12th VMO 1974 Problem A2 Find all positive integers n and b with 0 < b < 10 such that if a n is the positive integer with n digits, all of them 1, then a 2n - b a n is a square.
b = 2 works for any n b = 7 works for n = 1 Solution a n = (10 n -1)/9, so we need 10 2n - b10 n + (b-1) to be a square. For b = 2, this is true for all n. Note that (10 n - c) 2 = 10 2n - 2c10 n + c 2 , so if b = 2c, we need c 2 = b-1 = 2c - 1 and hence c = 1. So b = 4, 6, 8 do not work. Equally 10 2n > 10 2n - 10 n > (10 n - 1) 2 = 10 2n - 2·10 n + 1, so it cannot be a square for b = 1. Similarly, (10 n - 1) 2 = 10 2n - 2·10 n + 1 > 10 2n - 3·10 n + 2 > 10 2n - 4·10 n + 4 = (10 n - 2) 2 , so b = 3 does not work. Similarly, (10 n - 2) 2 = 10 2n - 4·10 n + 4 > 10 2n - 5·10 n + 4 > 10 2n - 6·10 n + 9 = (10 n - 3) 2 , so b = 5 does not work. Similarly, (10 n - 3) 2 = 10 2n - 6·10 n + 9 > 10 2n - 7·10 n + 6 > 10 2n - 8·10 n + 16 = (10 n - 4) 2 for n > 1, so b = 7 does not work, except possibly for n = 1. Since 11 - 7 = 2 2 , b = 1 does work for n = 1. Finally, (10 n - 4) 2 = 10 2n - 8·10 n + 16 > 10 2n - 9·10 n + 8 > 10 2n - 10·10 n + 25 = (10 n - 4) 2 for n > 1, so b = 9 does not work, except possibly for n = 1. It is easy to check it does not work for n = 1. 12th VMO 1974 © John Scholes jscholes@kalva.demon.co.uk 6 March 2004 Last corrected/updated 6 Mar 04