Prove or Disprove: There exists positive integers $x, y, z$, such that $x^8-y^5=z^3$

Question

$x^8-y^5=z^3$

I believe it is some form of a Diophantine equation but since each variable is to a different power, I am unable to solve it.

Answer 1

We can get examples using powers of $2$, and exploiting the fact that $2^n-2^{n-1}=2^{n-1}$. We just need $n$ such that $8\,|\,n$ and $15\,|\,(n-1)$.

We see that $$2^{16}-2^{15}=2^{15}\implies (2^2)^8-(2^3)^5=(2^5)^3$$

for instance.

In this way we get an infinite family of solutions, given by $n\equiv 16\pmod {120}$. Of course, these are not all the solutions. If $(a,b,c)$ is a solution then so is $(m^{15}a, m^{24}b, m^{40}c)$ for any $m\in \mathbb N$. I don't know if there are any solutions in which the three integers are relatively prime.