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On the extension of \(D(-8k^2)\)-pair \(\{8k^2, 8k^2 +1\}\). (English) Zbl 1433.11028

Summary: Let \(k\) be a positive integer. The triple \(\{1, 8k^2, 8k^2 + 1\}\) has the property that the product of any two of its distinct elements subtracted by \(8k^2\) is a perfect square. By elementary means, we show that this triple can be extended to at most a quadruple retaining this property, i.e., if \(\{1, 8k^2, 8k^2 + 1, d\}\) has the same property, then \(d\) is uniquely determined \((d = 32k^2 + 1)\). Moreover, we show that even the pair \(\{8k^2, 8k^2 + 1\}\) can be extended in the same manner to at most a quadruple (the third and fourth element can only be \(1\) and \(32k^2 + 1)\). At the end, we suggest considering a similar problem of extending the triple \(\{1, 2k^2, 2k^2 + 2k + 1\}\) with a similar property as possible future research direction.

MSC:

11D09 Quadratic and bilinear Diophantine equations
11A99 Elementary number theory
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