简介
2008年11月northwolves在csdn要求 :
四个不同自然数a < b < c < d a\lt b\lt c\lt d a < b < c < d a 2 + b 2 + c 2 + d 2 = a b c d a^2+b^2+c^2+d^2=abcd a 2 + b 2 + c 2 + d 2 = a b c d
n个不同自然数a 1 < a 2 < ⋯ < a n a_1\lt a_2\lt\dots\lt a_n a 1 < a 2 < ⋯ < a n a 1 2 + a 2 2 + ⋯ + a n 2 = a 1 a 2 … a n a_1^2+a_2^2+\dots+a_n^2=a_1a_2\dots a_n a 1 2 + a 2 2 + ⋯ + a n 2 = a 1 a 2 … a n
2013年2月wayne在数学研发论坛寻求下面不定方程的解 :x 2 + y 2 + z 2 = 3 x y z x^2+y^2+z^2=3xyz x 2 + y 2 + z 2 = 3 x y z
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更进一步，假如 0 < x ≤ y ≤ z ≤ 1 0 10 0\lt x\le y\le z\le 10^{10} 0 < x ≤ y ≤ z ≤ 1 0 1 0
litaoye(绿色夹克衫)给出了一种利用韦达定理进行递降的巧妙算法，mathe利用这种方法彻底解决了更大范围的解。a = 1155707276902 a=1155707276902 a = 1 1 5 5 7 0 7 2 7 6 9 0 2 b = 52631648121718437290225329446159132818342 b=52631648121718437290225329446159132818342 b = 5 2 6 3 1 6 4 8 1 2 1 7 1 8 4 3 7 2 9 0 2 2 5 3 2 9 4 4 6 1 5 9 1 3 2 8 1 8 3 4 2 c = 121653557459230956410906311942093212906353571240214806 c=121653557459230956410906311942093212906353571240214806 c = 1 2 1 6 5 3 5 5 7 4 5 9 2 3 0 9 5 6 4 1 0 9 0 6 3 1 1 9 4 2 0 9 3 2 1 2 9 0 6 3 5 3 5 7 1 2 4 0 2 1 4 8 0 6 d = 54756951556826666523961130929258828162755166803423037863993739741969018235335921416820498267983420590064684413368261305133980613412835260713754668464533252084098231022109546996215107093857821479530279109208302608 d=54756951556826666523961130929258828162755166803423037863993739741969018235335921416820498267983420590064684413368261305133980613412835260713754668464533252084098231022109546996215107093857821479530279109208302608 d = 5 4 7 5 6 9 5 1 5 5 6 8 2 6 6 6 6 5 2 3 9 6 1 1 3 0 9 2 9 2 5 8 8 2 8 1 6 2 7 5 5 1 6 6 8 0 3 4 2 3 0 3 7 8 6 3 9 9 3 7 3 9 7 4 1 9 6 9 0 1 8 2 3 5 3 3 5 9 2 1 4 1 6 8 2 0 4 9 8 2 6 7 9 8 3 4 2 0 5 9 0 0 6 4 6 8 4 4 1 3 3 6 8 2 6 1 3 0 5 1 3 3 9 8 0 6 1 3 4 1 2 8 3 5 2 6 0 7 1 3 7 5 4 6 6 8 4 6 4 5 3 3 2 5 2 0 8 4 0 9 8 2 3 1 0 2 2 1 0 9 5 4 6 9 9 6 2 1 5 1 0 7 0 9 3 8 5 7 8 2 1 4 7 9 5 3 0 2 7 9 1 0 9 2 0 8 3 0 2 6 0 8
