Problem source

By considering the coefficient of $x^{n}$ in the identity $(1-x)^{n}(1+x)^{n}=(1-x^{2})^{n},$ or otherwise, simplify 
\[
\binom{n}{0}^{2}-\binom{n}{1}^{2}+\binom{n}{2}^{2}-\binom{n}{3}^{2}+\cdots+(-1)^{n}\binom{n}{n}^{2}
\]
in the cases \textbf{(i) }when $n$ is even, \textbf{(ii) }when $n$
is odd.

Solution source

The coefficient of $x^n$ on the LHS is

\begin{align*}
 && (1-x^2)^n  &= (1-x)^n(1+x)^n \\
[x^n]: && \begin{cases} (-1)^{\lfloor \frac{n}2 \rfloor}\binom{n}{\lfloor \frac{n}2 \rfloor} &\text{if } n\text{ even} \\
0 & \text{otherwise}
\end{cases} &= \sum_{i=0}^n \underbrace{(-1)^i\binom{n}{i}}_{\text{take }(-x)^i\text{ from first bracket}} \cdot \underbrace{\binom{n}{n-i}}_{\text{take }x^{n-i}\text{ from second bracket}} \\
&&&=  \sum_{i=0}^n (-1)^i\binom{n}{i}\binom{n}{i} \\
&&&=  \sum_{i=0}^n (-1)^i\binom{n}{i}^2\\
\end{align*}