19th Balkan 2002 Problem 2 The sequence a n is defined by a 1 = 20, a 2 = 30, a n+1 = 3a n - a n-1 . Find all n for which 5a n+1 a n + 1 is a square. Solution
: n = 3, 1 + 5 a 3 a 4 = 63001 = 251 2 . We show that 1 + 5a n+1 a n = (a n + a n+1 ) 2 + 501. Induction on n. It is true for n = 1. Suppose it is true for n-1. Then we have a n+1 = 3a n - a n-1 . Hence 5a n = 2a n + a n+1 + a n-1 . Multiplying by (a n+1 - a n-1 ) gives 5a n a n+1 - 5a n-1 a n = 2a n a n+1 - 2a n-1 a n + a n+1 2 - a n-1 2 . Adding to 1 + 5a n-1 a n = (a n-1 + a n ) 2 + 501 gives 1 + 5a n a n+1 = (a n + a n+1 ) 2 + 501, which completes the induction. The difference between m 2 and (m+1) 2 is 2m+1. So m 2 + 501 cannot be a square for m > 250. A trivial induction shows that a n+1 > a n and a 4 = 180, so for n ≥ 4, we have a n + a n+1 > 360. Thus the only possible solutions are n = 1, 2 or 3. But for n = 1 we have 3001, which is not a square and for n = 2 we have 10501 which is not a square. 19th Balkan 2002 © John Scholes jscholes@kalva.demon.co.uk 16 June 2002