Let $n$ be a positive integer and $N=\displaystyle\sum_{k=0}^{n}(-1)^k\binom{n}{k}^2$. Show that $N=0$ if $n$ is odd, and $N=(-1)^{\frac{n}{2}}\displaystyle\binom{n}{\frac{n}{2}}$ if $n$ is even.
Using binomial expansion can give $$(1-x^2)^n=\sum_{k=0}^{n}(-1)^k\binom{n}{k}x^{2k}$$
Meanwhile $$(1-x^2)^n=(1-x)^n(1+x)^n=\left(\sum_{k=0}^{n}\binom{n}{k}x^k\right)\left(\sum_{k=0}^{n}(-1)^k\binom{n}{k}x^k\right)$$
The coefficient of the $x^n$ term in the second expansion is always $N=\displaystyle\sum_{k=0}^{n}(-1)^k\binom{n}{k}^2$. However, the first expansion does not have any term with odd power to $x$. Therefore, when $n$ is odd, it must hold that $N=0$. Meanwhile, when $n$ is even, the coefficient of the $x^n$ term in the first expansion equals $(-1)^{\frac{n}{2}}\displaystyle\binom{n}{\frac{n}{2}}$.
Thus, the claim holds.