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arXiv 0803.3956math.GM
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丢番图方程X^2+ bX +c=Y^2与一个问题的解

A solution to a problem and the Diophantine equation X^2+bX+c=Y^2

Konstantine "Hermes" Zelator

摘要:

证明了对于给定的整数b和c,丢番图方程x^2+bx+c=y^2,有n个整数解(即ZxZ中的解对),实际上是偶数个整数解(包括零解或无解的情况),并给出了解集的明确描述.这种描述取决于整数b^2-4c的形式。一些推论确实遵循。进一步证明了当b^2-4c= 1,4,16,-4,或-16时,该方程有两个整数解.在每一种情况下,我们用系数b和c列出两个解。

英文摘要:

We prove that for given integers b and c, the diophantine equation x^2+bx+c=y^2, has finitely many integer solutions(i.e. pairs in ZxZ),in fact an even number of such solutions(including the zero or no solutions case).We also offer an explicit description of the solution set. Such a description depends on the form of the integer b^2-4c. Some Corollaries do follow. Furthermore, we show that the said equation has exactly two integer solutions, precisely when b^2-4c= 1,4,16,-4,or-16. In each case we list the two solutions in terms of the coefficients b and c.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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