斜率为整数的直线
已直线`y=k_ix,(i=1,2,3)`的斜率`k_i`都是正整数，并且2是1,3的角平分线。试求`(k_1,k_2,k_3)`.易知直线$y=k_1x,y=k_2x$之间的夹角为$\arctan\left(\frac{k_1-k_2}{k_1k_2+1}\right)$
所以，不妨设 $k_1 > k_2 >k_3 >0$ 解方程
sol=Values@Solve[(k1-k2)/(1+k1 k2)==(k2-k3)/(1+k2 k3)&&10^7>k1>k2>k3>0,{k1,k2,k3},Integers];{Length@sol,sol}
返回：
{163,{{7,2,1},{38,4,2},{41,12,7},{117,6,3},{239,70,41},{268,8,4},{515,10,5},{682,72,38},{882,12,6},{1393,14,7},{1393,408,239},{2072,16,8},{2943,18,9},{4030,20,10},{4443,228,117},{5357,22,11},{6948,24,12},{8119,2378,1393},{8827,26,13},{11018,28,14},{12238,1292,682},{13545,30,15},{16432,32,16},{17684,528,268},{19703,34,17},{23382,36,18},{27493,38,19},{32060,40,20},{37107,42,21},{42658,44,22},{47321,13860,8119},{48737,46,23},{52525,1020,515},{55368,48,24},{62575,50,25},{70382,52,26},{78813,54,27},{87892,56,28},{97643,58,29},{108090,60,30},{119257,62,31},{128766,1752,882},{131168,64,32},{143847,66,33},{157318,68,34},{168717,8658,4443},{171605,70,35},{186732,72,36},{202723,74,37},{219602,76,38},{219602,23184,12238},{237393,78,39},{256120,80,40},{275807,82,41},{275807,2772,1393},{275807,80782,47321},{296478,84,42},{318157,86,43},{340868,88,44},{364635,90,45},{389482,92,46},{415433,94,47},{442512,96,48},{470743,98,49},{500150,100,50},{530757,102,51},{534568,4128,2072},{562588,104,52},{595667,106,53},{630018,108,54},{665665,110,55},{702632,112,56},{740943,114,57},{780622,116,58},{821693,118,59},{864180,120,60},{908107,122,61},{953498,124,62},{959409,5868,2943},{1000377,126,63},{1048768,128,64},{1098695,130,65},{1150182,132,66},{1166876,34840,17684},{1203253,134,67},{1257932,136,68},{1314243,138,69},{1372210,140,70},{1431857,142,71},{1493208,144,72},{1556287,146,73},{1607521,470832,275807},{1620050,8040,4030},{1621118,148,74},{1687725,150,75},{1756132,152,76},{1826363,154,77},{1898442,156,78},{1972393,158,79},{2048240,160,80},{2126007,162,81},{2205718,164,82},{2287397,166,83},{2371068,168,84},{2456755,170,85},{2544482,172,86},{2603491,10692,5357},{2634273,174,87},{2726152,176,88},{2820143,178,89},{2916270,180,90},{3014557,182,91},{3115028,184,92},{3217707,186,93},{3322618,188,94},{3429785,190,95},{3539232,192,96},{3650983,194,97},{3765062,196,98},{3881493,198,99},{3940598,416020,219602},{4000300,200,100},{4015932,13872,6948},{4121507,202,101},{4245138,204,102},{4371217,206,103},{4499768,208,104},{4630815,210,105},{4764382,212,106},{4900493,214,107},{5039172,216,108},{5180443,218,109},{5324330,220,110},{5357035,104030,52525},{5470857,222,111},{5620048,224,112},{5771927,226,113},{5926518,228,114},{5984693,17628,8827},{6083845,230,115},{6243932,232,116},{6406803,234,117},{6406803,328776,168717},{6572482,236,118},{6740993,238,119},{6912360,240,120},{7086607,242,121},{7263758,244,122},{7443837,246,123},{7626868,248,124},{7812875,250,125},{8001882,252,126},{8193913,254,127},{8388992,256,128},{8587143,258,129},{8660134,22008,11018},{8788390,260,130},{8992757,262,131},{9200268,264,132},{9369319,2744210,1607521},{9410947,266,133},{9624818,268,134},{9841905,270,135}}}
