7th Iberoamerican 1992 Problem B1 Let a n , b n be two sequences of integers such that: (1) a 0 = 0, b 0 = 8; (2) a n+2 = 2 a n+1 - a n + 2, b n+2 = 2 b n+1 - b n , (3) a n 2 + b n 2 is a square for n > 0. Find at least two possible values for (a 1992 , b 1992 ).
(1992·1996, 4·1992+8), (1992·1988, -4·1992+8) Solution a n satisfies a standard linear recurrence relation with general solution a n = n 2 + An + k. But a n = 0, so k = 0. Hence a n = n 2 + An. If you are not familiar with the general solution, then you can guess this solution and prove it by induction. Similarly, b n = Bn + 8. Hence a n 2 +b n 2 = n 4 + 2An 3 + (A 2 +B 2 )n 2 + 16Bn + 64. If this is a square, then looking at the constant and n 3 terms, it must be (n 2 + An + 8). Comparing the other terms, A = B = ±4. Thanks to Juan Ignacio Restrepo 7th Ibero 1992 © John Scholes jscholes@kalva.demon.co.uk 29 March 2004 Last corrected/updated 29 Mar 04