Proving a theorem by Polya

I've been trying to figure out where to start with this problem and am quite lost. Any help would be appreciated.

Theorem:

Let Ck(x)= [x(x-1)(x-2)...(x-(k-1))] / k!

Then any polynomial p(x) with the property that p(n) is in **Z**, for all n in **Z**

can be written as an integer linear combination

p(x) = (summation from 1 to k) MiCi(x)

with each Mi in **Z**, i = 1,...,k

Any help or place to start would be greatly appreciated. Thanks!

Re: Proving a theorem by Polya

Seems that there may be something missing. If x = 0, then $\displaystyle \sum_{i=1}^k M_i C_i(0) = 0$ since $\displaystyle C_k(0) = 0, \forall k$. But, what if $\displaystyle p(0)\ne 0$? Is the theorem (which I tried to find online with no luck) perhaps this?:

$\displaystyle p(x) = M_0 + \sum_{i=1}^k M_i C_i(x)$

Re: Proving a theorem by Polya

Ah, yes I forgot to say that we define C_0(x) = 1.

So I think they are actually equivalent, as C_0 (x) = 1, so the first term in the sequence would just be M_0.

Re: Proving a theorem by Polya

Quote:

Originally Posted by

**Magnechu** Ah, yes I forgot to say that we define C_0(x) = 1.

So I think they are actually equivalent, as C_0 (x) = 1, so the first term in the sequence would just be M_0.

I've been trying to think about the series given above. One thing is bothering me that I don't now how to tackle. It seems that the C_k(x) vanish beyond a certain term l where x>l, which is great if you want to show that certain M_i exist. You can construct the M_i in this way by starting at p(0), then p(1) (which requires p(0), computed previously), and p(3) similarly. However, when one considers the negative integers the C_k(x) do not vanish, and also, the M_i's would be completely determined by the procedure I mentioned above (for positive integers). I'm not sure how one reconciles this to also include negative integers. I can't help you beyond this (sorry!).