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Diophantine triples with largest two elements in common

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Abstract

In this paper we prove that if \(\{a,b,c\}\) is a Diophantine triple with \(a<b<c\), then \(\{a+1,b,c\}\) cannot be a Diophantine triple. Moreover, we show that if \(\{a_1,b,c\}\) and \(\{a_2,b,c\}\) are Diophantine triples with \(a_1<a_2<b<c < 16b^3\), then \(\{a_1,a_2,b,c\}\) is a Diophantine quadruple. In view of these results, we conjecture that if \(\{a_1,b,c\}\) and \(\{a_2,b,c\}\) are Diophantine triples with \(a_1<a_2<b<c\), then \(\{a_1,a_2,b,c\}\) is a Diophantine quadruple.

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Acknowledgements

The authors thank the referee for useful comments.

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Correspondence to Yasutsugu Fujita.

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Andrej Dujella acknowledges support from the QuantiXLie Center of Excellence, a project cofinanced by the Croatian Government and European Union through the European Regional Development Fund—the Competitiveness and Cohesion Operational Programme (Grant KK.01.1.1.01.0004). Andrej Dujella was supported also by the Croatian Science Foundation under the project no. IP-2018-01-1313. Yasutsugu Fujita was supported by JSPS KAKENHI Grant Number 16K05079.

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Cipu, M., Dujella, A. & Fujita, Y. Diophantine triples with largest two elements in common. Period Math Hung 82, 56–68 (2021). https://doi.org/10.1007/s10998-020-00331-4

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