解丢番图方程x^2+y^2+z^2=3xyz

hujunhua

1. f[x_] := {{x[[1]], x[[3]], 3 x[[1]] x[[3]] - x[[2]]}, {x[[2]], x[[3]],3 x[[2]] x[[3]] - x[[1]]}};
2. g[x_] := Flatten[f /@ x, 1];
3. h[x_] := Select[g[x], #[[3]] < 10^10 &]
4. Answer = NestWhileList[h, {{1, 2, 5}}, Length[#] > 0 &];
5. Flatten[Answer, 1]
6. (Flatten[Answer, 1] // Length) + 2

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1. {{1, 2, 5}, {1, 5, 13}, {2, 5, 29}, {1, 13, 34}, {5, 13, 194}, {2, 29,169}, {5, 29, 433}, {1, 34, 89}, {13, 34, 1325}, {5, 194, 2897}, {13, 194, 7561}, {2, 169, 985},{29,169,14701}, {5, 433, 6466}, {29, 433, 37666}, {1, 89, 233}, {34, 89, 9077}, {13, 1325, 51641}, {34, 1325, 135137}, {5, 2897, 43261}, {194, 2897, 1686049},{13,7561,294685}, {194, 7561, 4400489}, {2, 985, 5741}, {169, 985, 499393}, {29, 14701, 1278818}, {169, 14701, 7453378}, {5, 6466, 96557}, {433, 6466, 8399329}, {29, 37666, 3276509}, {433, 37666, 48928105}, {1, 233, 610}, {89, 233, 62210}, {34, 9077, 925765}, {89, 9077, 2423525}, {13, 51641, 2012674}, {1325,51641,205272962}, {34, 135137, 13782649}, {1325, 135137, 537169541}, {5, 43261, 646018}, {2897, 43261, 375981346}, {194, 1686049, 981277621}, {13,294685, 11485154}, {7561, 294685, 6684339842}, {194, 4400489,2561077037}, {2, 5741, 33461}, {985, 5741, 16964653}, {169, 499393, 253191266}, {985,499393, 1475706146}, {29, 1278818, 111242465}, {169, 7453378, 3778847945}, {5, 96557, 1441889}, {6466,96557, 1873012681}, {29, 3276509, 285018617}, {1,610, 1597}, {233,610, 426389}, {89, 62210, 16609837}, {233, 62210, 43484701}, {34,925765, 94418953}, {89, 2423525, 647072098}, {13, 2012674, 78442645}, {34,13782649, 1405695061}, {5, 646018, 9647009}, {13,11485154, 447626321}, {2, 33461, 195025}, {5741, 33461,576298801}, {29, 111242465, 9676815637}, {5,1441889, 21531778}, {1,1597, 4181}, {610, 1597, 2922509}, {233, 426389, 298045301}, {610,426389, 780291637}, {89, 16609837, 4434764269}, {34,94418953,9629807441}, {13, 78442645, 3057250481}, {5, 9647009,144059117}, {2, 195025, 1136689}, {5, 21531778, 321534781}, {1, 4181, 10946}, {1597,4181,20031170}, {610, 2922509,5348189873}, {5, 144059117, 2151239746}, {2, 1136689, 6625109}, {5, 321534781, 4801489937}, {1, 10946, 28657}, {4181,10946, 137295677}, {2, 6625109, 38613965}, {1, 28657, 75025}, {10946,28657, 941038565}, {2, 38613965, 225058681}, {1, 75025,196418}, {28657,75025,6449974274}, {2, 225058681, 1311738121}, {1,196418, 514229}, {2, 1311738121, 7645370045}, {1, 514229,1346269}, {1, 1346269, 3524578}, {1,3524578, 9227465}, {1, 9227465,24157817}, {1, 24157817, 63245986}, {1, 63245986, 165580141}, {1, 165580141, 433494437}, {1, 433494437, 1134903170}, {1,1134903170, 2971215073}, {1, 2971215073, 7778742049}}
2. 107

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10^10以内的解除{1,1,1}和{1，1，2}外都在此处了，共计107个。

hujunhua

由于对称性，可以定义满足方程的 x 的递增序列，它将包含方程的解集｛{x,y,z}｝中的所有数值。这个序列称为马尔科夫数（Markov numbers）,见A002559： {1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610, 985, 1325, 1597, 2897, 4181, 5741, 6466, 7561, 9077,...}

