If $a$ and $b$ are prime to each other and $n$ is prime then$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor. Question is:

If $a$ and $b$ are prime to each other and $n$ is prime then prove that $\frac{a^n+b^n}{a+b}$ and
  $a+b$ have no common factor unless $a+b$ is a multiple of $n$.

This is what I thought so far:
 $a^n$ is also prime to $b^n$ because of the theorem (If $a$ is prime to $b$ then $a^n$ is also prime to $b^n$), Also $a^n$ and $b^n$ is a divisor of $(a+b)$ therefore by theorem (If $a$ is prime to $b$, and each of these numbers is a divisors of $N$, then $ab$ is a divisor of $N$) $a^nb^n$ is a divisor of $(a+b)$.
Thereafter no idea, How should I proceed further if I'm thinking in a right way and If not, How should I prove this?
Source: Higher Algebra by Barnard and Child
 A: First, when is $\frac {a^n + b^n}{a+b}$ an integer.
Or when does $(a+b)|(a^n + b^n)$
$n$ is odd
$\frac {a^n + b^n}{a+b} = a^{n-1} - a^{n-2}b + a^{n-3}b^2 -\cdots ab^{n-2} + b^{n-1}$
$n$ is even:
Apply the euclidean algorithm.
$(a+b)|(a^n + b^n) \iff  (a+b)| (a^n + b^n)+k(a+b)$
$(a^n + b^n) - a^{n-1} (a+b) + a^{n-2}b (a+b)\cdots$
$(a^n+b^n) - (a+b)\sum_\limits{i=0}^{n-1} (-1)^i a^{n-1-i} b^i = 2b^n$
$(a+b)|(a^n + b^n) \iff (a+b)|2b^n$
$(a+b) = 2,$  or $a = 0$ or $b = 0$ or $a|b^n$
since $a,b$ are co-prime.
$a\ne0, b\ne0$ and $a$ does not divide $b$
$a = b = 1$ is a degenerate case.
In this case, 2 does not divide 1.
n is odd:
again apply the Euclidean algorithm.
$\sum_\limits{i=0}^{n-1} (-1)^i a^{n-1-i}b^i + (a+b)\sum_\limits{i=1}^{n-1} (-1)^i ia^{n-1-i}b^{i-1} = nb^{n-1}$
$(a+b)|\frac {a^n + b^n}{a+b} \implies (a+b)|nb^{n-1}$
as dicussed above $(a+b)$ does not divide $b^{n-1}$
$(a+b)|\frac {a^n + b^n}{a+b} \implies (a+b)|n$
A: Counterexample: $a=1, b=2, n=9$.
$a$ and $b$ are coprime, and $a+b$ is not a multiple of $n$, but $(a^n + b^n)/(a+b) = 171$ and $a+b$ have a common factor of $3$.  I would guess that the question should read "unless $a+b$ has a common factor with $n$", or else $n$ should be prime, or "has no common factor with" should be replaced by "is not divisible by" as in DougM's answer.
Edit: Assuming $n$ is prime, then the statement holds, by a routine calculation rather similar to DougM's answer:
We rule out the possibility that $n$ is even, since if $a+b \mid a^n+b^n$ then $a+b$ divides both $2a^n$ and $2b^n$, hence it divides $(2a^n,2b^n) = 2$, forcing $a=b=1$ which is trivial to verify.
If $n$ is odd, then $(a^n+b^n)/(a+b)$ may be written as $\sum_{i=1}^n a^{n-i} (-b)^{i-1}$.  Evaluating this mod $a+b$, we see that every term is congruent to $a^{n-1}$ so the sum simplifies to $na^{n-1}$.  We know that $a+b$ and $a$ are coprime so $(a+b, (a^n+b^n)/(a+b)) = (a+b, na^{n-1}) = (a+b,n)$.  The claim follows immediately from the fact that if $n$ is prime then $(a+b,n) = 1$ iff $n$ does not divide $a+b$.
