## On the Exponential Diophantine Equation $(6m^{2}+1)^{x}+(3m^{2}-1)^{y}=(3m)^{z}$

Fundamental journal of mathematics and applications (Online), vol.5, no.3, pp.174-180, 2022 (Peer-Reviewed Journal)

• Publication Type: Article / Article
• Volume: 5 Issue: 3
• Publication Date: 2022
• Doi Number: 10.33401/fujma.1038699
• Journal Name:
• Journal Indexes: TR DİZİN (ULAKBİM)
• Page Numbers: pp.174-180
• Yıldız Technical University Affiliated: Yes

#### Abstract

Let m" role="presentation" >m$m$ be a positive integer. In this paper, we consider the exponential Diophantine equation (6m2+1)x+(3m2&#x2212;1)y=(3m)z" role="presentation" >(6m2+1)x+(3m21)y=(3m)z$\left(6{m}^{2}+1{\right)}^{x}+\left(3{m}^{2}-1{\right)}^{y}=\left(3m{\right)}^{z}$ and we show that it has only unique positive integer solution (x,y,z)=(1,1,2)" role="presentation" >(x,y,z)=(1,1,2)$\left(x,y,z\right)=\left(1,1,2\right)$ for all m&gt;1." role="presentation" >m>1.$m>1.$ The proof depends on some results on Diophantine equations and the famous primitive divisor theorem.