Find $P(-\frac{1}{2})$ given a infinite points of a polynomial The problem
Given $P(x)$ a polynomial with real coefficients and satisfies
\begin{align} P(n)= 1^{2010} +2^{2010}+ \cdots + n^{2010}, \forall n \in \mathbb{N*}
\end{align},
then calculate the value of $P(-\frac{1}{2})$.
Attempts
The first flash that came to my mind was that I have to use the Lagrange Interpolation, which I assume should apply for P(1) to P(2012). Note that using $\lim_{n \to + \infty}$ it's very easy to prove that $degP =2011$.
But then the normal binomial does not appear ( precisely, only the odd factors appear).
Then what should I do now? Any help is appreciated
 A: The below is a simple argument which works under the assumption the polynomial $P(x)$ with given property $P(n)=1^{p} +2^{p}+ \cdots + n^{p}$ exists (here $p=2010$, but we really only need that $p$ is even). (Without such assumption we could show it exists using the argument similar to derive  Faulhaber polynomials, but as currently stated we are given the polynomial $P(x)$)
Since $P(n)-P(n-1)=n^{p}$ for infinitely many values of $n$ (positive integers), and $P(x)$ is a polynomial, we must have equality of polynomials, i.e. $P(x)-P(x-1)=x^p$ for all $x$ real. This is because $P(x)-P(x-1)-x^p$ is a polynomial with infinitely many zeroes, hence must a be a zero polynomial.
Using $P(x-1)=P(x)-x^p$ we can get values of $P(x)$ at negative integers. For example $P(0)=P(1)-1=0, P(-1)=0, P(-2)=-1, P(-3)=-1-2^p$ and so on (note $(-n)^p=n^p$ since $p$ is even). It is not hard to extend the above by induction into
\begin{align}
P(-n)&=-1-2^p-\dots-(n-1)^p=-(1+2^p+\dots+(n-1)^p)=-P(n-1).
\end{align}
As before, this holds for infinitely many values so it must be equality of polynomials $P(-x)=-P(x-1)$ for all $x$ real. Now just set $x=\frac{1}{2}$ to get
$$
P\Big(-\frac{1}{2}\Big)=-P\Big(\frac{1}{2}-1\Big)=-P\Big(-\frac{1}{2}\Big)
$$
and hence $P(-\frac{1}{2})=0$.
Note: The above implies also $x(x+1)(2x+1) \mid P(x)$ for even $p$, and similarly $x(x+1) \mid P(x)$ for odd $p$.
