史上最贱的数学题

liangbch

见 https://www.sohu.com/a/223262966_115479

mathe

又见方程$\sum a/(b+c)=4$的正整数解。其实没那么恐怖，原理就是这是一个齐次三次曲线。而我们用计算机容易找到一组整数解。然后分别通过过已知点做切线交曲线于第三点或连接俩已知点交曲线于第三点找到更多整数解，碰运气总能得到正整数解。不需要先转化为标准曲线

wayne

https://bbs.emath.ac.cn/thread-15284-1-1.html

mathe

方程展开后就是$F(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(a+c)(b+c)-4(a+b)(b+c)(c+a)=0$，代表了一条三次曲线的在射影平面上的齐次方程。
显然这条曲线上有三个(1:-1:0),(1:0:-1),(0:1:-1)。
我们知道三次曲线上任意两个有理点的连线交曲线于第三个有理点，同样任意一个有理点的切线也交曲线于另外一个有理点。
所以首先想到的是这三个点之间用上面规则来派生。很可惜，计算表明它们的切线都是三重的也就是切线不能产生新的有理点，而且这三点共线，也不能产生新交点。不过链接表明已经有人找到特殊点(4:-1:11)。
我们先计算这个点的切线，为此计算关于a的偏导数$d(a,b,c)=F_a(a,b,c)=3a^2+2a(-3c-3b)+(-3c^2-5bc-3b^2)$
于是我们得出这个点处切线方程为$d(4,-1,11)(a-4)+d(-1,4,11)(b+1)+d(11,4,-1)(c-11)=-503a +134c - 538b=0$
我们可以用上面切线方程代入曲线方程，消去c得到$a^3 + 84776/9499ba^2 + 222256/9499b^2a + 140544/9499b^3=0$
由于我们知道点(4:-1:11)是切点，所以$a/b=-4$是上面的重根，根据韦达定理知道$a/b$还可以等于$-84776/9499-2*(-4)=-8784/9499$
于是我们可以选择$a=-8784,b=9499$,代入切线方程得到$c=5165$。
由此我们得到曲线上新点(-8784:9499:5165)。当然这个还不是正整数解。
此后我们可以选择点(1:-1:0)和(-8784:9499:5165)连线类似计算曲线第三个交点，得到交点在和(4:-1:11)相连等等不同方法得到更多的点，直到找到符合条件的一个解

liangbch

wayne 发表于 2019-4-4 17:55
https://bbs.emath.ac.cn/thread-15284-1-1.html

你们一年前就在讲这个问题了，我属于后知后觉。

mathe

F(a,b,c)=a*(a+b)*(a+c)+b*(a+b)*(b+c)+c*(c+a)*(c+b)-4*(a+b)*(b+c)*(c+a)
d(a,b,c)=3*a^2+2*a*(-3*b-3*c)+(-3*c^2-5*b*c-3*b^2)
msquare(data)={local(a,b,u,v,h); a=data[1];b=data[2];u=d(a,b,1)*(x-a)+d(b,a,1)*(y-b)+d(1,a,b)*(z-1);u=subst(u,z,1);h=-polcoeff(u,0,y)/polcoeff(u,1,y);u=F(x,y,1);u=subst(u,y,h);v=-polcoeff(u,2,x)/polcoeff(u,3,x)-2*a;u=subst(h,x,v);[v,u]}

