Does $ x^p+y^p=kz^p$ have any solutions when $x,y,z,k,p>2, gdc(x,y,z)=1$?

Question

Does the Diophantine equation

$$\displaystyle x^p+y^p=k(z^p)$$

have any solutions when $x,y,z,k,p>2, $ and $ x,y,z$ are co-primes?

Answer 1

Let $a,b$ be two positive integers, and $p$ odd prime.

Then $$(a^{p}-b)^{p}+(a^{p}+b)^{p} \equiv 0 \pmod{a^{p}}$$ Thus $k=\frac{(a^{p}-b)^{p}+(a^{p}+b)^{p}}{a^{p}}$ is integer and

$x=(a^{p}-b)^{p}, y=(a^{p}+b)^{p}, z=a, k, p$ is a solution.

And it is easy to generate relatively prime solutions( for example $a$ is odd prime, $b=2$).

In the case $p=3, a=3, b=2$ for example we get

$$25^3+29^3 =1482 \cdot 3^3$$

Answer 2

Like

$$3^3+5^3=19\cdot 2^3\ldots?$$

Oops! $\;2\rlap{\,\,/}>2\;$...or only $\;p>2\;$ ?