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 $\mu =\pm 1$ 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 $ABC=2^m{p}^n{q}^r$ and which satisfy $C>\mathrm{rad}(ABC{)}^{1+\epsilon }$ for any real $\epsilon >0$, 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, $\mathrm{rad}(ABC{)}^{1+\epsilon }>C$ holds for any $\epsilon >0$. Furthermore, finiteness of the number of solutions of (iii) when $n$ is even, is demonstrated elsewhere (Ref. [64]). All other solutions are enumerated.

arXiv comment39 pages

References in corpus (3)

• Elementary treatment of $p^a\pm p^b+1=x^2$
• An Elementary Proof for Ljunggren Equation
• The Nagell-Ljunggren equation via Runge's method