If $(a,b)=1$ and $n$ is a prime, prove that $(a^n+b^n)/(a+b)$ and $(a+b)$ have no common factor, unless $a+b$ is a multiple of $n$ I have no hints to get started. This is a really tough problem involving prime powers and it's the first time I have seen such question.
I need some good hints to get started. Thanks!
Please do not use congruence, limits or derivatives because they are out of scope of the chapter which has this question.
 A: Assume $a+b$ is not a multiple of $n$.
For $n=2$, a common prime factor of $a+b$ and $\frac{a^2+b^2}{a+b}=a+b-\frac{2ab}{a+b}$ must be $2(=n)$, which is excluded, or must divide one of $a,b$ and hence both.
Thus we may assume that $n$ is odd.
Let $q$ be a prime with $q^k\| a+b$, $k\ge 1$. So $q\nmid n$ and we have $b=-a+rq^k$ with $q\nmid r$.
We need to show that $q^{k+1}\nmid a^n+b^n$:
$$a^n+b^n=a^n+(-a)^n+n(-a)^{n-1}rq^k+{n\choose 2}(-a)^{n-2}r^2q^{2k}+\ldots\equiv na^{n-1}rq^k\pmod{q^{k+1}}$$
So we want to show that $q\nmid na^{n-1}r$. We already know $q\nmid r$ and $q\nmid n$. Remains the possibility that  $q\mid a$, but then also  $q\mid (a+b)-a=b$, contradicting $\gcd(a,b)=1$. Hence indees $q\nmid na^{n-1}r$, $q^{k+1}\nmid a^n+b^n$ and ultimately $q\nmid \frac{a^n+b^n}{a+b}$.
A: well , the problem may seem confusing but it can be easily approached using fermat theorem 
    as, a^n==a mod n .....(1)
 and b^n== b mod n ....(2)
 now, adding both
    a^n+b^n==a+b mod n 
 now if (a+b) is a multiple of n,
   a^n+b^n==0 mod n,
  dividing by (a+b) on both the sides we can imply that ,
 a^n+b^n/a+b and a+b have no common factors unless a+b is a multiple of n
 thanks...waiting for response!
