English

On the Exponential Diophantine Equation $(a^2-2)(b^2-2)=x^2$

Number Theory 2018-01-16 v1

Abstract

In this paper, we consider the equation (an2m)(bn2m)=x2(a^n-2^{m})(b^n-2^{m})=x^2. By assuming the abc conjecture is true, in [8], Luca and Walsh gave a theorem, which implies that the above equation has only finitely many solutions n,xn,x if a and b are different fixed positive integers. We solve the above equation when m=1m=1 and (a,b)=(2,10),(4,100),(10,58),(3,45)(a,b)=(2,10),(4,100),(10,58),(3,45). Moreover, we show that (a22)(b22)=x2(a^2-2)(b^2-2)=x^2 has no solution n,x if 2|n and gcd(a,b)=1(a,b)=1. We also give a conjecture which says that the equation (222)((2Pk)n2)=x2(2^2-2)((2P_k)^n-2)=x^2 has only the solution (n,x)=(2,Qk)(n,x)=(2,Q_k), where k>3k>3 is odd and Pk,QkP_k,Q_k are Pell and Pell Lucas numbers, respectively. We also conjecture that if the equation (a22)(b22)=x2(a^2-2)(b^2-2)=x^2 has a solution n,xn,x, then n<7n<7, where 2<a<b2<a<b.

Keywords

Cite

@article{arxiv.1801.04770,
  title  = {On the Exponential Diophantine Equation $(a^2-2)(b^2-2)=x^2$},
  author = {Zafer Şiar and Refik Keskin},
  journal= {arXiv preprint arXiv:1801.04770},
  year   = {2018}
}
R2 v1 2026-06-22T23:45:13.391Z