5
$\begingroup$

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...

$\endgroup$
4
  • $\begingroup$ 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 $\endgroup$ Jan 10 at 13:25
  • $\begingroup$ Note that since $a_i^2 =1$, you can rewrite this as $a_2/a_1 + \ldots + a_n/a_{n-1} + a_1/a_n =0$ $\endgroup$ Jan 10 at 13:25
  • 1
    $\begingroup$ @AlbusDumbledore yes $\endgroup$
    – Tran Tu
    Jan 10 at 13:26
  • $\begingroup$ Strongly related: math.stackexchange.com/questions/3019953/… $\endgroup$ Jan 10 at 19:04
4
$\begingroup$

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$.

$\endgroup$
2
  • 2
    $\begingroup$ On the other hand, if $4 \mid n$ by taking the pattern $+, +, -, -$ you will get 0 :) $\endgroup$ Jan 10 at 13:38
  • 1
    $\begingroup$ @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.) $\endgroup$
    – Calvin Lin
    Jan 10 at 13:41
8
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ Nice(+1) it is ingenious i must say! $\endgroup$ Jan 10 at 13:33
  • 2
    $\begingroup$ Oh, I thought the question was asking for the $28,30$ cases equal to $0$ separately...? $\endgroup$
    – Derek Luna
    Jan 10 at 13:35
  • $\begingroup$ @DerekLuna so did i see my comment and the reply by OP $\endgroup$ Jan 10 at 13:36
  • $\begingroup$ @AlbusDumbledore yep, after I already wrote the comment here :). $\endgroup$
    – Derek Luna
    Jan 10 at 13:36
  • $\begingroup$ It is a fascinating answer but the 28 and 30 cases are separate. It is my fault for bad phrasing $\endgroup$
    – Tran Tu
    Jan 10 at 14:00
1
$\begingroup$

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 $$ n = 4m + 4 = 4(m+1). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.