详细内容
northwolves先给出了一组问题1的解: 2，6，22，262a [ 0 ] = a [ 1 ] = a [ 2 ] = a [ 3 ] = 2 a[0]=a[1]=a[2]=a[3]=2 a [ 0 ] = a [ 1 ] = a [ 2 ] = a [ 3 ] = 2 a [ n ] = a [ n − 1 ] × a [ n − 2 ] × a [ n − 3 ] − a [ n − 4 ] a[n]=a[n-1]\times a[n-2]\times a[n-3]-a[n-4] a [ n ] = a [ n − 1 ] × a [ n − 2 ] × a [ n − 3 ] − a [ n − 4 ] a [ n − 3 ] 2 + a [ n − 2 ] 2 + a [ n − 1 ] 2 + a [ n ] 2 = a [ n − 3 ] × a [ n − 2 ] × a [ n − 1 ] × a [ n ] a[n-3]^2+a[n-2]^2+a[n-1]^2+a[n]^2=a[n-3]\times a[n-2]\times a[n-1]\times a[n] a [ n − 3 ] 2 + a [ n − 2 ] 2 + a [ n − 1 ] 2 + a [ n ] 2 = a [ n − 3 ] × a [ n − 2 ] × a [ n − 1 ] × a [ n ]
对a 2 + b 2 + c 2 = a b c a^2+b^2+c^2=abc a 2 + b 2 + c 2 = a b c a [ 0 ] = a [ 1 ] = a [ 2 ] = 3 a[0]=a[1]=a[2]=3 a [ 0 ] = a [ 1 ] = a [ 2 ] = 3 a [ n ] = a [ n − 1 ] × a [ n − 2 ] − a [ n − 3 ] a[n]=a[n-1]\times a[n-2]-a[n-3] a [ n ] = a [ n − 1 ] × a [ n − 2 ] − a [ n − 3 ] a 2 + b 2 + c 2 = a b c a^2+b^2+c^2=abc a 2 + b 2 + c 2 = a b c
一般的,对k = a k − 2 k=a^{k-2} k = a k − 2 a 0 a_0 a 0 a [ 0 ] = a [ 1 ] = ⋯ = a [ k − 1 ] = a 0 a[0]=a[1]=\dots=a[k-1]=a_0 a [ 0 ] = a [ 1 ] = ⋯ = a [ k − 1 ] = a 0 a [ n ] = a [ n − 1 ] × ⋯ × a [ n − ( k − 1 ) ] − a [ n − k ] a[n]=a[n-1]\times\dots\times a[n-(k-1)]-a[n-k] a [ n ] = a [ n − 1 ] × ⋯ × a [ n − ( k − 1 ) ] − a [ n − k ] ( a [ n − ( k − 1 ) ] , × , a [ n − 1 ] , a [ n ] ) (a[n-(k-1)],\times,a[n-1],a[n]) ( a [ n − ( k − 1 ) ] , × , a [ n − 1 ] , a [ n ] ) a 1 2 + a 2 2 + ⋯ + a k 2 = a 1 a 2 … a k a_1^2+a_2^2+\dots+a_k^2=a_1a_2\dots a_k a 1 2 + a 2 2 + ⋯ + a k 2 = a 1 a 2 … a k
northwolves发现medie2005构造的解就是A061292 .A086326 A002559 Markoff numbers对应wayne的问题。
但是northwolves发现还有解不在medie2005的系列解之中，比如
northwolves搜索了部分解，medie2005根据northwolves陆续搜索出来的结果总结出
