x = (-28 (a + b + 2 c))/(6 a + 6 b - c), \quady = (364 (a - b))/(
6 a + 6 b - c)这个二元变换 ...
https://www.math.tamu.edu/~rojas/cubic2weierstrass.pdf
Transforming a general cubic elliptic curve equation to Weierstrass form
95% 的人解不出这道题《史上最贱的数学题》能否用数学软件计算出来？ - Nemesis XX的回答 - 知乎
https://www.zhihu.com/question/267427508/answer/323883323
如何得到那个变换，答案也许在这里面！
nyy 发表于 2023-3-28 16:01
https://www.math.tamu.edu/~rojas/cubic2weierstrass.pdf
Transforming a general cubic elliptic cu ...
数学中最著名的曲线和曲面分别是什么？ - 刘醉白的回答 - 知乎
https://www.zhihu.com/question/23697239/answer/2961227001
这个带图片的回答，似乎还不错！ nyy 发表于 2023-4-3 08:08
数学中最著名的曲线和曲面分别是什么？ - 刘醉白的回答 - 知乎
https://www.zhihu.com/question/2369723 ...
Clear["Global`*"];(*删除所有变量*)
(*通分,合并同类项,取得分子*)
f=a/(b+c)+b/(a+c)+c/(a+b)-4//Together//Numerator
(*找到适合方程的整数解*)
Do,Print[{a,b,c}]],{a,-12,12},{b,-12,12},{c,-12,12}]
先找部分整数解
{-11,-9,5}
{-11,-4,1}
{-11,1,-4}
{-11,5,-9}
{-9,-11,5}
{-9,5,-11}
{-5,9,11}
{-5,11,9}
{-4,-11,1}
{-4,1,-11}
{-1,4,11}
{-1,11,4}
{1,-11,-4}
{1,-4,-11}
{4,-1,11}
{4,11,-1}
{5,-11,-9}
{5,-9,-11}
{9,-5,11}
{9,11,-5}
{11,-5,9}
{11,-1,4}
{11,4,-1}
{11,9,-5}
先搞部分解再说，这只是一小步！ https://link.zhihu.com/?target=http%3A//ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf
看不懂
An unusual cubic representation problem
但是转发过来，希望有人能看懂
补充内容 (2023-4-4 09:27):
https://ami.uni-eszterhazy.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf nyy 发表于 2023-4-3 15:10
https://link.zhihu.com/?target=http%3A//ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf
...
The general procedure to transform a genus one cubic projective curve into Weierstrass form is outlined in the book by Silverman/Tate "Rational points on elliptic curves", Chapter I, Section 3, pages 22-23.
Some software packages can do this for you, for instance, Magma. There is an online Magma calculator http://magma.maths.usyd.edu.au/calc/ which is free to use as long as the calculation does not go over 120 seconds. Here is some code that verifies the equations in the paper.
F<N>:=FunctionField(Rationals());
PS<a,b,c>:=ProjectiveSpace(F,2);
C:=Curve(PS, N*(a+b)*(b+c)*(c+a)-a*(a+b)*(a+c)-b*(b+c)*(b+a)-c*(c+a)*(c+b));
P0:=C!;
F, CtoF:=EllipticCurve(C,P0);
E<X,Y,Z>:=EllipticCurve();
test,FtoE:=IsIsomorphic(F,E);
assert test eq true;
CtoE:=Expand(CtoF*FtoE);
CtoE;
Good wishes,
Andrew Bremner
这个作者给我的回信！
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