Diophantine triples with largest two elements in common
A set of \(m\) positive integers \(\{a_{1},a_{2},\cdots, a_{m}\}\) is called a Diophantine \(m\)-tuple if \(a_{i}a_{j}+1\) is a perfect square for all \(i\) and \(j\) with \(1 \leq i <j \leq m\). \textit{A. Dujella} [J. Reine Angew. Math. 566, 183--214 (2004; Zbl 1037.11019)] proved that that there exist only finitely many Diophantine quintuples whereas there is no Diophantine sextuple. \textit{B. He} et al. [Trans. Am. Math. Soc. 371, No. 9, 6665--6709 (2019; Zbl 1430.11044)] proved that there does not exist a Diophantine quintuple. A strong conjecture asserting that all Diophantine quadruples are regular is still open. For Diophantine quadruples \(\{a,b,c,d\}\) with \(a<b<c<d\) containing various pairs \(\{a,b\}\) or triples \(\{a,b,c\},\) such as\par (i) \(\{k-1,k+1\}\) with \(k\geq 2\) an integer \par (ii) \(\{k,4k\pm4\}\) with \(k\) a positive integer \par (iii) \(\{k,a^{2}k\pm2a,(a+1)^{2}k\pm2(a+1)\}\) with \(a,k\) positive integers \par (iv) \(\{a,b,c\}\) with \(c \geq 200b^{4}\) \par it is known that \(d\) must be equal to \[a+b+c+2abc+2(\sqrt{ab+1})(\sqrt{ac+1})(\sqrt{bc+1}).\] In proving each of these results, firstly the conditions that \(ad+1=x^{2}, bd+1=y^{2}, cd+1=z^{2}\) for some positive integers \(x,y, z\) are to be transformed into a system of Pellian equations \(az^{2}-cx^{2}=a-c, bz^{2}-cy^{2}=b-c, \) and an upper bound for \(z\) is deduced by using Baker's method or hypergeometric method. In this paper, it is proved that if \(\{a,b,c\}\) is a Diophantine triple then \(\{a+1,b,c\}\) is not a Diophantine triple. It is also proved that if \(c<16b^{3}\) and \(\{a_{1},b,c\}\) 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. It is proved using the properties of regular Diophantine quadruples.
scientific article; zbMATH DE number 7364769
scientific article