Question regarding the proof of $\sum_{k=0}^{n} (-1)^k \binom{n}{k}^{2}$ Let $n$ be a positive integer.  Prove that
$$\sum_{k=0}^{n} (-1)^k \binom{n}{k}^{2} = \begin{cases}
0   & \text{if $n$ is odd} \\
(-1)^m \binom{2m}{m}, & \text{if $n$ = 2m}
\end{cases}$$
So I wrote out a first few terms to get a feel of the problem
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$  
When n is odd (eg n =3): $$\binom{3}{0}^2 -\binom{3}{1}^2 +\binom{3}{2}^2 -\binom{3}{3}^2 = 0$$
When n is even (eg n = 4): $$\binom{4}{0}^2 - \binom{4}{1}^2 + \binom{4}{2}^2 - \binom{4}{3}^2 + \binom{4}{4}^2 = \binom{4}{2}^2 = \binom{2(2)}{2}^2 (-1)^2$$
So there is a hint that the textbook gave me, it is

For $n = 2m$, consider the coefficient of $x^n$ in $(1-x^2)^n = (1 +
 x)^n(1-x)^n$

So I tried writing out what $(1+x)^n$ and $(1-x)^n$ are 
$$(1+x)^n = \sum_{k=0}^{n} \binom {n}{k} x^k$$ 
$$(1-x)^n = \sum_{k=0}^{n} \binom {n}{k} (-1)^k x^k$$
And this is where I've been lost for the last few hours....

If you have polynomials $f(x)=\sum_k a_kx^k$ and $g(x)=\sum_k b_kx^k$
then the $x^n$-coefficient of $f(x)g(x)$ is
$$c_n=a_0b_n+a_1b_{n-1}+\cdots+a_nb_0=\sum_{k=0}^n a_{k}b_{n-k}.$$
Here $a_k=\binom nk$ and $b_k=(-1)^k\binom nk$, and here
$$c_n=\sum_{k=0}^n (-1)^{n-k}\binom nk\binom{n}{n-k}.$$
Can you relate this to the sum you're studying?
