4th Balkan 1987 Problem 2 Find all real numbers x ≥ y ≥ 1 such that √(x - 1) + √(y - 1) and √(x + 1) + √(y + 1) are consecutive integers. Solution
: there is a unique solution for each n > 3. Namely, x = √(1+ ( (7+4k-4k 2 )/(4-8k) ) 2 ), y = √(1+ ( (7-4k-4k 2 )/(4+8k) ) 2 ), where k = √( (2n-7)/(8n+4) ). This gives the integers n and n+1. Put f(x) = √(x+1) - √(x-1). Then f(x) is strictly monotonic decreasing. f(4 1/16) = 1/2 and f(x) tends to 0 as x tends to infinity. So since x ≥ y, we have f(x) ≤ 1/2 and f(y) ≥ 1/2. For any k in the range 0 ≤ k ≤ 1/2, there is a unique x such that f(x) = 1/2 - k. Indeed we may find this explicitly. Squaring etc gives x = √(1+ ( (7+4k-4k 2 )/(4-8k) ) 2 ). Then since √(x - 1) + √(y - 1) and √(y - 1) + √(y + 1) are consecutive integers we must have f(y) = 1/2 + k. Solving we get y = √(1+ ( (7-4k-4k 2 )/(4+8k) ) 2 ). Now √(x - 1) + √(y - 1) = (4k 2 + 7)/(2 - 8k 2 ). This must be some integer n, so k 2 = (2n - 7)/(8n + 4). Clearly k 2 must be non- negative, so n > 3. 4th Balkan 1987 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002 Last corrected/updated 19 Oct 03