<s>[INST] 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.
 [/INST] 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.
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