litaoye(绿色夹克衫)总结出可以利用二次方程的韦达定理来构造出所有的解:2 2 + 2 2 + 2 2 + X 2 = 8 X 2^2 + 2^2 + 2^2 + X^2 = 8X 2 2 + 2 2 + 2 2 + X 2 = 8 X 2 2 + 6 2 + 2 2 + X 2 = 24 X 2^2 + 6^2 + 2^2 + X^2 = 24X 2 2 + 6 2 + 2 2 + X 2 = 2 4 X 2 2 + 6 2 + 2 2 2 + X 2 = 264 X 2^2 + 6^2 + 22^2 + X^2 = 264X 2 2 + 6 2 + 2 2 2 + X 2 = 2 6 4 X
然后他给出了证明思路:a < b < c < d a\lt b\lt c\lt d a < b < c < d a 2 + b 2 + c 2 + d 2 = a b c d a^2+b^2+c^2+d^2=abcd a 2 + b 2 + c 2 + d 2 = a b c d a b c e a b c e a b c e a 2 + b 2 + c 2 + e 2 = a b c e a^2+b^2+c^2+e^2=abce a 2 + b 2 + c 2 + e 2 = a b c e e < d e \lt d e < d a b c e a b c e a b c e a b f e a b f e a b f e a 2 + b 2 + f 2 + e 2 = a b f e a^2+b^2+f^2+e^2=abfe a 2 + b 2 + f 2 + e 2 = a b f e f < c … f \lt c \dots f < c … w 2 + x 2 + y 2 + z 2 = w x y z w^2 + x^2 + y^2 + z^2 = wxyz w 2 + x 2 + y 2 + z 2 = w x y z w = x = y = z w=x=y=z w = x = y = z w = x = y = z = 2 w=x=y=z=2 w = x = y = z = 2
n个不同自然数a 1 < a 2 < ⋯ < a n a_1\lt a_2\lt\dots\lt a_n a 1 < a 2 < ⋯ < a n a 1 2 + a 2 2 + ⋯ + a n 2 = a 1 a 2 … a n a_1^2+a_2^2+\dots+a_n^2=a_1a_2\dots a_n a 1 2 + a 2 2 + ⋯ + a n 2 = a 1 a 2 … a n
但是随即发现对于5个数的情况的分析存在问题:2 × a 5 > a 1 a 2 a 3 a 4 2\times a_5 \gt a_1a_2a_3a_4 2 × a 5 > a 1 a 2 a 3 a 4 2 × a 4 > a 1 a 2 a 3 2\times a_4 \gt a_1a_2a_3 2 × a 4 > a 1 a 2 a 3 2 a 3 > a 1 a 2 2a_3 \gt a_1a_2 2 a 3 > a 1 a 2 2 a 4 > a 1 a 2 a 3 2a_4 \gt a_1a_2a_3 2 a 4 > a 1 a 2 a 3 a 1 ≥ a 2 ≥ a 3 ≥ a 4 a_1\ge a_2\ge a_3\ge a_4 a 1 ≥ a 2 ≥ a 3 ≥ a 4 2 a 4 < a 1 a 2 a 3 2a_4 \lt a_1a_2a_3 2 a 4 < a 1 a 2 a 3 2 a 4 2 < a 1 a 2 a 3 a 4 2a_4^2 \lt a_1a_2a_3a_4 2 a 4 2 < a 1 a 2 a 3 a 4 2 a 4 2 < a 1 2 + a 2 2 + a 3 2 + a 4 2 2a4^2 \lt a_1^2 + a_2^2 + a_3^2 + a_4^2 2 a 4 2 < a 1 2 + a 2 2 + a 3 2 + a 4 2 a 4 2 < a 1 2 + a 2 2 + a 3 2 a_4^2 \lt a_1^2 + a_2^2 + a_3^2 a 4 2 < a 1 2 + a 2 2 + a 3 2 a 1 a 2 a 3 a 4 < 2 ( a 1 2 + a 2 2 + a 3 2 ) a_1a_2a_3a_4 \lt 2(a_1^2 + a_2^2 + a_3^2) a 1 a 2 a 3 a 4 < 2 ( a 1 2 + a 2 2 + a 3 2 ) a 1 a 2 a 3 a 3 < 2 ( 3 a 3 2 ) a_1a_2a_3a_3 \lt 2(3a_3^2) a 1 a 2 a 3 a 3 < 2 ( 3 a 3 2 ) a 1 a 2 < 6 a_1a_2 \lt 6 a 1 a 2 < 6 a 3 < a 4 a_3 \lt a_4 a 3 < a 4 a 3 = a 4 a_3 = a_4 a 3 = a 4
