IMO 1983 SL 7

Origin: CAN

Problem

Let a be a positive integer and let {an} be defined by a0 = 0 and an+1 = (an + 1)a + (a + 1)an + 2 a(a + 1)an(an + 1) (n = 1, 2 . . .). Show that for each positive integer n, an is a positive integer.

Solution

Clearly, each an is positive and \sqrtan+1 = \sqrtan \sqrta + 1+\sqrtan + 1\sqrta. Notice that \sqrtan+1 + 1 = \sqrta + 1\sqrtan + 1 + \sqrta\sqrtan. Therefore ( \sqrt a + 1 −\sqrta)(\sqrtan + 1 −\sqrtan) = ( \sqrt a + 1 \sqrt an + 1 + \sqrta\sqrtan) −(\sqrtan \sqrt a + 1 + \sqrt an + 1\sqrta)

an+1 + 1 −\sqrtan+1. By induction, \sqrtan+1−\sqrtan =
\sqrta + 1 −\sqrta n. Similarly, \sqrtan+1+\sqrtan =
\sqrta + 1 + \sqrta n. Hence, \sqrtan = 1 
\sqrt a + 1 + \sqrta n −
\sqrt a + 1 −\sqrta n , from which the result follows.