Question involving the sum $\sum_{k=0}^n(-1)^k\binom nk^2$ I've been tasked to prove the following equation is true:

$$\sum_{k=0}^n(-1)^k\binom nk^2=\begin{cases}0&\text{if }n\ \text{is odd}\\\displaystyle(-1)^m\binom{2m}m&\text{if }n=2m,m\in\mathbb Z^+\end{cases}$$

I pretty much have no clue where to start on this question so if you guys have any hints or pointers on how to solve this, that would be great. I'm assuming that a lot of terms will cancel out because the first case is $0$, and the second case looks like the "middle" term of the summation.
 A: Besides the approach suggested in the comments, you could try to analyze it combinatorially. Note that
$$\sum_{k=0}^n(-1)^k\binom{n}k^2=\sum_{k=0}^n(-1)^k\binom{n}k\binom{n}{n-k}\,.\tag{1}$$
Suppose that you have a pool of $n$ women and $n$ men, and from this pool you want to choose a team of $n$ people. There are $\binom{n}k\binom{n}{n-k}$ ways to choose a team with $k$ women and $n-k$ men, so $(1)$ is counting each possible team that has an even number of women positively and each team that has an odd number of women negatively. That is, that sum is the number of possible teams with an even number of women minus the number with an odd number of women. You’re suppose to show that this is $0$ when $n$ is odd and $(-1)^m\binom{2m}m$ when $n=2m$.
HINT: When $n$ is odd, try to pair up each possible team that has an even number of women with one that has an odd number of women. When $n$ is even, use the same basic idea to pair up each possible team that has a minority of women with one that has a majority of women.
A: $$(1+X)^n(1-X)^n=\left(1-X^2\right)^{n}=\sum_{m=0}^{n}\binom{n}{m}(-1)^mX^{2m}=\sum_{m=0}^{2n}a_mX^m$$
where $a_m=0$ if $m$ is odd and $a_m=(-1)^{m/2}\binom{n}{m/2}$ if $m$ is even. But
$$ (1+X)^n(1-X)^n=\left(\sum_{k=0}^n\binom{n}{k}X^k\right)\left(\sum_{k=0}^n\binom{n}{k}(-1)^kX^k\right)=\sum_{m=0}^{2n}\left(\sum_{k=0}^m (-1)^k\binom{n}{k}\binom{n}{m-k}\right)X^m $$
The coefficients before $X^n$ are the same in the two expressions, we thus have, using $\binom{n}{n-k}=\binom{n}{k}$,
$$ a_n=\sum_{k=0}^n(-1)^k\binom{n}{k}^2 $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}^{2} & =
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{n \choose n - k} =
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\bracks{z^{n - k}}\pars{1 + z}^{n}
\\[5mm] & =
\bracks{z^{n}}\pars{1 + z}^{n}\sum_{k = 0}^{n}{n \choose k}\pars{-z}^{k} =
\bracks{z^{n}}\pars{1 + z}^{n}\pars{1 - z}^{n}
\\[5mm] & =
\bracks{z^{n}}\pars{1 - z^{2}}^{n} =
\bbx{\bracks{n\ \mbox{even}}\pars{-1}^{n/2}{n \choose n/2}}
\\ &
\end{align}