1、a 3 < a 4 a_3 \lt a_4 a 3 < a 4 a 4 2 < a 1 2 + a 2 2 + a 3 2 a_4^2 \lt a_1^2 + a_2^2 + a_3^2 a 4 2 < a 1 2 + a 2 2 + a 3 2 ( a 4 − a 3 ) ( a 4 + a 3 ) < a 1 2 + a 2 2 (a_4-a_3)(a_4+a_3) \lt a_1^2 + a_2^2 ( a 4 − a 3 ) ( a 4 + a 3 ) < a 1 2 + a 2 2 ( a 4 − a 3 ) (a_4-a_3) ( a 4 − a 3 ) a 4 + a 3 < a 1 2 + a 2 2 a_4+a_3 \lt a_1^2 + a_2^2 a 4 + a 3 < a 1 2 + a 2 2
2、a 3 = a 4 a_3 = a_4 a 3 = a 4 a 1 2 + a 2 2 + a 3 2 + a 4 2 = a 1 a 2 a 3 a 4 a_1^2 + a_2^2 + a_3^2 + a_4^2 = a_1a_2a_3a_4 a 1 2 + a 2 2 + a 3 2 + a 4 2 = a 1 a 2 a 3 a 4 a 1 2 + a 2 2 + 2 a 4 2 = a 1 a 2 a 4 2 a_1^2 + a_2^2 + 2a_4^2 = a_1a_2a_4^2 a 1 2 + a 2 2 + 2 a 4 2 = a 1 a 2 a 4 2 a 1 2 + a 2 2 = ( a 1 a 2 − 2 ) a 4 2 a_1^2 + a_2^2 = (a_1a_2 - 2)a_4^2 a 1 2 + a 2 2 = ( a 1 a 2 − 2 ) a 4 2 2 a 4 < a 1 a 2 a 3 2a_4 \lt a_1a_2a_3 2 a 4 < a 1 a 2 a 3
mathe指出:x 1 x 2 … x n − 2 ≤ n x_1x_2\dots x_{n-2}\le n x 1 x 2 … x n − 2 ≤ n
再加一个对x n − 1 x_{n-1} x n − 1 x n − 1 ≤ x 1 2 + x 2 2 + ⋯ + x n − 2 2 x 1 x 2 … x n − 2 − 2 x_{n-1}\le\sqrt{\frac{x_1^2+x_2^2+\dots+x_{n-2}^2}{x_1x_2\dots x_{n-2}-2}} x n − 1 ≤ x 1 x 2 … x n − 2 − 2 x 1 2 + x 2 2 + ⋯ + x n − 2 2 x 1 2 + x 2 2 + x 3 2 = x 1 x 2 x 3 x_1^2+x_2^2+x_3^2=x_1x_2x_3 x 1 2 + x 2 2 + x 3 2 = x 1 x 2 x 3 x 1 ≤ 3 x_1\le 3 x 1 ≤ 3 x 1 ≤ 2 x_1\le 2 x 1 ≤ 2 x 1 = 3 x_1=3 x 1 = 3 x 2 ≤ x 1 2 x 1 − 2 = 3 x_2\le \sqrt{\frac{x_1^2}{x_1-2}}=3 x 2 ≤ x 1 − 2 x 1 2 = 3 x n x_n x n n = 4 n=4 n = 4 x 1 x 2 ≤ 4 x_1x_2\le 4 x 1 x 2 ≤ 4 n = 4 n=4 n = 4 x 1 = x 2 = 2 x_1=x_2=2 x 1 = x 2 = 2 x 3 ≤ 8 4 − 2 = 2 x_3\le\sqrt{\frac8{4-2}}=2 x 3 ≤ 4 − 2 8 = 2 x 3 = 2 x_3=2 x 3 = 2 n = 4 n=4 n = 4 ( 2 , 2 , 2 , x 4 ) (2,2,2,x_4) ( 2 , 2 , 2 , x 4 )
