we need to show gcd is $1$ I need to show if $(a,b)=1,n$ is an odd positive integer then $\displaystyle \left(a+b,{a^n+b^n\over a+b}\right)\mid n.$ 
let $\displaystyle \left(a+b,{a^n+b^n\over a+b}\right)=d$
$\displaystyle d\mid {a^n+b^n\over a+b}=(a+b)^2(a^{n-3}-a^{n-4}b\dots+b^{n-3})-2ab(a^{n-3}\dots+b^{n-3})-ab^{n-2}$
$d\mid (a+b)$ so from the rest can I conclude $d=1$ as $(a,b)=1$?

Modulo $a+b$, you have $a^kb^{n-k}\equiv (-1)^kb^n$, so
$$ \frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+a^{n-3}b^2\mp\cdots -ab^{n.2}+b^{n-1}\equiv n$$
so $n=\frac{a^n+b^n}{a+b}+(\ldots)\cdot (a+b)$.
