# Can $S_n=a_1a_2+a_2a_3+\cdots+a_na_1=0$ if $a_i=±1$ for $n=28$ and $n=30$?

Let $$S_n=a_1a_2+a_2a_3+\cdots+a_na_1$$ If $$a_i=\pm 1$$, can $$S_{28}=S_{30}=0$$?

My approach: Start from small $$n$$ if we can see a pattern

For $$n=1$$:

$$S_1=a_1=±1 \neq0$$

For $$n=2$$:

$$S_2=a_1a_2+a_2a_1=2a_1a_2=±2 \neq0$$

For $$n=3$$:

$$S_3=a_1a_2+a_2a_3+a_3a_1$$

Because $$a_1a_2$$, $$a_2a_3$$ and $$a_3a_1$$ are odd numbers and $$0$$ is an even number, $$S_3$$ cannot be $$0$$. (this can be applied to other $$S_n$$ if $$n$$ is an odd number)

For $$n=4$$:

$$S_4=a_1a_2+a_2a_3+a_3a_4+a_4a_1=(a_1+a_3)(a_2+a_4)$$

However, I can't do the same with $$n=6$$ and above...

• Do the conditions have to hold simultaneously, i mean the a_1,a_2...used in S_28 have to be same while in S_30 Jan 10 at 13:25
• Note that since $a_i^2 =1$, you can rewrite this as $a_2/a_1 + \ldots + a_n/a_{n-1} + a_1/a_n =0$ Jan 10 at 13:25
• @AlbusDumbledore yes Jan 10 at 13:26
• Strongly related: math.stackexchange.com/questions/3019953/… Jan 10 at 19:04

Hint: Show that if $$S_n = 0$$, then $$4 \mid n$$.
Your work supports this hypothesis, so prove it.

Corollary: $$S_{30}$$ is never 0.

Clearly we need $$2 \mid n$$. (Your work hints at this strongly.)

Suppose that there are $$a$$ terms of the form $$+1$$ and $$b$$ terms of the form $$-1$$.
Then, $$n = a + b$$, and $$0 = a - b$$.

What is $$1^a (-1)^b$$ in 2 different ways?
Thus, show that $$b$$ is even, so $$4 \mid n$$.

• On the other hand, if $4 \mid n$ by taking the pattern $+, +, -, -$ you will get 0 :) Jan 10 at 13:38
• @AndreaMarino Right, this was meant to guide towards the much more general case of any $n$, whereas cosmos' solution works mainly because the difference between $n$ is 2. IE Solution exists iff $4 \mid n$ for all the given values. The "and only if" follows from your observation (and we have a lot of patterns, in fact, any 2 + 2 - sequence works.) Jan 10 at 13:41

Write $$S_{30}=S_{28} -a_{28}a_1+a_{28}a_{29}+a_{29}a_{30}+a_{30}a_{1}$$

If $$S_{30}=0=S_{28}$$,

$$0=-a_{28}a_1+a_{28}a_{29}+a_{29}a_{30}+a_{30}a_{1}$$

$$a_{28}(a_{1}-a_{29})=(a_{29}+a_{1})a_{30}$$

But one of $$(a_{1}-a_{29})$$, $$(a_{29}+a_{1})$$ is zero, while other is not.

Hence impossible.

• Nice(+1) it is ingenious i must say! Jan 10 at 13:33
• Oh, I thought the question was asking for the $28,30$ cases equal to $0$ separately...? Jan 10 at 13:35
• @DerekLuna so did i see my comment and the reply by OP Jan 10 at 13:36
• @AlbusDumbledore yep, after I already wrote the comment here :). Jan 10 at 13:36
• It is a fascinating answer but the 28 and 30 cases are separate. It is my fault for bad phrasing Jan 10 at 14:00

I would like to suggest one more approach to show that existence of the solution implies $$4|n$$. Define $$x_i = a_i a_{i+1}$$, obviously $$x_i = \pm 1$$. For any choice of signs of $$x_i,\dots,x_{n-1}$$ you can always find appropriate $$a_1,\dots,a_n$$. Now what about the last term $$a_n a_1$$? See that $$x_1 x_2 \dots x_{n-1} = a_1 a_2^2\dots a_{n-1}^2 a_n = a_1 a_n$$ and obviously $$x_1 \dots x_{n-1} = \pm 1$$. Now the problem reads $$x_1 + \dots + x_{n-1} + x_1 x_2 \dots x_{n-1} = 0.$$ Solution exists if and only if ($$x_1+ \dots + x_{n-1} = 1$$ and $$x_1 \dots x_{n-1} = -1$$) or ($$x_1+ \dots + x_{n-1} = -1$$ and $$x_1 \dots x_{n-1} = 1$$). Obviously, considering only one case suffices. Suppose, $$x_1 \dots x_{n-1} = -1$$. This means, we have odd number of $$-1$$-s, say $$2m+1$$. Then the number of $$+1$$-s is $$n - 1 - (2m + 1)$$. The latter means $$1 = x_1+ \dots + x_{n-1} = [n - 1 - (2m + 1)] - (2m + 1) = n - 3 - 4m,$$ thus $$n = 4m + 4 = 4(m+1).$$