周长、面积相等的本原三角形

wayne

我不确定是否要捡起 之前的椭圆曲线的方法,  因为之前的椭圆曲线的方法有一个很好的特征,就是 给定周长和面积, 唯一确定一条椭圆曲线, 对于该椭圆曲线, 其秩基本上跟解的个数 相当. 周长L,面积S对应的椭圆曲线是 $Y^2= -\frac{1}{48} L^4 X \left(L^4-24 S^2\right)+\frac{1}{864} L^4 \left(L^8-36 L^4 S^2+216 S^4\right)+X^3$
比如,拿那个8组解的为例, $L=721786, S=13338605280$

1. E=ellinit([-12693394974858489906001/3,2855319653729016837546676404344498/27])
   
2. %11 = [0, 0, 0, -12693394974858489906001/3, 2855319653729016837546676404344498/27, 0, -25386789949716979812002/3, 11421278614916067350186705617377992/27, -161122275987762763592862656596653605815812001/9, 203094319597735838496016, -91370228919328538801493644939023936, 16535957613113937235108867376892012004454523086428107177984000000, 120564527471242839032838375966526083395690401/237988279160426028378126972616704000000, Vecsmall([1]), [Vecsmall([128, 1])], [0, 0, 0, 0, 0, 0, 0, 0]]
   
3. 
   
4. ellrank(E,1)
   
5. %12 = [7, 7, 0, [[-166027186751/3, 13054191586359000], [-32496620351/3, 12260263348733400], [93823415689/3, 2003871483863040], [320862387547/9, 12679875082388800/27], [11173263416731/9, 37297965934570781440/27], [72976429912633/1875, 2789962258736068416/15625], [1095660492472393/28227, 17053651951448555520/912673]]]
   

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