TITLE: Polynomial: $p(x) = p(x+3)$.
QUESTION [15 upvotes]: Determine polynomial $p(x)$ s.t. $p(x) = p(x+3)$. 

Just by looking at the above equation, it immediately appears that p has got to be some kind of constant function. I thought it might also be a periodic function but, correct me as I am wrong, polynomials are not generally periodic functions and a constant function is the closest we get to a sort of periodic function (without the period).
My problem now is figuring out how to show this. Someone hinted that I might be able to do so by considering the polynomial 
$$q(x) = p (x) - p(x+3)$$
This expands out to something like
$$q (x) = a_n(x^n - (x+3)^n) + a_{n-1}(x^{n-1}-(x+3)^{n-1})...+ 3 a_1$$
Anyway, thoughts on where to go from here are appreciated.

REPLY [4 votes]: All of the above answers are algebraic. Let's get some calculus argument:
Every non-constant polynomial has $\lim\limits_{x\to\infty} p(x)=\pm\infty$. But for a polynomial with $p(x)=p(x+3)$, you can take the sequence $x_n=x_0+3\cdot n$, which clearly advances to $\infty$ and yields $p(x_n)=p(x_{n-1}+3)=p(x_{n-1})=\ldots=p(x_0)$, which is a constant sequence. So for this polynomial $p(x)$ $\lim\limits_{x\to\infty} p(x)=\pm\infty$ is clearly wrong and the polynomial has to be constant.