When does $a + b$ divide $a^p + b^p$?

Question

I came across a problem in Niven's number theory text (problem 51 on page 20) that asks the following:

Show that if $(a, b) = 1$ and $p$ is an odd prime, then $$\left(a + b, \frac{a^p + b^p}{a + b}\right) = 1 \text{ or } p.$$

I am not asking for a solution to this problem; instead, I'm trying to understand why $a^p + b^p$ would always be divisible by $a + b$ given the above conditions. Does anyone have any insights as to why this would be true? Where (if at all) do we use the conditions that $(a, b) = 1$ and $p$ is an odd prime?

Answer 1

$x^p+1$ has a zero at $x=-1$, so a factorization with factor $(x+1)$ exists (and can be given explicitly).

Now replace $x$ on both sides by $a/b$ and multiply everything with $b^p$.

Answer 2

Explicitly, if $p$ is any odd positive integer, $$(a + b) \sum_{k=0}^{p-1} (-1)^k a^{p-1-k} b^k = a^p + b^p$$ You don't need $(a,b) = 1$, in fact $a$ and $b$ don't need to be integers (it works in any commutative ring), and you don't need $p$ to be prime.

Answer 3

Hint $\rm\: \left(x\!-\!a,\frac{f(x)-f(a)}{x-a}\!\right) = (x\!-\!a,\,f\:\!'(a))\:$ by $\rm\: \frac{f(x)-f(a)}{x-a} \equiv\ f\:\!'(a)\pmod{\!x\!-\!a}\ $ for $\rm\ f\in \mathbb Z[x]$

For further details see here, which elaborates on how this result is a number-theoretical analog of a well-known result about functions (polynomials), viz. about multiplicity of roots.