Alternating binomial-square sum
= Alternating binomial-square sum
Coefficient extraction from $(1-x)^n(1+x)^n=(1-x^2)^n$ gives
$$
\sum_{r=0}^n(-1)^r\binom nr^2
=\begin{cases}0,&n\text{ odd},\\(-1)^{n/2}\binom n{n/2},&n\text{ even}.
\end{cases}
$$
= Alternating binomial-square sum
Coefficient extraction from $(1-x)^n(1+x)^n=(1-x^2)^n$ gives
$$
\sum_{r=0}^n(-1)^r\binom nr^2
=\begin{cases}0,&n\text{ odd},\\(-1)^{n/2}\binom n{n/2},&n\text{ even}.
\end{cases}
$$