Let $X$ be a projective scheme over algebraically closed field $k$, $L$ is invertible sheaf on $X$ and $\mathcal F \in \operatorname{Coh}(X)$ , we define Hilbert polynomial $P_{\mathcal F}(n)=\chi(\mathcal F \otimes L^n).$ Is there any way to prove that $P$ is polynomial not using RiemannRoch and Chern classes. I think that there should be some argument similar to mdeland's argument(Why is the Euler characteristic of powers of a line bundle a polynomial in the power?), which works for locally free $\mathcal F$.

2$\begingroup$ Why should something be proved without using very simple and standard techniques? $\endgroup$ – Alex Degtyarev Jan 14 '15 at 20:01

$\begingroup$ Do you know how to compute the Hilbert polynomial of a projective variety, say as outlined in Hartshorne I.7? $\endgroup$ – Ben Lim Jan 14 '15 at 20:07

$\begingroup$ @BenLim Yes, I'm able to prove it in the case of very ample $L$. $\endgroup$ – David Jan 14 '15 at 21:33
Prove that $\chi (\mathcal F \otimes L_1^{n} \otimes L_2^m)$ is a polynomial function of $n$ and $m$ of degree $d$ when $L_1$ and $L_2$ are very ample. You can do this by exactly the same induction argument. (If you have a function of two variables whose successive differences in both variables are degree $d1$ polynomials, then it is a degree $d$ polynomial.)
Then write your divisor as a difference of two ample divisors and set $m=n$.