Ibero 00/A3 solution

15th Iberoamerican 2000

Problem A3

Find all solutions to (m + 1)^a = m^b + 1 in integers greater than 1.

Answer

(m, a, b) = (2, 2, 3).

Solution

Thanks to José Nelson Ramirez

Taking equation mod m+1 we get (-1)^b = -1, so b is odd. Hence we can divide the rhs by m+1 to get m^b-1 - m^b-2 + ... - m + 1. This has an odd number of terms. If m is odd, then each term is odd and so the total is odd, but (m+1)^a-1 is even (note that a > 1). Contradicton, so m is even.

We have m^b = (m+1)^a - 1. Expanding the rhs by the binomial theorem, and using b > 1, we see that m must divide a. So a is even also. Put a = 2A, m = 2M. We can factorise (m+1)^a - 1 as ( (m+1)^A + 1) ( (m+1)^A - 1). The two factors have difference 2, so their gcd divides 2, but both factors are even, so their gcd is exactly 2.

If M = 1 or a power of 2, then the smaller factor 3^A - 1 must be 2, so A = 1 and we have 3^A + 1 = 4, so (2M)^b = 8. Hence M = 1 and b = 3 and we have the solution (m, a, b) = (2, 2, 3).

If M is not a power of 2, then M^b > 2^b, so we must have the larger factor 2·M^b and the smaller factor 2^b-1. But the larger factor is now > 2^b+1, so the difference between the factors is at least 3·2^b-1 > 2. Contradiction.

15th Ibero 2000