可表示为连续正整数平方和的四次方数

wayne

wayne 发表于 2025-5-2 08:52
由 $m^4=\sum _{k=0}^{n-1} (k+p)^2 , 2 p + n - 1 = q_1$,得到$12 m^4-3 n q_1^2=n^3-n$,

这个可以稍微变换一下,  由 $M=m^4=\sum _{k=0}^{n-1} (k+p)^2 , 2 p + n - 1 = q_1$,得到$12 M-3 n q_1^2=n^3-n$,其中$n|6M$
也就是$4 M n-\frac{n^4}{3}+\frac{n^2}{3} = (nq_1)^2=q^2$,  这个四次曲线双有理等价于一个椭圆曲线. $(x,y)\to(-\frac{36M + n}{n}, \frac{108Mq}{n^2})$,  或者$(n,q)\to(-\frac{36M}{x+1},\frac{12 M y}{(x+1)^2})$
对应的椭圆曲线$x^3 + 3888*M^2 + y^2 - 3*x - 2=0$,  四次方就是$M=m^4$,三次方就是$M=m^3$,平方就是$M=m^2$
首先$n|M$,其次根据表达式$x=-\frac{36M + n}{n}$ 得知,x是整数, 所以, 我们就是要求椭圆曲线 $x^3 + 3888*M^2 + y^2 - 3*x - 2=0$ 的整数解. 这就开始变得有趣了.

也就是求方程$Y^2 = X^3+ 3X^2- 3888M^2$的整数解.其中$X=\frac{36M}{n}, Y=\frac{108Mq_1}{n}$,
或者求方程$Y^2 = X^3+ \frac{X^2 }{12}- \frac{M^2 }{12}$的有理解,其中$X=\frac{M}{n}, Y=\frac{Mq_1}{2n}$, 因为$n|6M$,所以需要$6X$是整数

Mathematica验证均满足恒等关系.

1. Factor[-3888 M^2 + 3 X^2 + X^3 - y^2 /. Thread[{X, y} -> {36 M/n, 108 M q/n}]]
   
2. Factor[-(M^2/12)+x^2/12+x^3-y^2/.Thread[{x,y}->{M/n,1/2M q/n}]]
   

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wayne

比如计算m=143^2的情况

1. m=143^2;
   
2. E=ellinit([0,1/12,0,0,-m^2/12]);
   
3. ellratpoints(E,[10^8,6])
   

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得到这么多有理解
格式是n,a,b, 这里的n,a,b都成了有理数.

1. {33,-39,-7}
   
2. {33,7,39}
   
3. {11,-48,-38}
   
4. {11,38,48}
   
5. {121/2,-(71/2),24}
   
6. {121/2,-24,71/2}
   
7. {1,-143,-143}
   
8. {1,143,143}
   
9. {1014/19,-(727/19),268/19}
   
10. {1014/19,-(268/19),727/19}
    
11. {99/26,-(5825/78),-(2803/39)}
    
12. {99/26,2803/39,5825/78}
    
13. {676/109,-(13079/218),-(11945/218)}
    
14. {676/109,11945/218,13079/218}
    
15. {338/181,-(19019/181),-(18862/181)}
    
16. {338/181,18862/181,19019/181}
    
17. {3718/427,-(22311/427),-(19020/427)}
    
18. {3718/427,19020/427,22311/427}
    
19. {81796/1321,-(89169/2642),71781/2642}
    
20. {81796/1321,-(71781/2642),89169/2642}
    
21. {122694/8263,-(361814/8263),-(247383/8263)}
    
22. {122694/8263,247383/8263,361814/8263}
    
23. {363/2702,-(1053001/2702),-(527670/1351)}
    
24. {363/2702,527670/1351,1053001/2702}

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再比如 109^3,得到
{{218, -153, 64}, {218, -64, 153}}

而 13^4 得到的解

1. {52,-(87/2),15/2}
   
2. {52,-(15/2),87/2}
   
3. {2,-120,-119}
   
4. {2,119,120}
   
5. {1,-169,-169}
   
6. {1,169,169}
   
7. {13182/367,-(16032/367),-(3217/367)}
   
8. {13182/367,3217/367,16032/367}
   
9. {85683/14702,-(532206/7351),-(993431/14702)}
   
10. {85683/14702,993431/14702,532206/7351}
    
11. {85683/499394,-(101773177/249697),-(203960065/499394)}
    
12. {85683/499394,203960065/499394,101773177/249697}

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只可惜m不能太大, 因为四次方就是m^8.这蹭蹭的往上涨.

m=2026的时候,

1. {3/2,-37,-(73/2)}
   
2. {3/2,73/2,37}
   
3. {4052/211,-(3746/211),95/211}
   
4. {4052/211,-(95/211),3746/211}
   
5. {961/91,-(51609/2821),-(24639/2821)}
   
6. {961/91,24639/2821,51609/2821}
   
7. {58081/2149,-(8689644/517909),4789968/517909}
   
8. {58081/2149,-(4789968/517909),8689644/517909}

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1. cases="[[184366/961, 77239224/29791], [184366/961, -77239224/29791], [4353874/58081, 3950371788/13997521], [4353874/58081, -3950371788/13997521], [211/2, 3651/4], [211/2, -3651/4], [4052/3, 49637], [4052/3, -49637]]";
   
2. SortBy[Block[{M=2026},Table[{n,(q-n+1)/2,(q+n-1)/2}/.Thread[{n,q}->({M/x,(2 y)/x}/.Thread[{x,y}->xy])],{xy,Union[Partition[ToExpression[StringCases[cases,RegularExpression["[\\d|\\-|/]+"]]],2]]}]],Max[Max@@Numerator[Abs[#]],Max@@Denominator[Abs[#]]]&]//Column

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