msquare([-4,-11])
[-8784/9499, 5165/9499]
msquare(%)
[-6696085890501216/6531563383962071, -6334630576523495/6531563383962071]
msquare(%)
[-2054217703980198940765993621567260834791816664149006217306067776/343258303254635343211175484588572430575289938927656972201563791, 2110760649231325855047088974560468667532616164397520142622104465/343258303254635343211175484588572430575289938927656972201563791]
mquare(%)
[19283381957291396433156955403945340327978628751799020388633390627682629903117233906742494815438084023999487548662834775007613535719589795660137011425408374888700039268757712301660891739488140506970109930721482543811429886659708462415655488918157551986304/6089467617785193345809614706133792707828070934976824628499538611505403395705375827452269659832018276089057383870476712524073526235843072072079742233379562470123829427973318197724308683085131372984770065416230293953776305562502686158993927036777047552991, -18316645005361901449615437412145592466671365181448584613140124608117563398709158807155147037787526755675191496681898108699821877038950214153486807444901712715631255553791733624142340929791420022387570726653856342185802266810783878674982862360530541901855/6089467617785193345809614706133792707828070934976824628499538611505403395705375827452269659832018276089057383870476712524073526235843072072079742233379562470123829427973318197724308683085131372984770065416230293953776305562502686158993927036777047552991]
的确直接平方下去好多下都找不到全正的解。

mathe

line3rd(d1,d2)={local(u,v,w);u=F(x,y,1);v=(d1[2]-d2[2])/(d1[1]-d2[1])*(x-d1[1])+d1[2];u=subst(u,y,v);w=-polcoeff(u,2,x)/polcoeff(u,3,x)-d1[1]-d2[1];v=subst(v,x,w);[w,v]}
(12:14) gp > line3rd([-4,-11],[0,-1])
%14 = [4/11, -1/11]
(12:15) gp > line3rd([-8784/9499, 5165/9499],[4/11,-1/11])
%15 = [883659076/679733219, -375326521/679733219]
(12:16) gp > line3rd([-4,-11],%15)
%16 = [6696085890501216/6334630576523495, -6531563383962071/6334630576523495]
(12:16) gp > line3rd([-4,-11],%16)
%17 = [883659076/679733219, -375326521/679733219]
(12:16) gp > line3rd([-4,-11],%15)
%18 = [6696085890501216/6334630576523495, -6531563383962071/6334630576523495]
(12:16) gp > line3rd(%18,[0,-1])
%19 = [-6696085890501216/6531563383962071, -6334630576523495/6531563383962071]
(12:16) gp > line3rd(%19,[-4,-11])
%20 = [-2798662276711559924688956/5824662475191962424632819, 5048384306267455380784631/5824662475191962424632819]
(12:16) gp > line3rd(%,[4/11,-1/11])
%21 = [434021404091091140782000234591618320/287663048897224554337446918344405429, -399866258624438737232493646244383709/287663048897224554337446918344405429]
(12:17) gp > line3rd(%,[-4,-11])
%22 = [3386928246329327259763849184510185031406211324804/1637627722378544613543242758851617912968156867151, -678266970930133923578916161648350398206354101381/1637627722378544613543242758851617912968156867151]
(12:17) gp > line3rd(%,[4/11,-1/11])
%23 = [-2054217703980198940765993621567260834791816664149006217306067776/2110760649231325855047088974560468667532616164397520142622104465, 343258303254635343211175484588572430575289938927656972201563791/2110760649231325855047088974560468667532616164397520142622104465]
(12:17) gp > line3rd(%,[-4,-11])
%24 = [4373612677928697257861252602371390152816537558161613618621437993378423467772036/36875131794129999827197811565225474825492979968971970996283137471637224634055579, 154476802108746166441951315019919837485664325669565431700026634898253202035277999/36875131794129999827197811565225474825492979968971970996283137471637224634055579]
于是找到了a=4373612677928697257861252602371390152816537558161613618621437993378423467772036,b=154476802108746166441951315019919837485664325669565431700026634898253202035277999,
c=36875131794129999827197811565225474825492979968971970996283137471637224634055579

mathe

如果限定a+b+c=1,做a,b的图可以如图
可以看出GeoGebra做的点不够多，在拐弯处漏了很多点。而如果要找到正整数解，需要我们这样反复做切线或两点连线，找到一个图中同时符合$0<a<1,0<b<1$的解，所以的确是比较困难