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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:

1. $text{rad}(x^p y^q z^r) < z^{frac{2r}{3}}$


2. 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?