I want to show that given $p,q,r \ge 3$ the diophantine equation $x^py^q=z^r-1$ has only finitely many solutions with $x,y,z \in \mathbb{N} = 1 ,2, \dots$ assuming the $abc$-conjecture.
The proof supposedly goes in two steps:
- $\text{rad}(x^p y^q z^r) < z^{\frac{2r}{3}}$
- Now apply the $abc$-conjecture to conclude that there are only finitely many solutions.
The statement of the $abc$-conjecture that we use is:
For every $\epsilon>0$ there exists only finitely many $ABC$-triples $(a,b,c)\in \mathbb{N}^3$ s.t. $$c > (\text{rad}(abc))^{1+\epsilon}.$$
For step 1, I made some observations:
- $\text{rad}(x^py^qz^r) = \text{rad}(xyz) \le xyz \le (z^r-1)^{\frac{1}{3}}z$
- $\text{rad}(x^py^qz^r) = \text{rad}{(z^r-1)z}$
Any help is appreciated; I'm mainly stuck at the first part.
EDIT:
For step 1 we use the further observations $(z^r-1) < z^r$, and $z \le z^{\frac{r}{3}}$, to find that $\text{rad}(x^py^qz^r) = \text{rad}(xyz) \le xyz \le (z^r-1)^{\frac{1}{3}}z < z^{\frac{2r}{3}}$.
EDIT: For step 2 take $\epsilon = \frac{1}{2}$, then the $abc$-conjecture states that there are only finite solutions with $z^r > \text{rad}(x^py^qz^r)^{1 + \epsilon} = \text{rad}(x^py^qz^r)^{1 + \frac{1}{2}} = \text{rad}(x^py^qz^r)^{\frac{3}{2}}$.
But from step 1 we have $z^{\frac{2r}{3}} > \text{rad}(x^py^qz^r)$ which implies $z^r > \text{rad}(x^py^qz^r)^{\frac{3}{2}}$, allowing us to conclude that we have only finitely many solutions.
Is my proof correct?