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Let $a_1,a_2…a_n$ be n integers each of them being either 1 or -1.If $a_1.a_2.a_3.a_4+a_2.a_3.a_4.a_5+… + a_n.a_1.a_2.a_3=0$ then show that$ 4|n$

My solution so far: I am aware of parity arguments using them along with consider number of terms in above sum is $n-4$ then half of them must be -1 other half must be +1 it is clear that $2|n$ further I called such consecutive 4 integer pairing as quadruple and at one point in quadruple in the ones where integers overlap there must be a sign change so I showed that for some$ k$ $a_k= -a_{k+4}$ but I am not able to go any further.I suspect this problem is from some math olympiad.

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    $\begingroup$ Think about what you get if you multiply all those products of four terms together. $\endgroup$ – Gerry Myerson Nov 12 '17 at 6:35
  • $\begingroup$ If $n$ was of the form $4k+2$ then $2k + 1$ terms in the sum will be negative. However, every negative term should contain odd number of negative numbers. Only two cases are possible, a single negative number or three negative numbers. Thus $2|2k+1$ leads to contradiction. $\endgroup$ – B2VSi Nov 12 '17 at 6:49
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Let $T = \{a_1a_2a_3a_4,a_2a_3a_4a_5,...,a_{n}a_1a_2a_3\}$, that is, it contains each of the terms summing to zero above. ($T$ is not a set, it is a multiset : it may contain elements more than once).

So, $|T| = n$, and $\sum_{x \in T} x = 0$. Moreover, each element of $T$ is either $1$ or $-1$, since each term is the product of $a_i$s, each of which is $1$ or $-1$.

Note that the number of terms which are $1$ and those which are $-1$, are equal. This is because, their sum is $0$, so each term which is $1$, must cancel each term that is $-1$, so the number of these terms muse be equal.

So, $n=2k$, where $k$ is the number of terms which evaluate to $-1$.

Now, we take the product of all elements in $T$: so this is $a_1a_2a_3a_4 \times a_2a_3a_4a_5 \times ... \times a_na_1a_2a_3$. In this, every $a_i$ comes four times. So we can rewrite this product as $\prod_{i \leq n} a_i^4$. However, since each $a_i$ is $1$ or $-1$, the above product is equal to $1$.

Now, let us divide the set of terms in $T$ into two parts $T_1$ and $T_{-1}$ : $T_1$ contains those terms which evaluate to $1$ ,and $T_{-1}$ contain those terms evaluating to $-1$. Note that $1 = \prod_{i \leq n} a_i^4 = \prod_{t \in T_1} t \times \prod_{t \in T_{-1}} t$. Further, $|T_1| = |T_{-1}| = k$.

Note that every element of $T_{1}$ is $1$, so the product of all these terms is going to be $1$. Therefore, the product of the terms in $T_{-1}$ is also going to be $1$. But then, each term in $T_{-1}$ is $-1$, so $(-1)^k = 1$. Therefore, $k$ is going to be an even number.

$n = 2k$, and $k$ is an even number. Now, you should be able to conclude.

If you have any questions, please ask them clearly, so that I can clarify them without having to rewrite the answer completely.

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