It's a hard knock life

if $n$ is an odd number then

$a^n+b^n=(a+b)(\sum_{i=0}^{n-1}a^ib^{n-1-i}(-1)^i)$

so

$a^p+b^p=(a+b)(\sum_{i=0}^{n-1}a^ib^{p-1-i}(-1)^i) \implies \frac{a^n+b^n}{a+b}=\sum_{i=0}^{n-1}a^ib^{p-1-i}(-1)^i$

by repeatedly applying Euclidean algorithm, we find the GCD to be

$GCD(\sum_{i=0}^{n-1}a^ib^{p-1-i}(-1)^i,a+b)=GCD(pb^{p-1},a+b)=1$

since $(a+b,p)=1$ and $(a+b,b^t)=1 \ \forall t\in \mathbb{N}$

Edit:

As requested by Sonu Gupra, I decided to make the proof more accurate, for the case $p=2$ .

The GCD would be defined if $\frac{a^2+b^2}{a+b}$ is a whole number. So, we check the condition, for which is a whole number.

$a+b | a^2+b^2=(a+b)^2-2ab \implies a+b | 2ab$

as $gcd(a,b)=1$ , this is only possible if $a+b=2$ . because, if $q | 2ab$ , for a prime $q | a+b \ , \ q>2$ , then either $q|a$ or $q|b$ . this is contradiction, since if $q|a$ and $gcd(a,b)=1$ then $q \nmid a+b$ . Therefore, $a+b$ can only be $2$ , meaning $a=1,b=1$ . Consequently, one realises that the final answer stays the same.

Logically, if the answer is a single number then that number must be 1.

If it were anything else it would have to be some expression in terms of $a$ and $b$ .