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.

"I'd like to avoid using generating functions, calculus, complex numbers, trigonometric functions, chabyshev polynomials and induction"-- what is left then? Abacus? $\endgroup$"...Abacus?"-- I wish! Please see the update to the OP which I'll post shortly. But seriously, I would like to use any binomial identities that can be proven with "counting arguments", recurrence, inclusion-exclusion, bijective arguments, coefficient extraction, Riordan arrays etc... $\endgroup$Divine Proportionsby Norman Wildberger is distibuted for free at researchgate.net: Rational Trigonometry $\endgroup$1more comment