注意到：
k1={1393, 219602, 6406803}分别有两组解
{1393,14,7},{1393,408,239}
{219602,76,38},{219602,23184,12238}
{6406803,234,117},{6406803,328776,168717}
k1={275807}有三组解
{275807,82,41},{275807,2772,1393},{275807,80782,47321} 本帖最后由 northwolves 于 2025-3-20 18:07 编辑
突然发现k2对于任意正偶数都有解。
sol=Flatten,{k,2,200,2}],1]
{{7,2,1},{38,4,2},{117,6,3},{268,8,4},{515,10,5},{41,12,7},{882,12,6},{1393,14,7},{2072,16,8},{2943,18,9},{4030,20,10},{5357,22,11},{6948,24,12},{8827,26,13},{11018,28,14},{13545,30,15},{16432,32,16},{19703,34,17},{23382,36,18},{27493,38,19},{32060,40,20},{37107,42,21},{42658,44,22},{48737,46,23},{55368,48,24},{62575,50,25},{70382,52,26},{78813,54,27},{87892,56,28},{97643,58,29},{108090,60,30},{119257,62,31},{131168,64,32},{143847,66,33},{157318,68,34},{239,70,41},{171605,70,35},{682,72,38},{186732,72,36},{202723,74,37},{219602,76,38},{237393,78,39},{256120,80,40},{275807,82,41},{296478,84,42},{318157,86,43},{340868,88,44},{364635,90,45},{389482,92,46},{415433,94,47},{442512,96,48},{470743,98,49},{500150,100,50},{530757,102,51},{562588,104,52},{595667,106,53},{630018,108,54},{665665,110,55},{702632,112,56},{740943,114,57},{780622,116,58},{821693,118,59},{864180,120,60},{908107,122,61},{953498,124,62},{1000377,126,63},{1048768,128,64},{1098695,130,65},{1150182,132,66},{1203253,134,67},{1257932,136,68},{1314243,138,69},{1372210,140,70},{1431857,142,71},{1493208,144,72},{1556287,146,73},{1621118,148,74},{1687725,150,75},{1756132,152,76},{1826363,154,77},{1898442,156,78},{1972393,158,79},{2048240,160,80},{2126007,162,81},{2205718,164,82},{2287397,166,83},{2371068,168,84},{2456755,170,85},{2544482,172,86},{2634273,174,87},{2726152,176,88},{2820143,178,89},{2916270,180,90},{3014557,182,91},{3115028,184,92},{3217707,186,93},{3322618,188,94},{3429785,190,95},{3539232,192,96},{3650983,194,97},{3765062,196,98},{3881493,198,99},{4000300,200,100}} 我晕，有一个通解：
{4 n^3 + 3 n, 2 n, n} 另外，k3任意正整数似乎都有解：
Table[{n,
Values@Solve[{k3 ==n, (k1 - k2)/(1 + k1 k2) == (k2 - k3)/(1 + k2 k3) && k1 > k2 > k3 > 0}, {k1, k2, k3}, Integers]}, {n, 100}] // Grid
1 {{7,2,1}}
2 {{38,4,2}}
3 {{117,6,3}}
4 {{268,8,4}}
5 {{515,10,5}}
6 {{882,12,6}}
7 {{41,12,7},{1393,14,7}}
8 {{2072,16,8}}
9 {{2943,18,9}}
10 {{4030,20,10}}
11 {{5357,22,11}}
12 {{6948,24,12}}
13 {{8827,26,13}}
14 {{11018,28,14}}
15 {{13545,30,15}}
16 {{16432,32,16}}
17 {{19703,34,17}}
18 {{23382,36,18}}
19 {{27493,38,19}}
20 {{32060,40,20}}
21 {{37107,42,21}}
22 {{42658,44,22}}
23 {{48737,46,23}}
24 {{55368,48,24}}
25 {{62575,50,25}}
26 {{70382,52,26}}
27 {{78813,54,27}}
28 {{87892,56,28}}
29 {{97643,58,29}}
30 {{108090,60,30}}
31 {{119257,62,31}}
32 {{131168,64,32}}
33 {{143847,66,33}}
34 {{157318,68,34}}
35 {{171605,70,35}}
