It is known that $a\_1, a\_2, a\_3, ... , a\_n \in \left\{-1, 1 \right\}$ and $S = a\_1a\_2a\_3a\_4 + a\_2a\_3a\_4a\_5 + ... + a\_na\_1a\_2a\_3 = 0$

Prove that $n \equiv 0\space(mod\space 4)$

I know this problem can be solved using number theory, but I am looking for a solution utilizing the concept of invariance. 

I was able to identify an invariant of sorts: even if you change the sign of any $a\_i$, the sum will still remain congruent to same number $mod \space 4$. It can be proved using this invariant that no matter what values you choose for each of the $a\_i$, $S \equiv 0 \space (mod \space 4)$. It is not yet clear how I can prove the required statement using this result.