Proof of the binomial identity $\displaystyle\binom{m}{n}=\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{1-\delta_{k,n-k}} \binom{m/2}{k} \binom{m/2}{n-k}$ Trying to prove some uncorrelated things, I came across the following identity:
$$\binom{m}{n}=\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{1-\delta_{k,n-k}} \binom{m/2}{k} \binom{m/2}{n-k}, $$
where $\delta_{i,j}$ is the Kronecker delta, equal to 1 if $i=j$ and vanishing otherwise.
This identity seems to hold for every $m$ and $n$ (I checked it with Mathematica for each pair of integers $n, m$ from 1 up to 100).
I've never seen such an identity, and it doesn't seem straightforward to prove.
Is this some known identity? And how could I go in proving (or disproving) it?
 A: It's true. The number of ways we can pick $n$ things from $m$ is:
Divide the $m$ things into two half-sized chunks. Then we need to do one of the following:


*

*pick at most $n/2$ things out of the first chunk (say we pick $k$ from the first chunk), and the remaining $n-k$ things out of the second chunk; 

*pick at most $n/2$ things out of the second chunk, and the remaining $n-k$ things out of the first chunk.


By symmetry, we take care of the second case by simply doubling all terms from the first case for which $k \not = n-k$.
A: Vandermonde Identity gives
$$\sum_{k=0}^nu_k=\sum_{k=0}^n\binom {m/2}k\binom {m/2}{n-k}=\binom mn\qquad\qquad (1) $$
If $n$ is even:
then number of terms is odd, with the middle term being the $n/2$-th, with symmetrical terms on both sides of this, as $k, n-k$ are symmetrical about $n/2$, i.e. 
$$\begin{align}
u_0&=u_n\\
u_1&=u_{n-1}\\
u_2&=u_{n-2}\\
&\vdots\\
u_{\frac n2-1}&=u_{\frac n2+1}\\
&u_{\frac n2}
\end{align}$$
Hence
$$\sum_{k=0}^n\binom {m/2}k\binom {m/2}{n-k}=\left[2^\color{lightgrey}{1-0}\sum_{k=0}^{\frac n2-1}\binom {m/2}k\binom {m/2}{n-k}\right]+\color{lightgrey}{2^{1-1}}\binom {m/2}{n/2}\binom{m/2}{n/2} \qquad\qquad(2)$$
If $n$ is odd:
then number of terms is even and there is no middle term as $n/2$ is not an integer, i.e. 
$$\begin{align}
u_0&=u_n\\
u_1&=u_{n-1}\\
u_2&=u_{n-2}\\
&\vdots\\
u_{ \big\lfloor \frac n2 \big\rfloor}&=u_{\big\lfloor \frac n2\big\rfloor +1}
\end{align}$$
Hence
$$\sum_{k=0}^n\binom {m/2}k\binom {m/2}{n-k}
=2^\color{lightgrey}{1-0}\sum_{k=0}^{\big\lfloor \frac n2 \big\rfloor}\binom {m/2}k\binom {m/2}{n-k} \qquad\qquad(3) $$
To cater for both odd and even $n$, the RHS of equations $(2),(3)$ can be expressed as 
$$\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{1-\delta_{k,n-k}} \binom{m/2}{k} \binom{m/2}{n-k}$$
As this is equal to $(1)$, we have
$$\binom{m}{n}=\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{1-\delta_{k,n-k}} \binom{m/2}{k} \binom{m/2}{n-k}\qquad\blacksquare$$
