Extensions of a Diophantine triple by adjoining smaller elements II

Pub Date : 2023-12-14 DOI:10.1007/s10998-023-00569-8
Mihai Cipu, Andrej Dujella, Yasutsugu Fujita
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Abstract

Let \(\{a_1,b,c\}\) and \(\{a_2,b,c\}\) be Diophantine triples with \(a_1<b<a_2<c\) and \(a_2\ne b+c-2\sqrt{bc+1}\). Put \(d_2=a_2+b+c+2a_2bc-2r_2st\), where \(r_2=\sqrt{a_2b+1}\), \(s=\sqrt{ac+1}\) and \(t=\sqrt{bc+1}\). In this paper, we prove that if \(c \le 16\mu ^2 b^3\), where \(\mu =\min \{a_1,d_2\}\), then \(\{a_1,a_2,b,c\}\) is a Diophantine quadruple. Combining this result with one of our previous results implies that if \(\{a_i,b,c,d\}\) \((i\in \{1,2,3\})\) are Diophantine quadruples with \(a_1<a_2<b<a_3<c<d\), then \(a_3=b+c-2\sqrt{bc+1}\). It immediately follows that there does not exist a septuple \(\{a_1,a_2,a_3,a_4,b,c,d\}\) with \(a_1<a_2<b<a_3<a_4<c<d\) such that \(\{a_i,b,c,d\}\) \((i \in \{1,2,3,4\})\) are Diophantine quadruples. Moreover, it is shown that there are only finitely many sextuples \(\{a_1,a_2,a_3,b,c,d\}\) with \(a_1<b<a_2<a_3<c<d\) such that \(\{a_i,b,c,d\}\) \((i \in \{1,2,3\})\) are Diophantine quadruples.

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通过邻接较小元素扩展 Diophantine 三元组 II
让 \(\{a_1,b,c\}\) 和 \(\{a_2,b,c\}\) 是二叉三元组,有 \(a_1<b<a_2<c\) 和 \(a_2\ne b+c-2sqrt{bc+1}\).把(d_2=a_2+b+c+2a_2bc-2r_2st),其中(r_2=sqrt{a_2b+1}),(s=sqrt{ac+1})和(t=sqrt{bc+1})。在本文中,我们证明如果(c \le 16\mu ^2 b^3),其中(\mu =\min \{a_1,d_2\}),那么(\{a_1,a_2,b,c\})就是一个二重四元数。把这个结果和我们之前的一个结果结合起来,就意味着如果 \(\{a_i,b,c,d\}\) \((i/in/{1,2,3/})\)是具有 \(a_1<a_2<b<a_3<c<d\) 的二重四次方,那么 \(a_3=b+c-2(sqrt{bc+1})。随即可以得出,不存在一个七元组 \(\{a_1,a_2,a_3,a_4,b,c,d\}) with \(a_1<a_2<;b<a_3<a_4<c<d\) such that \(\{a_i,b,c,d\}\) \((i \in \{1,2,3,4\})\) are Diophantine quadruples.此外,还证明了只有有限多个六次元 \(\{a_1,a_2,a_3,b,c,d\}) with \(a_1<b<a_2<a_3<c<d\) such that \(\{a_i,b,c,d\}) \((i \in \{1,2,3\})\) are Diophantine quadruples.
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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