如何找到一般双四次曲线的有理点
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 用eclib的工具来搜索,跑了一晚上.找到的点是$=$, 用到的算法应该是 二次筛法
cc@Mini:~/mathematics/eclib/progs$ ./quartic_points
eclib version 20250530, using NTL bigints and NTL real and complex multiprecision floating point
Verbose? 1
Enter quartic coefficients a,b,c,d,e ?
18422369 -81234596 29670080 102943032 -193447548
Limit on height? 18
I = -16797224843002928, J = 19297954620658708000436864
Minimal model for Jacobian:
Checking local solublity in R:
Checking local solublity at primes [ 2 3 17 281 433 3359 3457 4337 4993 ]:
new_qpsoluble with p<1000 passing to old qpsoluble.
new_qpsoluble with p<1000 passing to old qpsoluble.
new_qpsoluble with p<1000 passing to old qpsoluble.
new_qpsoluble with p<1000 passing to old qpsoluble.
Using new_qpsoluble with p = 3359
---------------------------------------------
LOCAL_SOL
-193447548 102943032 29670080 -81234596 18422369 p=3359
f is not 0 mod p: Case I
Factorization of f = [[ 1] [ 1] [ 2]]
Using new_qpsoluble with p = 3457
---------------------------------------------
LOCAL_SOL
-193447548 102943032 29670080 -81234596 18422369 p=3457
f is not 0 mod p: Case I
Using new_qpsoluble with p = 4337
---------------------------------------------
LOCAL_SOL
-193447548 102943032 29670080 -81234596 18422369 p=4337
f is not 0 mod p: Case I
Using new_qpsoluble with p = 4993
---------------------------------------------
LOCAL_SOL
-193447548 102943032 29670080 -81234596 18422369 p=4993
f is not 0 mod p: Case I
Factorization of f = [[ 1] [ 2]]
Everywhere locally soluble.
Searching for points on (18422369,-81234596,29670080,102943032,-193447548) up to height 18
Entering qsieve::search: y^2 = 18422369x^4 + -81234596x^3 + 29670080x^2 + 102943032x^1 + -193447548
Using speed ratios 1000 and 6.5
8 primes used for first stage of sieving
52 primes used for both stages of sieving together.
Sieving primes:
First stage: 13, 31, 53, 107, 29, 179, 67, 229,
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,
Probabilities: Min(13) = 0.3609467456, Cut1(229) = 0.4541484716, Cut2(7) = 0.7551020408, Max(17) = 1
Forbidden divisors of the denominator:
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,
Try to find the points up to height 65659969
x = 57878913/12642040 gives a rational point.
(x:y:z) = (57878913:5531870046610133823:12642040)
Point =
height = 83.99200507
Curve =
Point =
height = 83.99200507
Enter quartic coefficients a,b,c,d,e ? 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) $
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