Problem: Let $p,q,r$, be positive integers satisfying $\frac {1}{p} + \frac {1}{q} + \frac {1}{r} < 1$ . If the ABC conjecture is true, then $x^p + y^q = z^r$ has finitely many positive integer solutions $(x,y,z)$ that are co-prime.
Thoughts:
1) $p,q,r> 1$ because of the inequality.
2) In the formulation of the ABC conjecture I am familiar with , it requires that $gcd(a,b,c)=1$, so in our case I might need to show that $gcd(x^p, y^q,z^r)=1$
3) I know that if $gcd(x,y,z)=1$ then $rad(x^p, y^q,z^r)=rad(x,y,z)$
other than that I am not sure how to proceed. Insights appreciated.
Beukers in his 2005 lecture gives the following:
The ABC-conjecture
Fermat-Catalan conjecture
Consequence of $ABC$-conjecture:
The set of triples $x^p$, $y^q$, $z^r$ with $x$, $y$, $z$ coprime positive integers
such that $x^p$ + $y^q$ = $z^r$ and $1/p + 1/q + 1/r < 1$, is finite.
Observation, $1/p + 1/q + 1/r < 1$ implies $1/p + 1/q + 1/r \lt 1 − 1/42$.
Apply $ABC$ with $\kappa = 1.01$ to $A = x^p$, $B = y^q$, $C = z^r$ . Notice that $rad(x^r y^q z^r ) \le xyz < z^{r/p} z^{r/q} z$.
Hence, with finitely many exceptions we get $z^r < z^{\kappa(r/p+r/q+1)}$
This implies $r < \kappa(r/p + r/q + 1)$ and hence $1 < \kappa(1/p + 1/q + 1/r )$. But this is impossible because $\kappa = 1.01$ and $1/p + 1/q + 1/r \le 1 − 1/42$.