由此他继续算出了n=4的前10000组解，其中第10000组为:点击下载他的C++代码
然后在数学研发论坛的对应帖子中 ,他给出哪些n在100以内有非平凡解，并提交的OEIS产生了A146974
而后来对于wayne提出的类似问题，hujunhua提供了一份Mathematica计算方案
f[ x_] := { { x[ [ 1 ] ] , x[ [ 3 ] ] , 3 x[ [ 1 ] ] x[ [ 3 ] ] -x[ [ 2 ] ] } ,{ x[ [ 2 ] ] ,x[ [ 3 ] ] ,3x[ [ 2 ] ] x[ [ 3 ] ] -x[ [ 1 ] ] } } ;
NestList[ Flatten[ f/@
f[ x_] := { { x[ [ 1 ] ] , x[ [ 3 ] ] , 3 x[ [ 1 ] ] x[ [ 3 ] ] - x[ [ 2 ] ] } , { x[ [ 2 ] ] , x[ [ 3 ] ] ,3 x[ [ 2 ] ] x[ [ 3 ] ] - x[ [ 1 ] ] } } ;
g[ x_] := Flatten[ f /@ x, 1 ] ;
h[ x_] := Select[ g[ x] ,
Answer = NestWhileList[ h, { { 1 , 2 , 5 } } , Length[
Flatten[ Answer, 1 ]
( Flatten[ Answer, 1 ] // Length) + 2 wayne把代码优化为
Union@Flatten[
NestList[ Union@Sort@Flatten[ Table[ Union[
Map[ Sort, { x, y, z, 4 x y z - t} /. Thread[ Rule[ { x, y, z, t} ,
NestList[ RotateLeft, ii, 4 ] ] ] , { ii, 数学星空找到了维基百科中对应的介绍
mathe做了总结:x 1 2 + x 2 2 + . . . + x n 2 = m x 1 x 2 . . . x n x_1^2+x_2^2+...+x_n^2=mx_1x_2...x_n x 1 2 + x 2 2 + . . . + x n 2 = m x 1 x 2 . . . x n x 1 ≤ x 2 ≤ ⋯ ≤ x n x_1\le x_2\le\dots\le x_n x 1 ≤ x 2 ≤ ⋯ ≤ x n x n x_n x n X 2 − m x 1 x 2 . . . x n − 1 X + ( x 1 2 + . . . + x n − 1 2 ) = 0 X^2-mx_1x_2...x_{n-1}X+(x_1^2+...+x_{n-1}^2)=0 X 2 − m x 1 x 2 . . . x n − 1 X + ( x 1 2 + . . . + x n − 1 2 ) = 0 x n − 1 ≤ x n ≤ 2 ( x 1 2 + . . . + x n − 1 2 ) m x 1 x 2 … x n − 1 ≤ 2 ( n − 1 ) x n − 1 2 m x 1 x 2 . . . x n − 1 x_{n-1}\le x_n\le \frac{2(x_1^2+...+x_{n-1}^2)}{mx_1x_2\dots x_{n-1}}\le\frac{2(n-1)x_{n-1}^2}{mx_1x_2...x_{n-1}} x n − 1 ≤ x n ≤ m x 1 x 2 … x n − 1 2 ( x 1 2 + . . . + x n − 1 2 ) ≤ m x 1 x 2 . . . x n − 1 2 ( n − 1 ) x n − 1 2 x 1 x 2 . . . x n − 2 ≤ 2 ( n − 1 ) m x_1x_2...x_{n-2}\le\frac{2(n-1)}m x 1 x 2 . . . x n − 2 ≤ m 2 ( n − 1 ) x 1 x 2 ≤ 3 2 x_1x_2\le\frac32 x 1 x 2 ≤ 2 3 x 1 = x 2 = 1 x_1=x_2=1 x 1 = x 2 = 1 2 + z 2 + u 2 = 4 z u 2+z^2+u^2=4zu 2 + z 2 + u 2 = 4 z u
