Solutions to $x^p+y^q = z^r$

Question

Is there any $(p,q,r)$ with $\gcd(p,q,r) = 1$ and $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} < 1$ for which we know that the only integer solutions (not necessarily primitive) to the equation $x^p+y^q=z^r$ have $xyz = 0$? More generally, is there any $(p,q,r)$ as above such that we can describe explicitly all the solutions (not necessarily primitive) to $x^p+y^q=z^r$?

Answer 1

There are no $p,q,r$ meeting the given conditions for which we know the only solutions have $xyz=0$, because there are no $p,q,r$ for which the only solutions have $xyz=0$.

Given $p,q,r$, define $c$ by $1+2^q=c$. Then $c^s+2^qc^s=c^{s+1}$. Choose $s$ so that $s=pqt$ for some $t$, and $s+1=ru$ for some $u$ --- this can be done, since $p,q,r$ are coprime. Then $$(c^{qt})^p+(2c^{pt})^q=(c^u)^r$$ So, you always have nonzero solutions.