36th Eötvös 1932

Show that if m is a multiple of aⁿ, then (a + 1)^m -1 is a multiple of aⁿ⁺¹.

Solution

We use induction on n. For n = 0 we have to show that (a+1)^m - 1 is a multiple of a. That is obvious if we expand by the binomial theorem. So suppose it is true for n. Put m = am', where m' is divisible by aⁿ. For convenience we also put A = a+1. Then we have A^m - 1 = (A^m' - 1)(A^m'(a-1) + A^m'(a-2) + ... + A^m' + 1). The first bracket is divisible by aⁿ⁺¹ by induction. The second bracket has a terms, so adding a and subtracting a 1s we can write it as a + (A^m'(a-1) - 1) + (A^m'(a-2) - 1) + ... + (A^m' - 1). Now each term is divisible by a and hence the sum is divisible by a. So A^m - 1 is divisible by aⁿ⁺² as required.

36th Eötvös 1932

© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03