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

Abstract

For $p$ and $q$ any two distinct Fermat or Mersenne primes, $m,n,r$ as positive integers and $μ= \pm 1$ satisfying any diophantine relation, $\mbox{(i)}\; 2^m + μ= p^nq^r, \mbox{(ii)} \; 2^mp^n + μ= q^r \mbox{ or } \mbox{(iii)} \; p^n + μq^r = 2^m$, 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.