described above with $m|P,$ and $P$ is of degree $r.$ For example, $r_1=1$ since $P(n)=n$ satisfies the above description with $1|P;$ $r_2=2$ since $P(n)=n^2+n$ is always even, while $Q(n)=n+a_0$ is never even for all $n;$ $r_3=3$ since $P(n)=n^3-n$ is always divisible by 3, and this is the smallest degree for which this is possible; $r_4=4$ - consider $P(n)=n^4-n^2$ $r_6=3$ - consider $P(n)=n^3-n$ $r_8=4$ - consider $P(n)=n(n+1)(n+2)(n+3).$ Can we find the general term for $r_m?$ Clearly $r_m\leq m,$ since $m|n(n+1)\cdots(n+m-1)$ for all $m\in\mathbb{Z}.$ I suspect that we will also find that $r_p=p$ for $p$ prime. (Note that this problem is equivalent to finding the largest $s$ such that $(1,1,\ldots,1)\ ,\ (0,1,2,\ldots ,m-1)\ ,\ (0^2,1^2,2^2,\ldots,(m-1)^2)\ ,\ \ldots\ ,\ (0^{s-1},1^{s-1},2^{s-1},\ldots,\ (m-1)^{s-1})$ are linearly independent modulo $m$)
 February 3rd, 2010, 06:07 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Integer-valued polynomials These are the Kempner numbers, Sloane's A002034. Your suspicion is right regarding primes.

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