On the exponential Diophantine equation (3pm2-1)x+(p(p-3)m2+1)y=(pm)z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(3pm^2-1)^x+(p(p-3)m^2+1)^y=(pm)^z$$\end{document}

Let m be a positive integer, and let p be a prime with p≡1(mod4).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p \equiv 1~(\mathrm{mod}~4).$$\end{document} Then we show that the exponential Diophantine equation (3pm2-1)x+(p(p-3)m2+1)y=(pm)z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(3pm^2-1)^x+(p(p-3)m^2+1)^y=(pm)^z$$\end{document} has only the positive integer solution (x,y,z)=(1,1,2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(x, y, z)=(1, 1, 2)$$\end{document} under some conditions. As a corollary, we derive that the exponential Diophantine equation (15m2-1)x+(10m2+1)y=(5m)z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(15m^2-1)^x+(10m^2+1)^y=(5m)^z$$\end{document} has only the positive integer solution (x,y,z)=(1,1,2).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(x, y, z)=(1, 1, 2).$$\end{document} The proof is based on elementary methods and Baker’s method.