36 {{186732,72,36}}
37 {{202723,74,37}}
38 {{682,72,38},{219602,76,38}}
39 {{237393,78,39}}
40 {{256120,80,40}}
41 {{239,70,41},{275807,82,41}}
42 {{296478,84,42}}
43 {{318157,86,43}}
44 {{340868,88,44}}
45 {{364635,90,45}}
46 {{389482,92,46}}
47 {{415433,94,47}}
48 {{442512,96,48}}
49 {{470743,98,49}}
50 {{500150,100,50}}
51 {{530757,102,51}}
52 {{562588,104,52}}
53 {{595667,106,53}}
54 {{630018,108,54}}
55 {{665665,110,55}}
56 {{702632,112,56}}
57 {{740943,114,57}}
58 {{780622,116,58}}
59 {{821693,118,59}}
60 {{864180,120,60}}
61 {{908107,122,61}}
62 {{953498,124,62}}
63 {{1000377,126,63}}
64 {{1048768,128,64}}
65 {{1098695,130,65}}
66 {{1150182,132,66}}
67 {{1203253,134,67}}
68 {{1257932,136,68}}
69 {{1314243,138,69}}
70 {{1372210,140,70}}
71 {{1431857,142,71}}
72 {{1493208,144,72}}
73 {{1556287,146,73}}
74 {{1621118,148,74}}
75 {{1687725,150,75}}
76 {{1756132,152,76}}
77 {{1826363,154,77}}
78 {{1898442,156,78}}
79 {{1972393,158,79}}
80 {{2048240,160,80}}
81 {{2126007,162,81}}
82 {{2205718,164,82}}
83 {{2287397,166,83}}
84 {{2371068,168,84}}
85 {{2456755,170,85}}
86 {{2544482,172,86}}
87 {{2634273,174,87}}
88 {{2726152,176,88}}
89 {{2820143,178,89}}
90 {{2916270,180,90}}
91 {{3014557,182,91}}
92 {{3115028,184,92}}
93 {{3217707,186,93}}
94 {{3322618,188,94}}
95 {{3429785,190,95}}
96 {{3539232,192,96}}
97 {{3650983,194,97}}
98 {{3765062,196,98}}
99 {{3881493,198,99}}
100 {{4000300,200,100}}
最小k3
1组解：k3=1: {{7,2,1}}
2组解：k3=7:{{41, 12, 7}, {1393, 14, 7}}
3组解：k3=1393:{{8119, 2378, 1393}, {275807, 2772, 1393}, {10812186007, 2786, 1393}}
4组解：k3=275807: {{1607521, 470832, 275807}, {54608393, 548842, 275807}, {1855077841, 551532, 275807}, {83922003724759193, 551614, 275807}} 可以求出通解。
这个还和佩尔方程有关系。
考虑(1+a^2)(1+b^2)=c^2 (2)的正整数解。
b=(2k-a+ak^2)/(1+2ka-k^2) (4)
(4)可以简单的看出，k=2a是一个a、b的正整数解，可以进一步得到其它正整数解。研究后发现有一个递推多项式都满足。
多项式数列f(n,x)
f(1,x)=x
f(2,x)=4x^3+3x
f(n+1,x)=(4x^2+2)*f(n,x)-f(n-1,x)
f(3,x)=16x^5+20x^3+5x
f(4,x)=64x^7+112x^5+56x^3+7x
f(5,x)=256x^9+576x^7+432x^5+120x^3+9x
x为任意正整数，任意f(n,x)就是(2)中a、b的一个解。相同项（k=a=b)或者相邻项可以得到a、b、k的正整数解。
不知道f(n,x)是否(2)的全部解的表达式？
f(n,a)
=1/2*((1+a^2)^0.5-a)*(2*a^2+1+2*a*(1+a^2)^0.5)^n
-1/2*((1+a^2)^0.5+a)*(2*a^2+1-2*a*(1+a^2)^0.5)^n
=1/2*(1+a^2)^0.5*((2*a^2+1+2*a*(1+a^2)^0.5)^n-(2*a^2+1-2*a*(1+a^2)^0.5)^n)-1/2*a*((2*a^2+1+2*a*(1+a^2)^0.5)^n+(2*a^2+1-2*a*(1+a^2)^0.5)^n)
1，7，41，239，1393,8119,47321,275807,1607521,…2，38，682，12238，219602,…3,117,4443,168717,6406803,…4,268,17684,1166876,…5,515,52525,5457035,…6,882,128766,18798954,…7,1393,275807,54608393,…
上面分别是a=1-7的结果，一行里相邻两个数为斜率过原点的直线，角平分线斜率也为整数。
如果a是f(n,b)的一个值，其整个f(n,a)就是f(n,b)的一个子序列。
yuange1975 发表于 2025-3-21 14:04
考虑(1+a^2)(1+b^2)=c^2 (2)的正整数解。
b=(2k-a+ak^2)/(1+2ka-k^2) (4)
写那么多干什么！
这样就可以了：
$f(n,x)=\lfloor \frac{1}{2} \left(\sqrt{x^2+1}+x\right)^{2 n-1}\rfloor$
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