finding a proof for a problem $(a+b,\frac{a^p+b^p}{a+b})=p$ or $1$

Question

How can I prove $(a+b,\frac{a^p+b^p}{a+b})=p$ or $1$ when $p$ is a prime number and $a$ and $b$ are coprime number? any help would be great thanks.

Answer 1

If $k\ge 3$ is an odd number, $a,b\in\mathbb Z$, $a+b\neq 0$, then: $$\frac{a^k+b^k}{a+b}\equiv a^{k-1}-a^{k-2}b\pm \cdots +b^{k-1}\equiv ka^{k-1}\equiv kb^{k-1}\pmod{a+b}$$

$$\gcd\left(\frac{a^k+b^k}{a+b},a+b\right)=\gcd\left(ka^{k-1},kb^{k-1},a+b\right)$$

$$=\gcd\left(\gcd\left(ka^{k-1},kb^{k-1}\right),a+b\right)=\gcd\left(k\gcd\left(a,b\right)^{k-1},a+b\right)$$

If $\gcd(a,b)=1$ and $k=p$ is prime, then this equals $\gcd(p,a+b)\mid p$, so $\gcd(p,a+b)\in\{1,p\}$.

One of the few properties I used can be proved by the following equivalence (for any $d\in\mathbb Z$), which is easy to see by the congruences in the beginning: $$d\mid \frac{a^k+b^k}{a+b}, a+b\iff d\mid ka^{k-1}, kb^{k-1}, a+b$$

Answer 2

You have to assume that $p$ is an odd prime (because $\frac{a^2+b^2}{a+b}$ need not be an integer).

Hint: Show that

$$ \frac{a^p+b^p}{a+b} = \sum_{i=0}^{p-1}(-1)^ia^{p-1-i}b^i = \sum_{i=0}^{p-2}(-1)^i (i+1)(a+b)a^{p-2-i}b^i + pb^{p-1}. $$ From this, you can argue, that $(a+b, \frac{a^p+b^p}{a+b})$ is either $1$ or $p$ (using $(a,b)=1$ of course).