## Kodai Mathematical Journal

- Kodai Math. J.
- Volume 41, Number 2 (2018), 413-420.

### 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 $\mathbf{Q}$ ($\sqrt {-q}$).

#### Article information

**Source**

Kodai Math. J., Volume 41, Number 2 (2018), 413-420.

**Dates**

First available in Project Euclid: 2 July 2018

**Permanent link to this document**

https://projecteuclid.org/euclid.kmj/1530496850

**Digital Object Identifier**

doi:10.2996/kmj/1530496850

**Mathematical Reviews number (MathSciNet)**

MR3824859

**Zentralblatt MATH identifier**

06936461

#### Citation

Zhang, Xin. On Terai's conjecture. Kodai Math. J. 41 (2018), no. 2, 413--420. doi:10.2996/kmj/1530496850. https://projecteuclid.org/euclid.kmj/1530496850