Elementary Number Theory : a property about gcd

Question

If $(a,b)=1$ and $p$ is a prime which does not divide $(a + b)$, prove that $$\gcd\left(\frac{a^p + b^p}{a + b},a+b\right)=1$$ I have no idea how to attack this question.

Answer 1

More generally, if $a,b,n\in\mathbb Z$, $a\neq -b$, $n\ge 3$ is odd, then

$$\frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b\pm\cdots +b^{n-1}$$

$$\equiv na^{n-1}\equiv nb^{n-1}\pmod{a+b}$$

$$\left(\frac{a^n+b^n}{a+b},a+b\right)$$

$$=\left(na^{n-1},a+b\right)$$

$$=\left(nb^{n-1},a+b\right)$$

$$=\left(na^{n-1},nb^{n-1},a+b\right)$$

$$=\left(n(a,b)^{n-1},a+b\right)$$

If $n=p$ is prime, $p\nmid a+b$, $(a,b)=1$, then this equals $$(p\cdot 1^{p-1},a+b)$$

$$=(p,a+b)=1$$

Answer 2

HINT: Use the Lifting the Exponent Lemma to get that if prime $p$ divides $a+b$ then the highest power of $p$ dividing $a+b$ is the same as the highest power of $p$ dividing $a^n + b^n$

Answer 3

Let $x=a+b$ and let $p$ be an odd prime then $$a^p+(x-a)^p=a^p+\sum_{k=0}^p \binom{p}{k}x^k(-a)^{p-k} =\sum_{k=1}^p \binom{p}{k}x^k(-a)^{p-k}=x\sum_{k=1}^p \binom{p}{k}x^{k-1}(-a)^{p-k}.$$ Hence $$\frac{a^p+(x-a)^p}{x}=\sum_{k=1}^p \binom{p}{k}x^{k-1}(-a)^{p-k}= p(-a)^{p-1}+x\sum_{k=2}^p \binom{p}{k}x^{k-1}(-a)^{p-k}$$ and $$\gcd\left(\frac{a^p + b^p}{a + b},a+b\right) =\gcd\left(\frac{a^p + (x-a)^p}{x},x\right) =\gcd(pa^{p-1},a+b).$$ Now if $(a,b)=1$ and $p$ is a prime which does not divide $(a + b)$ then
$$\gcd(pa^{p-1},a+b)=\gcd(a^{p-1},a+b)=1.$$