the original question was to show that a^n - b^n is divisible by
(a-b) for n >= 1
For this question, you will have to use strong induction.
Let P(n): (a - b) l (a^n - b^n), where n is a positive
integer.
When n = 1, a^n - b^n = a - b, and (a - b) l (a - b).
Thus, P(1) holds true.
Assume that P(n) holds true for n < k, where k is an integer
more than 1.
Now, we need to show that P(n) holds true for n = k as well.
Let us consider two scenarios: k is odd and k is even.
If k is odd, k must be of the form 2g + 1, where g is some
positive integer
a^n - b^n
= a^(2g + 1) - b^(2g + 1)
= a(a^(2g)) - b(b^(2g))
= a(a^(2g)) - a(b^(2g)) + a(b^(2g)) - b(b^(2g))
= a(a^(2g) - b^(2g)) + (a - b)(b^(2g))
= a(a^g + b^g)(a^g - b^g) + (a - b)(b^(2g))
By the induction hypothesis, a^g - b^g is divisible by (a - b),
since g is a positive integer less than k. Thus, a^g - b^g = f(a -
b), where f is some integer.
a(a^g + b^g)(a^g - b^g) + (a - b)(b^(2g))
= a(a^g + b^g)(f(a - b)) + (a - b)(b^(2g))
Since both a(a^g + b^g)(f(a - b)) and (a - b)(b^(2g)) are
both divisible by (a - b), (a^k - b^h) must be divisible by (a - b)
as well.
If k is even instead, it must be of the form 2h, where h is some
positive integer.
a^(2h) - b^(2h)
= (a^h + b^h)(a^h - b^h)
Similarly to above, h is a positive integer less than k. Thus,
by the induction hypothesis, (a^h - b^h) must be divisible by (a -
b).
We have (a^h - b^h) = e(a - b), where e is some integer.
(a^h + b^h)(a^h - b^h)
= (a^h + b^h)(e(a - b))
(a^h + b^h)(e(a - b)) is clearly divisible by (a - b), so a^k -
b^k must be divisible by (a - b) as well.
Thus, P(n) holds true for n = k, where k is an integer greater
than 1.
Hence, by strong induction, P(n) holds true for all positive
integer values of n.