How do I prove that a-b is a factor of a^k-b^k with mathematical induction?

Question

The only other information I was given is: a^(k+1)-b^(k+1)=a(a^k-b^k)+b^k(a-b). Thanks!

Answer 1

I believe you meant that (a-b) is a factor of (a^k-b^k) for k an integer . It is better to show this because if you show it is true for (a^k-b^k) you will show that it is true for all number so it will be true for a²-b² , a³-b³, ...a¹⁰⁰- b¹⁰⁰, etc. The steps for mathematical induction are:

1. Show it is true for the base case k=1
2. Assume it is true for the kth case
3. Show that is true for the (k+1)th case

Lets start!

1. Let k=1 (then (a¹-b¹) = (a-b) which is clearly a factor of (a-b) .That is because a number is clearly divisible by itself.
2. Assume (a-b) is a factor of (a^k-b^k)
3. Lastly we show that (a-b) is a factor of (a^k+1-b^k+1)

We have (a^k+1- b^k+1) = a (a^k-b^k) + b^k(a-b) . We ASSUME that (a-b) is a factor of (a^k-b^k) so it will be a factor of the first term of you given info , a(a^k-b^k) which is a multiple of (a^k-b^k). The second term in your given information, b^k(a-b) is a multiple of (a-b) since it is a number (b^k) multiplied by (a-b) . Since both terms have (a-b) as a factor the sum of both terms will have (a-b) as a factor so that proves that (a-b) is a factor of (a^k+1- b^k+1). Thus by the principles of mathematical induction (a-b) is a factor of a^k-b^k for all integers k and therefore (a-b) is a factor of (a²-b²)

Q.E.D