IMO 1994 SL A1

Origin: USA | Category: Algebra

Problem

Let a0 = 1994 and an+1 = a2 n an+1 for each nonnegative integer n. Prove that 1994 −n is the greatest integer less than or equal to an, 0 \leqn \leq998.

Solution

Obviously a0 > a1 > a2 > \cdot \cdot \cdot . Since ak −ak+1 = 1 − ak+1, we have an = a0 +(a1 −a0)+\cdot \cdot \cdot+(an −an−1) = 1994−n+ a0+1 +\cdot \cdot \cdot+ an−1+1 > 1994 −n. Also, for 1 \leqn \leq998, a0 + 1 + \cdot \cdot \cdot + an−1 + 1 < n an−1 + 1 < a997 + 1 < 1 because as above, a997 > 997. Hence \lflooran\rfloor= 1994 −n.