Proving $\gcd( m,n)$=1 If $a$ and $b$ are co prime and $n$ is a prime, show that:
$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a+b$ is a multiple of $n$
Also enlighten me why $n$ has to be prime so that $\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a$+$b$ is a multiple of $n$?

If you like modular arithmetic, note that $$\tag 1  a\equiv -b\mod a+b$$  
Now, since $n$ is odd $$\frac{{{a^n} + {{b }^n}}}{{a + b}}=\frac{{{a^n} - {{\left( { - b} \right)}^n}}}{{a - \left( { - b} \right)}} = \sum\limits_{k = 0}^{n - 1} {{a^k}{{\left( { - b} \right)}^{n - k-1}}} $$ and using $(1)$ $$\frac{{{a^n} + {{b }^n}}}{{a + b}} = \sum\limits_{k = 0}^{n - 1} {{a^k}{a^{n - k - 1}}}  = \sum\limits_{k = 0}^{n - 1} {{a^{n - 1}}}  = n{a^{n - 1}}$$ 
Now since $(a,b)=1$ we have that $$(a,a+b)=(a,a+b\mod a)=(a,b)=1\implies (a^{n-1},a+b)=1$$ so it all reduces to $$(n,a+b)$$  
