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• Author or Editor: Alain Togbé
Clear All Modify Search  # On a family of diophantine triples {K,A 2 K + 2A, (A + 1)2 K + 2(A + 1)} with two parameters II

Periodica Mathematica Hungarica
Authors: Bo He and Alain Togbé

## Abstract

Let A and k be positive integers. In this paper, we study the Diophantine quadruples

\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\{ k,A^2 k + 2A,(A + 1)^2 k + 2(A + 1)d\} .$$ \end{document}
If d is a positive integer such that the product of any two distinct elements of the set increased by 1 is a perfect square, then
\documentclass{aastex} \usepackage{amsbsy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{bm} \usepackage{mathrsfs} \usepackage{pifont} \usepackage{stmaryrd} \usepackage{textcomp} \usepackage{upgreek} \usepackage{portland,xspace} \usepackage{amsmath,amsxtra} \pagestyle{empty} \DeclareMathSizes{10}{9}{7}{6} \begin{document} $$\begin{gathered} d = (4A^4 + 8A^3 + 4A^2 )k^3 + (16A^3 + 24A^2 + 8A)k^2 + \hfill \\ + (20A^2 + 20A + 4)k + (8A + 4) \hfill \\ \end{gathered}$$ \end{document}
for A ≥ 52330 and any k. This extends our result obtained in .

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# The extension of the D(−k 2)-pair {k 2,k 2 + 1}

Periodica Mathematica Hungarica
Authors: Yasutsugu Fujita and Alain Togbé

## Abstract

Let n be a nonzero integer. A set of m distinct positive integers is called a D(n)-m-tuple if the product of any two of them increased by n is a perfect square. Let k be a positive integer. In this paper, we show that if {k 2, k 2+1, c, d} is a D(−k 2)-quadruple with c < d, then c = 1 and d = 4k 2+1. This extends the work of the first author  and that of Dujella .

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# On the family of diophantine triples { k + 1, 4 k , 9 k + 3}

Periodica Mathematica Hungarica
Authors: Bo He and Alain Togbé

## Abstract

We prove that if k is a positive integer and d is a positive integer such that the product of any two distinct elements of the set {k + 1, 4k, 9k + 3, d} increased by 1 is a perfect square, then d = 144k 3 + 192k 2 + 76k + 8.

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