<s>[INST] Hard elementary combinatorics problem How does one compute (without brute force) the smallest integer $n$ such that
$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?
 [/INST] Let's replace an explicit number $-3$ with a symbol $z$, so that we consider:
$$
   \sum_{k=1}^n z^{k-1} \binom{2n}{2k-1} = \sum_{m=0}^{2n} \left( \frac{1-(-1)^m}{2} \right)z^{(m-1)/2} \binom{2n}{m} = \frac{\left(1+\sqrt{z}\right)^{2n} - \left(1-\sqrt{z}\right)^{2n}}{2 \sqrt{z}}
$$
Now substitute $z=-3$. Notice that $1 \pm \sqrt{-3} = 2 \exp\left( \pm i \frac{\pi}{3} \right)$, so it follows that the smallest such an integer is $n=3$.
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