Why is $\sum\limits_{k=0}^{n}(-1)^k\binom{n}{k}^2=(-1)^{n/2}\binom{n}{n/2}$ if $n$ is even? 
Why is $\displaystyle\sum\limits_{k=0}^{n}(-1)^k\binom{n}{k}^2=(-1)^{n/2}\binom{n}{n/2}$ if $n$ is even ?

The case if $n$ is odd, is clear, since $\displaystyle(-1)^k\binom{n}{k}+(-1)^{n-k}\binom{n}{n-k}=0$ (we have $(n+1)/2$ such pairs.)
but if $n$ is even we have no symmetry, I tried to consider the odd and even terms separately but with no success. 
Do you have an idea ? Thanks in advance.
 A: $$\sum_k (-1)^k{n\choose k}^2= \sum_k (-1)^k{n\choose k}{n\choose n-k}$$ is the coefficient of $x^n$ in 
$$(1-x)^n(1+x)^n=\sum_k{(-1)^k{n\choose k}}x^k\cdot \sum_k{{n\choose k}}x^k.$$
As $(1-x)^n(1+x)^n=(1-x^2)^n$, the claim follows (and we additionally see that $\sum_k (-1)^k{n\choose k}^2=0$ if $n$ is odd).
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}^{2}
     =\pars{-1}^{n/2}{n \choose n/2}}$

$$
\mbox{Hereafter we'll use widely the identity}\quad
{s \choose \ell} = \oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{s} \over z^{\ell + 1}}
\,{\dd z \over 2\pi\ic}\tag{1}
$$

\begin{align}&\color{#c00000}{\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}^{2}}
=\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\
\overbrace{\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}}
^{\ds{\mbox{See identity}\ \pars{1}}}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}
\sum_{k = 0}^{n}{n \choose k}\pars{-\,{1 \over z}}^{k}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z}
\bracks{1 + \pars{-\,{1 \over z}}}^{n}{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{n}\oint_{\verts{z}\ =\ 1}{\pars{1 - z^{2}}^{n} \over z^{n + 1}}
{\dd z \over 2\pi\ic}
=\pars{-1}^{n}\oint_{\verts{z}\ =\ 1}{1 \over z^{n + 1}}\
\overbrace{\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}z^{2k}}
^{\ds{=\ \pars{1 - z^{2}}^{n}}}\
{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{-1}^{n}\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}
\oint_{\verts{z}\ =\ 1}{1 \over z^{\color{#00f}{\Large n\ -\ 2k}\ +\ 1}}
\,{\dd z \over 2\pi\ic}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pars{2}
\end{align}

$$
\mbox{However,}\quad
\oint_{\verts{z}\ =\ 1}{1 \over z^{\color{#00f}{\Large n\ -\ 2k}\ +\ 1}}
\,{\dd z \over 2\pi\ic}
=\left\lbrace\begin{array}{lcl}
1 & \mbox{if} & n\ \mbox{is even and}\ n = 2k
\\
0 && \mbox{otherwise}
\end{array}\right.\qquad\qquad\,\quad\pars{3}
$$

With $\pars{2}$ and $\pars{3}$, we'll find:
$$\color{#00f}{\large%
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}^{2}}
=\color{#00f}{\large\left\lbrace\begin{array}{lcl}
\pars{-1}^{n/2}{n \choose n/2} & \mbox{if} & n\ \mbox{is even}
\\
0 && \mbox{otherwise}
\end{array}\right.}
$$
A: If $n=2m$ the problem becomes 
$\displaystyle\sum\limits_{k=0}^{2m}(-1)^k\binom{2m}{k}^2=(-1)^m\binom{2m}m$
Now for non-zero finite $x,$
$\displaystyle(1+x)^n\left(1-\frac1x\right)^n=\frac{(1-x^2)^n}{x^n}(-1)^n$
Now compare the constant terms of the above identity.
Here $n=2m$
