IMO 2004 SL A2

Origin: ROM | Category: Algebra

Problem

An infinite sequence a0, a1, a2, . . . of real numbers satisfies the condition an = |an+1 −an+2| for every n \geq0 with a0 and a1 positive and distinct. Can this sequence be bounded?

Solution

We claim that the sequence {an} must be unbounded. The condition of the sequence is equivalent to an > 0 and an+1 = an+an−1 or an −an−1. In particular, if an < an−1, then an+1 > max{an, an−1}. Let us remove all an such that an < an−1. The obtained sequence (bm)m\inN is strictly increasing. Thus the statement of the problem will follow if we prove that bm+1 −bm \geqbm −bm−1 for all m \geq2. Let bm+1 = an+2 for some n. Then an+2 > an+1. We distinguish two cases: (i) If an+1 > an, we have bm = an+1 and bm−1 \geqan−1 (since bm−1 is either an−1 or an). Then bm+1 −bm = an+2 −an+1 = an = an+1 − an−1 = bm −an−1 \geqbm −bm−1. (ii) If an+1 < an, we have bm = an and bm−1 \geqan−1. Consequently, bm+1−bm = an+2−an = an+1 = an−an−1 = bm−an−1 \geqbm−bm−1.