Can $\frac{a^n+b^n}{a+b}=c^n$ have a solution?

Question

Can $\frac{a^n+b^n}{a+b}=c^n$ have a solution for $a,b,c,n\in \mathbb N$, $a^n+b^n\neq a+b$, and $\gcd(a+b,c)=\gcd(a,b)=1$?

I notice that $n$ must be odd for the LHS to be integer. Also, as $\frac{a^n+b^n}{a+b}<(a+b)^{n-1}$, it follows that $c<a+b$; but little more progress...

Thanks in advance for your hints and advice!

Answer 1

For, $n=2$, we have:

$a^2+b^2=(a+b)c^2$ -------$(1)$

Above has numerical solution:

$(a,b,c)=(4,28,5)$

Also in eqn(1) if we take:

$a^2+b^2=c^2$, then the condition needed is:

a+b=1

We take:

$(a,b,c)=[(2mn),(m^2-n^2),(m^2+n^2)]$

Hence the condition is:

$m^2+2mn-n^2=1$

Above has solution at, $(m,n)=(1,2)$

And we get:

$(a,b,c)=[(4),(-3),(5)]$