If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$. If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$.
How does even even begin to prove this? I have proved that if $a+b$ is a multiple of $n$, then so is $a^n+b^n$, using FLT, but how to progress next eludes me.

When $n=2$ you have $\frac{a^n+b^n}{a+b}=a-b+\frac{2b}{a+b}$ which is not an integer in general.
So I assume $n=2k+1$ is an odd prime should be part of the statement.
In this case
$$\frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+...+b^{n-1}\\=
a^{n-1}+a^{n-2}b-2a^{n-2}b-2a^{n-3}b^2+3a^{n-3}b^2+3a^{n-4}b^3-...\\
=(a+b)(a^{n-2}-2a^{n-3}b+3a^{n-4}b^-...)+nb^{n-1}$$
Therefore
$$gcd(\frac{a^n+b^n}{a+b}, a+b)=gcd(nb^{n-1}, a+b)=gcd(n, a+b)$$
with the last equality following from $gcd(a+b, b^{n-1})=1$.
