Find all positive integers n and b with 0 < b < 10 such that if an is the positive integer with n digits, all of them 1, then a2n - b an is a square.
Answer
b = 2 works for any n
b = 7 works for n = 1
Solution
an = (10n-1)/9, so we need 102n - b10n + (b-1) to be a square. For b = 2, this is true for all n. Note that (10n - c)2 = 102n - 2c10n + c2, so if b = 2c, we need c2 = b-1 = 2c - 1 and hence c = 1. So b = 4, 6, 8 do not work. Equally 102n > 102n - 10n > (10n - 1)2 = 102n - 2·10n + 1, so it cannot be a square for b = 1. Similarly, (10n - 1)2 = 102n - 2·10n + 1 > 102n - 3·10n + 2 > 102n - 4·10n + 4 = (10n - 2)2, so b = 3 does not work. Similarly, (10n - 2)2 = 102n - 4·10n + 4 > 102n - 5·10n + 4 > 102n - 6·10n + 9 = (10n - 3)2, so b = 5 does not work. Similarly, (10n - 3)2 = 102n - 6·10n + 9 > 102n - 7·10n + 6 > 102n - 8·10n + 16 = (10n - 4)2 for n > 1, so b = 7 does not work, except possibly for n = 1. Since 11 - 7 = 22, b = 1 does work for n = 1. Finally, (10n - 4)2 = 102n - 8·10n + 16 > 102n - 9·10n + 8 > 102n - 10·10n + 25 = (10n - 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.
© John Scholes
jscholes@kalva.demon.co.uk
6 March 2004
Last corrected/updated 6 Mar 04