ABC Conjecture: $ABC = 2^m p^n q^r$ with Fermat or Mersenne Primes

TL;DR

This paper investigates the finiteness of certain triplets related to the ABC conjecture involving Fermat and Mersenne primes, providing new results on the number of solutions under specific Diophantine conditions.

Contribution

It establishes the finiteness of solutions for specific forms of triplets involving Fermat and Mersenne primes within the ABC conjecture framework.

Findings

Finiteness of triplets with specific Diophantine relations involving Fermat and Mersenne primes.

Identification of a particular triplet satisfying the relation with primes of the form 2^y+1 and 2^{2y}+1.

Enumeration of all solutions under the given conditions.

Abstract

For $p$ and $q$ any two distinct Fermat or Mersenne primes, $m,n,r$ as positive integers and $\mu = \pm 1$ satisfying any diophantine relation, $\mbox{(i)}\; 2^m + \mu = p^nq^r, \mbox{(ii)} \; 2^mp^n + \mu = q^r \mbox{ or } \mbox{(iii)} \; p^n + \mu q^r = 2^m$, it is shown that the number of triplets $\{A, B, C \}$ with $\gcd(A,B) = 1$ and $C = A + B$, for which their product is of the form $ABC = 2^mp^nq^r$ and which satisfy $C > \mathrm{rad}(ABC)^{1 + \varepsilon}$ for any real $\varepsilon > 0$, is finite. For the triplet $\{2^{y+1}, 2^{2y}+1, (2^y+1)^2\}$, a solution to (iii) with positive integer $y$ such that $2^y+1$ and $2^{2y}+1$ are primes, $\mathrm{rad}(ABC)^{1 + \varepsilon} > C$ holds for any $\varepsilon > 0$. Furthermore, finiteness of the number of solutions of (iii) when $n$ is even, is demonstrated elsewhere (Ref. [64]). All other solutions are enumerated.