Any Diophantine quintuple contains a regular Diophantine quadruple (Q1024538)

Suppose \(\{a,b,c,d\}\) is a Diophantine quadruple with \(a<b<c<d\) and put \(d^+=a+b+c+2abc+2rst\), with \(r=\sqrt{ab+1}\), \(s=\sqrt{ac+1}\) and \(t=\sqrt{bc+1}\). Then the Diophantine quadruple \(\{a,b,c,d\}\) is called regular if \(d=d^+\). The two most famous conjectures concerning Diophantine tuples are the following. First, there exists no Diophantine quintuple. Second, every Diophantine quadruple is regular. The author proves the following. Let \(\{a,b,c,d,e\}\) be a Diophantine quintuple, with \(a<b<c<d<e\), then \(d=d^+\), i.e. the Diophantine quadruple \(\{a,b,c,d\}\) is regular. Using the congruence method \textit{A. Dujella} and \textit{A. Pethő} [Q. J. Math. Oxf. Ser. (2) 49, 291--306 (1998; Zbl 0911.11018)] to get lower bounds for the solutions and \textit{M. A. Bennett}'s theorem on simultaneous approximations [J. Reine Angew. Math. 498, 173--199 (1998; Zbl 1044.11011)] to obtain upper bounds for the solutions the author is able to prove his results.