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.
 A: Note that the problem does not make sense if $n=2$, as $\dfrac{a^2+b^2}{a+b}$ is not generally an integer.  So we assume $n\ne2$, and hence $n$ is odd.
If $\dfrac{a^n+b^n}{a+b}$ and $a+b$ have a common factor then they have a common prime factor $p$.  So
$$p\mid a+b\quad\hbox{and}\quad p\mid a^{n-1}-a^{n-2}b+a^{n-3}b^2-\cdots+b^{n-1}\ .$$
This gives $b\equiv-a\pmod p$ and so the second expression above can be written
$$p\mid na^{n-1}\ .$$
Now $p\mid a+b$ and $n\not\mid a+b$, so $p\ne n$, so $p\mid a$ and hence $p\mid b$.  Since $a,b$ are coprime this is a contradiction.
Hence $\dfrac{a^n+b^n}{a+b}$ and $a+b$ are coprime.
A: 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 $gcd(a+b, b^{n-1})=1$.
