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Four at a time (Posted on 2019-05-16) Difficulty: 3 of 5
a1a2a3a4 + a2a3a4a5 + ... + ana1a2a3 = 0 with ai = +1 or -1 for i = 1, 2, 3, ..., n. Show that n is divisible by 4.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 1 of 1
Each term in the sum is odd, which means that the sum of any four terms is even.

Let S_i be the sum of the four terms that have a_i as a member.  Then if a_i is inverted between 1 and -1 then S_i changes to -S_i.  Then the total sum changes by -2S_1.

But S_i must be even, which means that the total sum changes by a multiple of 4 when a single element is inverted.

Any possible sum can be arrived at by starting with a_i=1 for all n=1 to n and then inverting the needed a_i's.

The sum when all a_i equal 1 is precisely n. 

Inverting elements changes the sum by a multiple of 4 therefore the sum is divisible by 4 exactly when n is a multiple of 4.

  Posted by Brian Smith on 2020-05-23 17:16:00
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