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I'm pretty sure this question has a duplicate, but since I can't seem to find it, I'm asking it again.

If $a_{1},a_{2}...a_{n}$ are $1$ or $-1$ and if,$$a_{1}a_{2}a_{3}a_{4}+a_{2}a_{3}a_{4}a_{5}+...+a_{n}a_{1}a_{2}a_{3}=0$$prove that $n$ is divisible by $4$.

What I've done so far is manage to prove that $n$ is even, which is pretty obvious. I haven't come up with any workable leads so far, so I'd be much obliged if you could give me some pointers.

sayantankhan
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1 Answers1

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Here's a nice way to do this problem, which relies on the following invariant: if one of the numbers, say $a_1$, is replaced by $-a_1$, then the sum remains the same modulo $4$ (namely $0$ modulo $4$). Given that this invariant is correct, replace each $+1$ with a $-1$ in the list of the $a_i$. Then the sum is $n$, which equals $0$ modulo $4$ by the invariant property.

Why does the invariant work? When we replace $a_1$ by $-a_1$ four of the terms of the sum change. There are three cases.

  1. If two of these four terms are positive and two negative then the sum does not change.
  2. If three of these four terms have the same sign then the sum changes by $\pm 4$.
  3. If four of these four terms have the same sign then the sum changes by $\pm 8$.

In every case the sum remains invariant modulo $4$.

user134824
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  • That is brilliant! I had in fact done the same thing, that is, replace an arbitrary a by -a, but I hadn't thought of checking modulo 4. Thank you so much for your insight. – sayantankhan May 07 '14 at 17:09
  • This idea isn't original; it comes from Arthur Engel's excellent book Problem-Solving Strategies. I highly recommend you read it. – user134824 May 07 '14 at 18:49