When do Co-Primes have a common factor?

Question

If a and b are co-primes and $n$ is a prime then prove that $\frac{a^n + b^n}{a+b}$ and $(a+b)$ have no common factors unless $(a+b)$ is a multiple of $n.$

I am unable to proceed, please help.

Answer 1

For $n=2$, the problem is unclear because $\frac{a^2 + b^2}{a+b}$ is not always an integer.

Suppose that $n$ is an odd prime and that $\frac{a^n + b^n}{a+b}$ and $(a+b)$ have some prime common factor $p$.

$$p \mid a+b\tag{1}$$ $$p \mid {a^n+b^n \over a + b}\tag{2}$$

Note that:

$$p \nmid a,\quad p\nmid b\tag{3}$$

(If $p \mid a$ and $p \mid a+b$ then $p\mid b$ and $a$ and $b$ would not be coprime)

For $n$ being an odd prime, you can write (3) as:

$$p|\sum_{i=0}^{n-1}(-1)^i a^{n-1-i}b^i\tag{4}$$

From (2) we have:

$$a+b=kp\implies a=kp-b\tag{5}$$

...for some $k\in N$

Replace (5) into (4):

$$p|\sum_{i=0}^{n-1}(-1)^i(kp-b)^ib^{n-1-i}\tag{6}$$

Because $p\mid kp$, (6) is equivalent to:

$$p|\sum_{i=0}^{n-1}(-1)^i(-b)^ib^{n-1-i}$$

$$p|\sum_{i=0}^{n-1}(-1)^{2i}b^{n-1}$$

$$p|\sum_{i=0}^{n-1}b^{n-1}$$

$$p|nb^{n-1}$$

But $p\nmid b$ and $p\nmid b^{n-1}$so it must be that $p|n$. But $n$ is a prime so this actually means that $n=p$ and $n\mid(a+b)$.