数学星空

http://en.wikipedia.org/wiki/Markov_number 对于这类不定方程最有趣的是如何构造递推解 例：$x^2+y^2+z^2+u^2=4*x*y*z*u$ 的递推解如何？

chyanog

Mathematica里面取下标[[]]看着有点碍眼，我更喜欢用

1. f[n_] := Block[{x, y, z}, {x, y, z} = n; {{x, z, 3 x z - y}, {y, z, 3 y z - x}}];

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wayne

24# chyanog 最好对n做一下模式判定

wayne

23# 数学星空 根据mathe的分析，我们可以这么递推， 如果存在解 {x,y,z,t},x<=y<=z<=t , 那么，那么必然同时存在解 {x,y,z,4xyz-t} 从{1,1,1,1},我们可以推出 {1,1,1,3} 从{1,1,1,3},我们推出{1,1,3,11} 从{1,1,3,11},我们推出{1, 3, 11, 131},{1, 1, 11, 41} ...... 继续同理，我们可以找到数列：http://oeis.org/A075276 不过该数列貌似第16项开始就不对了，应该是29681。 {1, 3, 11, 41, 131, 153, 571, 1561, 1803, 2131, 5761, 7953, 17291, \ 18601, 25091, 29681, 79291, 110771, 221651, 253353, 295681, 349451, \ 413403, 817961, 2282281, 2453891, 2641211, 3018753, 3252611, 3487001, \ 4114771, 4867203}

wayne

21# hujunhua 我也写了程序：

2. Union@Flatten[
3. NestList[Union@Sort@Flatten[Table[Union[
4. Map[Sort, {x, y, z, 4 x y z - t} /. Thread[Rule[{x, y, z, t}, #]] & /@
5. NestList[RotateLeft, ii, 4]]], {ii, #}], 1] &, {{1, 1, 1, 1}}, 5], 1]

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2. 1 1 1 1
3. 1 1 1 3
4. 1 1 3 11
5. 1 1 11 41
6. 1 1 41 153
7. 1 1 153 571
8. 1 1 571 2131
9. 1 3 11 131
10. 1 3 131 1561
11. 1 3 1561 18601
12. 1 3 18601 221651
13. 1 11 41 1803
14. 1 11 131 5761
15. 1 11 1803 79291
16. 1 11 5761 253353
17. 1 11 79291 3487001
18. 1 11 253353 11141771
19. 1 41 153 25091
20. 1 41 1803 295681
21. 1 41 25091 4114771
22. 1 41 295681 48489881
23. 1 131 1561 817961
24. 1 131 5761 3018753
25. 1 131 817961 428610003
26. 1 131 3018753 1581820811
27. 1 153 571 349451
28. 1 153 25091 15355651
29. 1 1561 18601 116144641
30. 1 1561 817961 5107348353
31. 1 1803 79291 571846681
32. 1 1803 295681 2132451331
33. 1 5761 253353 5838266521
34. 1 5761 3018753 69564144001
35. 3 11 131 17291
36. 3 11 17291 2282281
37. 3 11 2282281 301243801
38. 3 131 1561 2453891
39. 3 131 17291 27181441
40. 3 131 2453891 3857515091
41. 3 131 27181441 42729207961
42. 3 1561 18601 348433931
43. 3 1561 2453891 45966286081
44. 3 17291 2282281 473555049241
45. 3 17291 27181441 5639931555841
46. 11 41 1803 3252611
47. 11 41 3252611 5867708441
48. 11 131 5761 33206403
49. 11 131 17291 99665321
50. 11 131 33206403 191401701131
51. 11 131 99665321 574470892953
52. 11 1803 79291 6290313611
53. 11 1803 3252611 258036135811
54. 11 5761 253353 64220931851
55. 11 5761 33206403 8417291857921

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解的个数呈指数膨胀，迭代10次，得到4927组

mathe

23# 数学星空 根据mathe的分析，我们可以这么递推， 如果存在解 {x,y,z,t},x wayne 发表于 2013-2-23 13:51

这样无法保证得到所有的解。无穷递降后不一定必然可以变换到全1，我们需要穷举一定范围内的解

wayne

28# mathe 恩，是的

mathe

为了找到所有解，我们先查看一般有两个正根的二次方程 $x^2-bx+c=0$以及两根$x_1<=x_2$,于是$x_1(x_1+x_2)<=2x_1x_2$,得出$x_1<={2c}/b$