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Izvje??a sa znanstvenih skupova - pregledi

机译:科学会议报告 - 评论

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Cassels [2] was challenged to determine when the sum of three consecutivecubes equals a square. He [2] reduced the problem to finding integral points on theelliptic curve y2 = 3x(x2+2). Using the arithmetic of certain quartic number fields,he obtained that the integral points on the above elliptic curve were (x, y) = (0, 0),(1, 3), (2, 6), and (24, 204).Using the classical work of Ljunggren [5] and its generalizations (see [1,4,10,11]), Luca and Walsh [6] considered the problem of finding the number of positiveinteger solutions to the Diophantine equation y2 = nx(x2 + 2), where n 1 isa positive integer. They proved that the number of positive integer solutions toy2 = nx(x2 + 2) is at most 3 · 2ω(n) ? 1, where ω(n) is the number of distinct primefactors of n. In [3], Chen considered the case where n is a prime number greaterthan 3. He proved, in particular, that the Diophantine equation y2 = nx(x2 + 2)has at most two positive integer solutions.
机译:Cassels [2]受到挑战,以确定三个连续内容的总和等于广场。他[2]减少了在典型曲线Y2 = 3x(x2 + 2)上找到整体点的问题。使用某些四个字段的算法,他获得了上述椭圆曲线上的积分点(x,y)=(0,0),(1,3),(2,6),(24,204 )。在Ljunggren [5]的经典作品及其概括(见[1,4,10,11]),Luca和Walsh [6]考虑了发现对衍生释氨酸方程Y2 = NX的积极体解数的问题(x2 + 2),其中n> 1是正整数。他们证明了正整数解决方案Toy2 = NX(X2 + 2)的数量最多为3·2Ω(n)? 1,其中ω(n)是n的不同初步镜的数量。在[3]中,陈考虑了n是素数大的情况。特别是,特别是二子antine方程Y2 = NX(X2 + 2)在最多两个正整数解决方案中。

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    《Geoadria》 |2017年第1期|共6页
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    Josip Ri?anovi?;

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  • 入库时间 2022-08-19 00:10:04

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