Divisibility of (a^n ± b^n) by (a ± b) for different values of 'n'

Question

I am currently studying Number Theory and have come across the general expression "(a ± b) | (a^n ± b^n)". I know that there are specific results based on the parity of 'n':

I understand the following results:

1. If n is odd, then (a+b)|(a^n + b^n) and (a-b)|(a^n - b^n).
2. If n is even, then (a-b)|(a^n - b^n) and (a+b)|(a^n - b^n).

However, I'm interested in understanding the reverse scenario. For instance, when we know that (a + b) divides (a^n + b^n), can 'n' also be an even integer, or is it restricted to only odd values? In other words, are there any even values of 'n' that satisfy (a + b) | (a^n + b^n)? If such even 'n' values exist, how can we determine them, and if not, what is the reason behind this restriction?

The way I see it, we have four different conditions:

1. (a+b)|(a^n + b^n)
2. (a+b)|(a^n - b^n)
3. (a-b)|(a^n + b^n)
4. (a-b)|(a^n - b^n)

As I mentioned earlier, except for the third condition, other conditions have some obvious answers for n. (odd or even numbers, or both for the fourth condition.) These answers don't depend on the values of a,b .

I wonder how we can find other answers for unique pairs of a,b . For example, if (a,b) = (3,1) then for any even value of n, (a-b)|(a^n + b^n) but this is not a general answer because for example that's not the case for (a,b)=(4,1).

Does such an algorithm exist that can give us the set of all integers n that satisfy each of the four conditions for each unique pair of (a,b)?

I would greatly appreciate any insights, explanations, and proofs related to the divisibility properties mentioned above. Thank you for your assistance in my understanding of this topic!

Answer 1

You may take help of remainder theorem: $R(f(x) /(x-a)) = f(a)$. If $f(a)=0$, then $(x-a)$ divides $f(x)$.

In the following let n be a natural number.

Let $f(x) =x^n-a^n, f(a) =0$, for any natural number $n$. So $(x-a)$ divides $f(x).$

Let $g(x) = x^{2n+1}+a^{2n+1}$ then $g(-a) =0.$ So $(x+a)$ divides $g(x).$

Let $h(x) =x^{2n} -a^{2n}, h(a)=0=h(-a),$ so both $(x-a)$ and $(x+a)$ are factors of $h(x).$

You can check that $k(x)=x^{2n}+a^{2n}$ does not vanish for any real value of $x$. So $k(x)$ does not have any real factor.