TITLE: Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd
QUESTION [8 upvotes]: I read somewhere that

$(a^n - b^n)$

It is always divisible by $a-b$.
When $n$ is even it is also divisible by $a+b$.
When $n$ is odd it is not divisible by $a+b$.


and

$(a^n + b^n)$

It is never divisible by $a-b$.
When $n$ is odd it is divisible by $a+b$.
When $n$ is even it is not divisible by $a+b$.


I wonder what's the proof for this.
First postulate is clear. $(a-b)$ would or would not be a factor. Any light on others?

REPLY [3 votes]: for $a^n+b^n$, to treat the odd and even cases in a single framework-- if you do the long division you will notice the following identity:
$$
a^n+b^n = (a+b)\left(\sum_{k=0}^{n-1} (-1)^k a^{n-1-k}b^k\right) +(1+(-1)^n)b^n
$$