The Complete Solution of the Diophantine Equation Fn+1(k)x-Fn-1(k)x=Fm(k)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\left( F_{n+1}^{(k)}\\right) ^x - \\left( F_{n-1}^{(k)}\\right) ^x = F_{m}^{(k)}$$\\end{document}

Abstract

The well-known Fibonacci sequence has several generalizations, among them, the k-generalized Fibonacci sequence denoted by F(k)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$F^{(k)}$$\\end{document}. The first k terms of this generalization are 0,…,0,1\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$0, \\ldots , 0, 1$$\\end{document} and each one afterward corresponds to the sum of the preceding k terms. For the Fibonacci sequence the formula Fn+12-Fn-12=F2n\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$F_{n+1}^2 - F_{n-1}^2 = F_{2n}$$\\end{document} holds for every n≥1\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$n \\ge 1$$\\end{document}. In this paper, we study the above identity on the k-generalized Fibonacci sequence terms, completing the work done by Bensella et al. (On the exponential Diophantine equation (Fm+1(k))x-(Fm-1(k))x=Fn(k)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$(F_{m+1}^{(k)})^x - (F_{m-1}^{(k)})^x = F_n^{(k)}$$\\end{document}, 2022. arxiv:2205.13168).