而同样，对于任意一组满足条件的x 1 , x 2 , . . . , x n − 2 x_1,x_2,...,x_{n-2} x 1 , x 2 , . . . , x n − 2 A + X 2 + Y 2 = C X Y A+X^2+Y^2=CXY A + X 2 + Y 2 = C X Y X ≤ Y X\le Y X ≤ Y X ≤ Y ≤ 2 ( A + X 2 ) C X X\le Y\le\frac{2(A+X^2)}{CX} X ≤ Y ≤ C X 2 ( A + X 2 ) C X 2 ≤ 2 X 2 + 2 A CX^2\le 2X^2+2A C X 2 ≤ 2 X 2 + 2 A C ≥ 3 C\ge 3 C ≥ 3 X ≤ 2 A C − 2 X\le\sqrt{\frac{2A}{C-2}} X ≤ C − 2 2 A x n − 1 x_{n-1} x n − 1 x n x_n x n z ≤ 2 z\le\sqrt{2} z ≤ 2 x = y = z = 1 x=y=z=1 x = y = z = 1 3 + u 2 = 4 u 3+u^2=4u 3 + u 2 = 4 u
进一步分析，我们可以得出如下算法:x 1 ≤ x 2 ≤ ⋯ ≤ x n − 2 x_1\le x_2\le\dots\le x_{n-2} x 1 ≤ x 2 ≤ ⋯ ≤ x n − 2 x n − 1 ≤ 2 A C − 2 x_{n-1}\le\sqrt{\frac{2A}{C-2}} x n − 1 ≤ C − 2 2 A x n − 2 ≤ x n − 1 x_{n-2}\le x_{n-1} x n − 2 ≤ x n − 1 x 1 ≤ x 2 ≤ ⋯ ≤ x n − 2 ≤ x n − 1 x_1\le x_2\le\dots\le x_{n-2}\le x_{n-1} x 1 ≤ x 2 ≤ ⋯ ≤ x n − 2 ≤ x n − 1 x n x_n x n x n − 1 x_{n-1} x n − 1 x 1 , x 2 , . . . , x n − 1 x_1,x_2,...,x_{n-1} x 1 , x 2 , . . . , x n − 1 x n x_n x n
hujunhua进一步指出:1 ≤ x 1 < x 2 < ⋯ < x n 1\le x_1\lt x_2\lt\dots\lt x_n 1 ≤ x 1 < x 2 < ⋯ < x n ( n − 1 ) (n-1) ( n − 1 ) ( x 1 , x 2 , … , x n ) → ( x 1 , x 2 , … , x i − 1 , x i + 1 , , ˙ x n , x i ′ ) (x_1, x_2, \dots, x_n)\to(x_1, x_2,\dots, x_{i-1}, x_{i+1}, \dot, x_n, x^{\prime}_i) ( x 1 , x 2 , … , x n ) → ( x 1 , x 2 , … , x i − 1 , x i + 1 , , ˙ x n , x i ′ ) ( i = 1 , 2 , 3 , … , n − 1 ) (i=1, 2, 3, \dots, n-1) ( i = 1 , 2 , 3 , … , n − 1 ) x i ′ = m x 1 x 2 … x ( i − 1 ) x ( i + 1 ) … x n − x i x^{\prime}_i=mx_1 x_2 \dots x_{(i-1)} x_{(i+1)} \dots x_n-x_i x i ′ = m x 1 x 2 … x ( i − 1 ) x ( i + 1 ) … x n − x i ( x 1 , x 2 , … , x n ) (x_1, x_2, \dots, x_n) ( x 1 , x 2 , … , x n ) x i x_i x i x i ′ x^{\prime}_i x i ′
在这样的递推关系下，从一个初解出发可以生成一棵树。如果这棵树不能囊括方程的全解，那就表明有多个初解，相应地可生成多棵树。
不同的树有不同的初解（根），不同的初解，即mathe所言基本解，是不会存在递推关系的。因为，存在递推关系的两个解，必有一个不是基本解。
