We want to prove that $$\sum_{k=0}^n (-1)^k{n \choose k}^2=\ \Bigg\{ \begin{array} \ \hspace{2mm}0 \hspace{20mm}\text{ if n is odd} \\ \ (-1)^{\frac{n}{2}} {n \choose \frac{n}{2}}\hspace{6mm}\text{ if n is even} \end{array}$$ The odd proof is pretty easy by creating a bijection between odd and even subsets, but I cannot figure out the even proof. Any help would be significant!
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Not my choice of dupe target, but yes, this is well-known. This is the perfect case for approach0. – darij grinberg Sep 13 '20 at 17:23
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The second answer here (by Mike Spivey) gives a very nice combinatorial argument that covers both the odd and the even case. – Brian M. Scott Sep 13 '20 at 17:37