Alternating Sum of Binomial Coefficients Squared

Question

I was asked to prove that $\sum_{k = 0}^n (-1)^k \binom{n}{k}^2 = (-1)^m \binom{2m}{m}$, when $n = 2m$, and $0$ when n is odd. I have seen other posts on this but I wasn't able to grasp this concept. I'm looking for a direct proof of the identity. I have tried using $(1 + X)^n (1 - X)^2 = (1 - X^2)^n$, and also brute force writing out the sum to try to cancel terms, but neither has quite got me to the final identity yet any help would be appreciated.

Answer 1

\begin{eqnarray*} \sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 &=& [x^k]: \sum_{k=0}^{n} (-1)^k \binom{n}{k} (1+x)^n \\ &=& [x^n]: \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{n-k} (1+x)^n \\ &=& [x^n]: (x-1)^n (1+x)^n =[x^n]: (x^2-1)^n. \end{eqnarray*} It is clear that this zero if $n$ is odd. And if $n=2m$ then let $y=x^2$ and we want \begin{eqnarray*} [y^m]: (y-1)^{2m}. \end{eqnarray*}

Edit : Explanation of notation ...

$[x^k]:f(x)$ means the coefficient of $x^n$ in the function $f(x)$. So for instance \begin{eqnarray*} \binom{n}{k}=[x^k]: (1+x)^n. \end{eqnarray*}