There are numbers $$a_{1}, a_{2}, a_{3}, ....,a_n$$ each of them being +1 or -1. If it is known that $$a_{1}a_{2} + a_{2}a_{3} + a_{3}a_{4} +⋯+ a_{n−1}a_{n} + a_{n}a_1= 0$$ then

The possible products for a pair of integers, all of which are chosen from either $$+1$$ or $$-1$$, are:

• $$+1$$ and $$+1$$ having the product $$+1$$
• $$+1$$ and $$-1$$ having the product $$-1$$
• $$-1$$ and $$-1$$ having the product $$+1$$

Therefore, the only way that many such pairs have the sum of their products as $$0$$ is that there are as many products of $$+1$$s as there are $$-1$$s. From $$a_1a_2$$ to $$a_{n}a_1$$ there are a total of $$n$$ pairs. Since these $$n$$ number of pairs have $$\frac{n}{2}$$ pairs having the product $$+1$$ and $$\frac{n}{2}$$ pairs having the product $$-1$$, we can definitively say that $$n$$ has to be divisible by $$2$$ for $$\frac{n}{2}$$ to be an integer. 

Furthermore, the pairs are constructed cyclicly, we start with $$a_1$$ and end in $$a_1$$. Each term, therefore, affects the value of another. There have to be $$\frac{n}{2}$$ pairs that have the product $$-1$$, and these have the two terms as $$+1$$ and $$-1$$ in some order. There will also be $$\frac{n}{2}$$ pairs having the product $$+1$$, since these products are also given by adjacent terms, the terms having the values $$+1$$ and $$+1$$ will occur in pairs and those having the value $$-1$$ and $$-1$$ will occur in pairs. The number of terms, therefore, will further be divisible by 2. 

In essence:

• There is the same number of pairs having the product $$+1$$ as the number of those that have the product $$-1$$.
• For each product $$+1$$, the adjacent terms occur in pairs of $$(+1,+1)$$ or $$(-1,-1)$$; these are then arranged just so, so that there are adjacent products of $$-1$$ possible.

Therefore, $$n$$ should be twice-divisible by $$2$$, or that $$n$$ should be divisible by $$4$$. $$n$$ can therefore be any multiple of $$4$$.