Show that if there are ai = ±1 such that a1a2a3a4 + a2a3a4a5 + ... + ana1a2a3 = 0, then n is divisible by 4.
Solution
Suppose we change the sign of ai. Then exactly four terms change sign. If they are initially + + + +, then the net change in the sum is - 8. Similarly, if they are initially + + + - (in some order), the net change is -4, if they are + + - -, the net change is nil, if they are - - - +, the net change is +4, and if they are - - - -, the net change is +8. So the change is always 0 mod 4. But by changing the sign of at most n terms we get all ai positive and hence sum n. So if the sum was initially 0, then we must have n = 0 mod 4.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002