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

TL;DR

The paper narrows the ABC conjecture to the setting where $A+B=C$, $\gcd(A,B)=1$, and $ABC$ factors as $2^m p^n q^r$ with $p,q$ Fermat or Mersenne primes. By analyzing the three Diophantine relations $2^m+\mu=p^n q^r$, $2^m p^n+\mu=q^r$, and $p^n+\mu q^r=2^m$ under sign choices $\mu=\pm1$, it derives strong structural constraints (notably gcd$(n,r)$ being a power of two) and carries out detailed case analyses using parity, factorization, and classical results (e.g., Bang/Zsigmondy, Szalay, Scott–Styer). The results yield finiteness of the number of $\{A,B,C\}$ with $C>\mathrm{rad}(ABC)^{1+\varepsilon}$ in the Fermat/Mersenne setting, and the paper enumerates all small, sporadic solutions, including explicit instances such as $2^9+1=3^3\cdot19$ and $2^5\cdot3^2+1=17^2$, among others. A notable exception is the triplet $\{2^{y+1},2^{2y}+1,(2^y+1)^2\}$, for which $rad(ABC)^{1+\varepsilon}>C$ holds for all $\varepsilon>0$, illustrating that the bound can be strict in special cases. Collectively, the work provides a finite, explicit catalog of solutions and reinforces finiteness results for the corresponding Diophantine systems under these prime restrictions.

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.