如何找到一般双四次曲线的有理点

wayne

如何找到 $y^2=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$的有理点, 已知$a_i$是整数,且$a_0,a_4$都不是完全平方数.
如果$a_0$或者$a_4$是完全平方数的话,可以双有理变换到$y^2=x^3+a_2x^2+a_1x+a_0$,网上很容易搜得到方法.


咱们就拿$y^2=18422369 x^4-81234596 x^3+29670080 x^2+102943032 x-193447548$ 练练手, 可以确定存在有理数点.
这个曲线是: $w^2=f(u,v)=u^4 v^4-4 u^4 v^3-8 u^4 v+4 u^4-4 u^3 v^4+8 u^3 v^3-16 u^3 v^2+48 u^3 v-16 u^3-16 u^2 v^3+48 u^2 v^2-64 u^2 v-8 u v^4+48 u v^3-64 u v^2+32 u v+32 u+4 v^4-16 v^3+32 v-48$
在$u=-5/44$化简而来

wayne

用eclib的工具来搜索,跑了一晚上.找到的点是$[x,y]=[57878913/12642040,5531870046610133823/159821175361600]$, 用到的算法应该是 二次筛法

1. cc@Mini:~/mathematics/eclib/progs$ ./quartic_points
   
2. eclib version 20250530, using NTL bigints and NTL real and complex multiprecision floating point
   
3. Verbose? 1
   
4. Enter quartic coefficients a,b,c,d,e ?
   
5. 18422369 -81234596 29670080 102943032 -193447548
   
6. Limit on height? 18
   
7. I = -16797224843002928, J = 19297954620658708000436864
   
8. Minimal model for Jacobian: [0,-1,0,349942184229228,-11167797929528591502588]
   
9. Checking local solublity in R:
   
10. Checking local solublity at primes [ 2 3 17 281 433 3359 3457 4337 4993 ]:
    
11. new_qpsoluble with p<1000 passing to old qpsoluble.
    
12. new_qpsoluble with p<1000 passing to old qpsoluble.
    
13. new_qpsoluble with p<1000 passing to old qpsoluble.
    
14. new_qpsoluble with p<1000 passing to old qpsoluble.
    
15. Using new_qpsoluble with p = 3359
    
16. ---------------------------------------------
    
17. LOCAL_SOL
    
18. -193447548 102943032 29670080 -81234596 18422369      p=3359
    
19. f is not 0 mod p: Case I
    
20. Factorization of f = [[[717 1] 1] [[922 1] 1] [[1636 1] 2]]
    
21. Using new_qpsoluble with p = 3457
    
22. ---------------------------------------------
    
23. LOCAL_SOL
    
24. -193447548 102943032 29670080 -81234596 18422369      p=3457
    
25. f is not 0 mod p: Case I
    
26. Using new_qpsoluble with p = 4337
    
27. ---------------------------------------------
    
28. LOCAL_SOL
    
29. -193447548 102943032 29670080 -81234596 18422369      p=4337
    
30. f is not 0 mod p: Case I
    
31. Using new_qpsoluble with p = 4993
    
32. ---------------------------------------------
    
33. LOCAL_SOL
    
34. -193447548 102943032 29670080 -81234596 18422369      p=4993
    
35. f is not 0 mod p: Case I
    
36. Factorization of f = [[[370 523 1] 1] [[1371 1] 2]]
    
37. Everywhere locally soluble.
    
38. Searching for points on (18422369,-81234596,29670080,102943032,-193447548) up to height 18
    
39. Entering qsieve::search: y^2 = 18422369x^4 + -81234596x^3 + 29670080x^2 + 102943032x^1 + -193447548
    
40. Using speed ratios 1000 and 6.5
    
41. 8 primes used for first stage of sieving
    
42. 52 primes used for both stages of sieving together.
    
43. Sieving primes:
    
44. First stage: 13, 31, 53, 107, 29, 179, 67, 229,
    
45. Second stage: 191, 211, 109, 103, 239, 227, 149, 137, 181, 173, 157, 83, 241, 79, 199, 163, 73, 251, 61, 5, 131, 59, 223, 113, 233, 193, 167, 139, 101, 151, 127, 41, 197, 47, 97, 37, 89, 71, 23, 43, 3, 11, 19, 7,
    
46. Probabilities: Min(13) = 0.3609467456, Cut1(229) = 0.4541484716, Cut2(7) = 0.7551020408, Max(17) = 1
    
47. 
    
48. Forbidden divisors of the denominator:
    
49. 3, 49, 29, 31, 37, 41, 43, 59, 71, 79, 107, 113, 127, 137, 151, 163, 181, 191, 193, 197, 199, 227, 229, 241, 251,
    
50. 
    
51. Try to find the points up to height 65659969
    
52. x = 57878913/12642040 gives a rational point.
    
53. (x:y:z) = (57878913:5531870046610133823:12642040)
    
54. Point = [501719473680977804952504580457666799500891138875111852557260236:4561260782270963696842009762159238536332935102669773182469489577190:6269777709190453801465699995883947113443032275408248621]
    
55.         height = 83.99200507
    
56. Curve = [0,-1,0,349942184229228,-11167797929528591502588]
    
57. Point = [501719473680977804952504580457666799500891138875111852557260236:4561260782270963696842009762159238536332935102669773182469489577190:6269777709190453801465699995883947113443032275408248621]
    
58. height = 83.99200507
    
59. Enter quartic coefficients a,b,c,d,e ?

复制代码

wayne

基本思路就是将四次曲线转化成 $y^2 = g(x)^2 + h(x)$的形式,其中$g(x)$是二次函数,$h(x)$是一次多项式. 然后再通过$T = y+g(x), S=x(y+g(x))$ 变换.
不过,不管有没有有理点,四次曲线$y^2=a*x^4 + b*x^3 + c*x^2 + d*x + f$ 都可以从代数的层面 转换成椭圆曲线的Weierstrass形式: $ Y^2 = X^3+ 1/3 (-c^2 + 3 b d - 12 a f) X +1/27 (2 c^3 - 9 b c d + 27 a d^2 + 27 b^2 f - 72 a c f) $