IMO 1981 SL 4

Origin: CAN

Problem

Let {fn} be the Fibonacci sequence {1, 1, 2, 3, 5, . . .}. (a) Find all pairs (a, b) of real numbers such that for each n, afn + bfn+1 is a member of the sequence. (b) Find all pairs (u, v) of positive real numbers such that for each n, uf 2 n + vf 2 n+1 is a member of the sequence.

Solution

We shall use the known formula for the Fibonacci sequence fn = \sqrt (\alphan −(−1)n\alpha−n), where \alpha = 1 + \sqrt . (1) (a) Suppose that afn + bfn+1 = fkn for all n, where kn > 0 is an integer depending on n. By (1), this is equivalent to a(\alphan −(−1)n\alpha−n) + b(\alphan+1 + (−1)n\alpha−n−1) = \alphakn −(−1)kn\alpha−kn, i.e., \alphakn−n = a + b\alpha −\alpha−2n(−1)n(a −b\alpha−1 −(−\alpha)n−kn) \toa + b\alpha (2) as n \to\infty. Hence, since kn is an integer, kn −n must be constant from some point on: kn = n + k and \alphak = a + b\alpha. Then it follows from (2) that \alpha−k = a −b\alpha−1, and from (1) we conclude that afn + bfn+1 = fk+n holds for every n. Putting n = 1 and n = 2 in the previous relation and solving the obtained system of equations we get a = fk−1, b = fk. It is easy to verify that such a and b satisfy the conditions. (b) As in (a), suppose that uf 2 n + vf 2 n+1 = fln for all n. This leads to u + v\alpha2 − \sqrt 5\alphaln−2n = 2(u −v)(−1)n\alpha−2n −(u\alpha−4n + v\alpha−4n−2 + (−1)ln\sqrt 5\alpha−ln−2n) \to0, as n \to\infty. Thus u + v\alpha2 = \sqrt 5\alphaln−2n, and ln −2n = k is equal to a constant. Putting this into the above equation and multiplying by \alpha2n we get u −v \to0 as n \to\infty, i.e., u = v. Finally, substituting n = 1 and n = 2 in uf 2 n + uf 2 n+1 = fln we easily get that the only possibility is u = v = 1 and k = 1. It is easy to verify that such u and v satisfy the conditions.