1. Prove by induction that b - a | b^n - a^n for all n >= 1.
To prove that b - a divides b^n - a^n for all n ≥ 1 by induction, we need to show two things:
1. The base case: When n = 1, we need to show that b - a divides b^1 - a^1.
2. The induction step: Assuming that b - a divides b^k - a^k for some positive integer k, we need to show that b - a divides b^(k+1) - a^(k+1).
Let's start with the base case:
When n = 1, b - a divides b - a, which is trivially true since any number divides itself. So, the base case holds.
Now, let's move on to the induction step:
Assuming that b - a divides b^k - a^k for some positive integer k, we need to show that b - a divides b^(k+1) - a^(k+1).
We can start by expanding b^(k+1) - a^(k+1) using the binomial theorem:
b^(k+1) - a^(k+1) = (b - a)(b^k + b^(k-1)a + b^(k-2)a^2 + ... + ab^(k-1) + a^k).
Now, let's look at each term in the expansion:
- The first term (b - a) is already a factor of b^(k+1) - a^(k+1).
- Each subsequent term is a product of powers of b and a, meaning it is divisible by (b - a) due to our assumption, since it is divisible by b^k - a^k.
Therefore, since each term in the expansion is divisible by (b - a), the entire expression b^(k+1) - a^(k+1) is also divisible by (b - a).
By proving the base case and the induction step, we can conclude that b - a divides b^n - a^n for all n ≥ 1.