Binomial Coefficient Identity, Double Series, Floor Function I came across this (supposed) identity for natural numbers $m$ and $n$: $$ \sum_{i=0}^{n}{ \sum_{j=0}^{m}{ \left(-1 \right)^{\lfloor \frac{i}{2} \rfloor+j}2^{n-i}\binom{n-i+\lfloor \frac{i}{2} \rfloor}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor-j}{\lfloor \frac{i}{2} \rfloor}\binom{i}{m-j}}}= \left(-1 \right)^{m} \binom{2n+1}{2m+1} $$
and would like to prove it.
Cross posting from MSE after getting no replies. 
I am trying to keep a certain tone to my work so I am looking for a human, non-analytic, combinatorial or algebraic proof to the above. To be more specific, I'd like to avoid using generating functions, calculus, complex numbers, trigonometric functions, chabyshev polynomials and induction; every other technique would do.
Update #1: tested to be true for $n,m \leq100$
Update #2: I would like to give the reason as to why I would like to avoid using the above techniques.  I am trying to generlize this identity (12): $$ \left(\sin \left(nx \right) \right)^{2}= \left(\sin \left(x \right) \right)^{2} \left(\sum_{m=0}^{\frac{n-1}{2}}{\left(-1 \right)^{m} \binom{n}{2m+1}} \left( \cos^{2}x\right)^{\frac{n-1}{2}-m} \left( \sin^{2}x\right)^{m}\right)^{2}$$
for an odd natural number $n$ and a number $x$ to a general field, using the framework of Rational Trigonometry and its Spread Polynomials. I want to show that the latter identity is essentially a polynomial or an "arithmetical" identity, meaning that it is independent of the framework of the classical trigonometric functions and the framework of complex numbers, and that the problem of proving said identity is essentially a (rather complicated) problem of arithmetic or of counting. In this endeavor, I would like to keep a "theme" of not invoking techniques which regard the classical trigonometric functions, complex numbers or calculus. The binomial coefficient identity I wanted to prove came up in the process of this work.
 A: Notice that
$$\binom{n-i+\lfloor \frac{i}{2} \rfloor}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor-j}{\lfloor \frac{i}{2} \rfloor} = \binom{n-i}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor}{\lfloor \frac{i}{2} \rfloor}.$$
This lets us easily sum up the terms depending on $j$:
$$\sum_{j=0}^m (-1)^j  \binom{n-i}{j}  \binom{i}{m-j} = [x^m]\ (1-x)^{n-i}(1+x)^i = [x^{2m}]\ (1-x^2)^{n-i}(1+x^2)^i.$$
(We will see later why dealing with squares of $x$ is preferable.)
So, it remains to evaluate the following sum:
$$\sum_{i=0}^n (-1)^{\lfloor \frac{i}{2} \rfloor} (2-2x^2)^{n-i} (1+x^2)^i
\binom{n-i+\lfloor \frac{i}{2} \rfloor}{\lfloor \frac{i}{2} \rfloor}.$$
Let $i=2s+t$ where $s=\lfloor \frac{i}{2} \rfloor$ and $t=i\bmod 2$. Then the sum becomes 
$$\sum_{s\geq 0} \sum_{t=0}^1 (-1)^{s} (2-2x^2)^{n-2s-t} (1+x^2)^{2s+t}
\binom{n-s-t}{s}$$
$$=\sum_{t=0}^1 \frac{(1+x^2)^{2n-t}}{(2-2x^2)^{n-t}} (-1)^{n-t} \sum_{s\geq 0} \binom{n-s-t}{s} \left(-\frac{4(1-x^2)^2}{(1+x^2)^2}\right)^{n-s-t}$$
$$(\star)\quad = \sum_{t=0}^1 \frac{(1+x^2)^{2n-t}}{(2-2x^2)^{n-t}} (-1)^{n-t} \frac{(1+x^2)^2}{4(x-x^3)}\Im\left( \frac{2(1-x^2)}{(I-x)^2}\right)^{n-t+1}$$
$$=\Im \sum_{t=0}^1 \frac{(1+x^2)^{2n-t+2}}{2x} (-1)^{n-t} (I-x)^{-2(n-t+1)}$$
$$=-\Im \sum_{t=0}^1 \frac{(1-Ix)^{2n-t+2}(1+Ix)^t}{2x}$$
$$=-\Im \frac{(1-Ix)^{2n+1}}{x},$$
where $I$ is the imaginary unit ($I^2 = -1$) and $\Im X$ denotes the imaginary part of $X$.
Now we are ready to take the coefficient of $x^{2m}$:
$$[x^{2m}]\ -\Im \frac{(1-Ix)^{2n+1}}{x} = -\Im \binom{2n+1}{2m+1} (-I)^{2m+1} = (-1)^m\binom{2n+1}{2m+1}$$
as expected.

UPDATE. To clarify step $(\star)$:
$$\sum_{p\geq 0} \binom{p}{r-p} u^p = \sum_{p\geq 0} [z^{r-p}]\ (1+z)^p u^p = [z^r]\ \sum_{p\geq 0} (z(1+z)u)^p $$
$$= [z^r]\ \frac{1}{1-z(1+z)u} =[z^r]\ \frac{1}{(z_1-z_2)z}\left(\frac{1}{1-z_1z}-\frac{1}{1-z_2z}\right) = \frac{z_1^{r+1}-z_2^{r+1}}{z_1-z_2},$$
where $z_{1,2} = \frac{u\pm\sqrt{u^2+4u}}{2}$ are the reciprocals of the zeros of $f(z)=1-z(1+z)u$, so that $f(z)=(1-z_1z)(1-z_2z)$. When $u^2+4u=-v^2$ like in the case $(\star)$, we further have
$$\frac{z_1^{r+1}-z_2^{r+1}}{z_1-z_2} = \frac{2}{v}\Im \left(\frac{u+Iv}{2}\right)^{r+1}.$$
