Fermat type functional equation with four terms f(z)n+g(z)n+h(z)n+k(z)n=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} f(z)<^>{n}+g(z)<^>{n}+h(z)<^>{n}+k(z)<^>{n}=1 \end{aligned}$$\end{document}is difficult to solve completely even if n=2,3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n=2,3$$\end{document}, in which the certain type of the above equation is also interesting and significant. In this paper, we first to consider the Bi-Fermat type quadratic partial differential equation f(z1,z2)2+partial derivative f(z1,z2)partial derivative z12+g(z1,z2)2+partial derivative g(z1,z2)partial derivative z12=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} f(z_{1},z_{2})<^>{2}+\left( \frac{\partial f(z_{1},z_{2})}{\partial z_{1}}\right) <^>{2}+g(z_{1},z_{2})<^>{2}+\left( \frac{\partial g(z_{1},z_{2})}{\partial z_{1}}\right) <^>{2}=1 \end{aligned}$$\end{document}in C2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {C}<^>{2}$$\end{document}. In addition, we consider the Bi-Fermat type cubic difference equation f(z)3+g(z)3+f(z+c)3+g(z+c)3=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} f(z)<^>{3}+g(z)<^>{3}+f(z+c)<^>{3}+g(z+c)<^>{3}=1 \end{aligned}$$\end{document}in C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {C}$$\end{document} and obtain partial meromorphic solutions on the above equation.
