proofs for combinatorial identities? by using the identity $(1-x^2)^n=(1+x)^n(1-x)^n$, show that for each $m \in \Bbb N$ with $m < n$,
summation of 
$$\sum_{i=0}^n(-1)^i\binom{n}i^2=\begin{cases}
0,&\text{if }n\text{ is odd}\\
(-1)^{n/2}\binom{n}{n/2},&\text{if }n\text{ is even}\;.
\end{cases}$$
that is summation of (−1)^i (n taken i )(n taken 2m−i) i=0 to 2m is (−1)^m (n taken m) if n is even and Summation of (−1)^i (n taken i )(n taken 2m+1−i) i from 0 to 2m+1 is 0 if n is odd.
please explain and how to prove the two cases.
 A: Consider $(1+x)^n (1-x)^n \equiv (1-x^2)^n$, and compare coefficient of $x^n$ term.
$$\begin{align}
\text{Left hand side} =& \sum_{i=0}^{n} \binom{n}{n-i} \left[\binom{n}{i} \left( -1 \right)^{i} \right] &\text{(From the expanded terms)}\\
=& \sum_{i=0}^{n} \binom{n}{i}^2 \left( -1 \right)^{i}\\
\end{align}$$
For right hand side, you can easily find the coefficient of $x^n$ from the binomial expansion
$$\left( 1-x^2 \right)^n = \sum_{i=0}^n \binom{n}{i} \left( -x^2 \right)^i$$
A: First,
$$
\begin{align}
&(1+x)^n(1-x)^n\\
&=\left(\sum_{k=0}^n\binom{n}{k}x^k\right)
\left(\sum_{j=0}^n\binom{n}{j}(-1)^jx^j\right)&&\text{binomial theorem}\\
&=\sum_{m=0}^{2n}\left(\sum_{i=0}^n\binom{n}{m-i}\binom{n}{i}(-1)^i\right)x^m&&\text{power series convolution}\\
&=\sum_{m=0}^n\left(\sum_{i=0}^n\binom{n}{2m-i}\binom{n}{i}(-1)^i\right)x^{2m}&&\text{even terms}\tag{1a}\\
&+\sum_{m=0}^{n-1}\left(\sum_{i=0}^n\binom{n}{2m+1-i}\binom{n}{i}(-1)^i\right)x^{2m+1}&&\text{odd terms}\tag{1b}
\end{align}
$$
Next, use $(1+x)(1-x)=1-x^2$ :
$$
\begin{align}
\left(1-x^2\right)^n
&=\sum_{m=0}^n\binom{n}{m}(-1)^mx^{2m}&&\text{binomial theorem}\tag{2}
\end{align}
$$
Just compare the coefficients from the even and odd terms $(1)$ to those from the $(2)$ to get
$$
\sum_{i=0}^n\binom{n}{2m-i}\binom{n}{i}(-1)^i=(-1)^m\binom{n}{m}\tag{3}
$$
and
$$
\sum_{i=0}^n\binom{n}{2m+1-i}\binom{n}{i}(-1)^i=0\tag{4}
$$
If $n$ is even, use $(3)$ and set $m=\dfrac n2$. If $n$ is odd, use $(4)$ and set $m=\dfrac{n-1}{2}$.
