On Terai's conjecture

Abstract

Let $p$ be an odd prime such that $b^r+1=2p^t$, where $r$, $t$ are positive integers and $b \equiv$ 3,5 (mod 8). We show that the Diophantine equation $x^2+b^m=p^n$ has only the positive integer solution $(x,m,n)=(p^t-1,r,2t)$. We also prove that if $b$ is a prime and $r=t=2$, then the above equation has only one solution for the case $b \equiv$ 3,5,7 (mod 8) and the case $d$ is not an odd integer greater than 1 if $b \equiv$ 1 (mod 8), where $d$ is the order of prime divisor of ideal ($p$) in the ideal class group of $\mathbf{Q}$ ($\sqrt {-q}$).