IMO 1977 LL ROM36

Origin: ROM

Problem

Consider a sequence of numbers (a1, a2, . . . , a2n). Define the operation S((a1, a2, . . . , a2n)) = (a1a2, a2a3, . . . , a2n−1a2n, a2na1). Prove that whatever the sequence (a1, a2, . . . , a2n) is, with ai \in{−1, 1} for i = 1, 2, . . . , 2n, after finitely many applications of the operation we get the sequence (1, 1, . . ., 1).

Solution

It can be shown by simple induction that Sm(a1, . . . , a2n) = (b1, . . . , b2n), where bk = m

i=0 a(m i ) k+i (assuming that ak+2n = ak). If we take m = 2n all the binomial coefficients
m i  apart from i = 0 and i = m will be even, and thus bk = akak+m = 1 for all k.