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Let $a_1,a_2,\cdots,a_n$ be $n$ numbers such that $a_i$ is either $1$ or $-1$. If $$a_1a_2a_3a_4+a_2a_3a_4a_5+\cdots+a_na_1a_2a_3=0$$ then prove that $4 \mid n$.

My work:
By multiplying all the terms, we get,
$$a_1^4a_2^4\ldots a_n^4=1.$$

I think that I will be able to represent $4n$ a power of $1$, but getting no clue. Please help! I also think that this problem can be done with invariance and extremal principal too. Please help with these approaches too!

Batominovski
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Hawk
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  • I am curious how many sequences $(a_1,a_2,\ldots,a_n)\in{-1,+1}^n$ are there with the required property for a given integer $n>4$ with $4\mid n$. – Batominovski Jul 08 '20 at 18:53

5 Answers5

4

Hint 1 Each term you add is either $+1$ or $-1$. Since they add up to $0$, it must be an even number of terms. This tells you that $n$ is even.

[If this is not clear, what happens if you add an one to each particular term].

Hint 2 You know already that $n$ is even, and that $\frac{n}{2}$ terms (not a's) have to be $1$ and the other half must be $-1$.

In order for $a_ia_{i+1}a_{i+2}a_{i+3}=1$ an even number of $a_i, a_{i+1}, a_{i+2}, a_{i+3}$ must be $-1$.

In order for $a_ia_{i+1}a_{i+2}a_{i+3}=-1$ an odd number of $a_i, a_{i+1}, a_{i+2}, a_{i+3}$ must be $-1$.

So in total, the number of $-1$ which appears as $a_i$ in your expression have the same parity as $\frac{n}{2}$.

How many times does each $a_i$ appear? Can you finish the problem from here?

N. S.
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  • I think that I can do it from here. But,the concept is still foggy for me. Can you please elaborate the hints without providing the solution? – Hawk Jan 12 '14 at 07:38
4

All summands are either $1$ or $-1$. Since their sum is equal to $0$ the number of summands $1$ is equal to the number of summands $-1$. The product of all summands is $1$. This means that the number of summands $-1$ is even.

WimC
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4

$$(-1)^x \cdot 1^{n-x}=a_1^4a_2^4\ldots a_n^4=1$$ As a result, $$2\mid x.$$

Since $$a_1a_2a_3a_4+a_2a_3a_4a_5+\cdots+a_na_1a_2a_3=0$$ we have, $x = n-x$

Thus $$4\mid n.$$

Martin Sleziak
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lennon310
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1

Another approach with realisation. The sum of four consecutive summands can be $0,\pm2,\pm4$. If we change the sign of one $a_i$, then the signs of all these summands will change, the other summands will saved. Therefore, one such change either does not change the sum or changes it to $4$ or $8$. From the sequence $++\ldots +$ with the sum $n$ we can go to any other with the help of several such changes, so the modified sum will be $n + 4k$. It can be equal to zero only if $n\equiv0\pmod 4.$

Realisation. Evidently for $n=4$ there is no example.

$n=8: A=++++----$

$n=12: B=++++-+-+----$

$n\equiv0\pmod 8: AA\ldots A$

$n\equiv4\pmod 8: BAA\ldots A$

Martin Sleziak
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Minz
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0

Consider changes in $S=a_1a_2a_3a_4+\cdots+a_na_1a_2a_3$ under the transformations $a_i \mapsto -a_i$.

$\bullet$ If one or three $a_i$'s have the same sign, then $S \mapsto S \pm 4$.

$\bullet$ If two of the $a_i$'s are positive and two negative, then $S \mapsto S$.

$\bullet$ If all four $a_i$'s have the same sign, then $S \mapsto S \pm 8$.

Since $S=0$ initially, after each transformation $4 \mid S$. Applying the transformation $a_i \mapsto -a_i$ whenever $a_i=-1$ leads to the sum $S=n$. Thus, $4 \mid n$. $\blacksquare$

AT1089
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