勾股数新公式

Treenewbee

蔡家雄 发表于 2023-12-5 00:05
求解方程：\(y^2=3*x^3+4x+5\) 的正整数解，

{5,20}

蔡家雄

设 \(a_{1}=1, a_{2}=32\), 且 \(a_{n}^2+a_{n+1}^2\) 可以被 \(a_{n}*a_{n+1}+9\) 整除，求它的通解公式，

设 \(b_{1}=1, b_{2}=73\),  且 \(b_{n}^2+b_{n+1}^2\)  可以被 \(b_{n}*b_{n+1}+9\) 整除，求它的通解公式，

Treenewbee

蔡家雄 发表于 2023-12-5 12:33
高难度计算

求：\(\sqrt{3}+\sqrt{7}\) 的渐近分数的 分子 的通解公式，

1. f=Convergents[Sqrt[3]+Sqrt[7],30]
   
2. Denominator@f
   
3. Numerator@f

复制代码

\[\left\{4,\frac{9}{2},\frac{13}{3},\frac{22}{5},\frac{35}{8},\frac{162}{37},\frac{197}{45},\frac{3905}{892},\frac{8007}{1829},\frac{51947}{11866},\frac{59954}{13695},\frac{291763}{66646},\frac{2685821}{613509},\frac{5663405}{1293664},\frac{25339441}{5788165},\frac{31002846}{7081829},\frac{56342287}{12869994},\frac{200029707}{45691811},\frac{456401701}{104253616},\frac{656431408}{149945427},\frac{2425695925}{554089897},\frac{5507823258}{1258125221},\frac{7933519183}{1812215118},\frac{13441342441}{3070340339},\frac{21374861624}{4882555457},\frac{12859733178465}{2937486169996},\frac{38600574397019}{8817341065445},\frac{51460307575484}{11754827235441},\frac{244441804698955}{55836650007209},\frac{295902112274439}{67591477242650}\right\}\]

\[\{1,2,3,5,8,37,45,892,1829,11866,13695,66646,613509,1293664,5788165,7081829,12869994,45691811,104253616,149945427,554089897,1258125221,1812215118,3070340339,4882555457,2937486169996,8817341065445,11754827235441,55836650007209,67591477242650\}\]

\[\{4,9,13,22,35,162,197,3905,8007,51947,59954,291763,2685821,5663405,25339441,31002846,56342287,200029707,456401701,656431408,2425695925,5507823258,7933519183,13441342441,21374861624,12859733178465,38600574397019,51460307575484,244441804698955,295902112274439\}\]

cz1

王守恩 发表于 2023-12-3 09:59
谢谢 Treenewbee！还得再来一题。
a,b是正整数,  满足 \(\bigg\lceil\frac{n-a/b}{\sqrt[n]{\sin(\pi/5 ...

请计算：MultiplicativeOrder[10, 66666666666666666666666666666666666666666659 ]，谢谢！

王守恩

Treenewbee 发表于 2023-11-29 13:02
这样的题目有什么实际意义呢？

这串分数实用性强, 比渐近分数分布均匀多了。

这个程应该怎样编？EulerGamma=0.57721566...

Abs[EulerGamma-1/2]<(1/10)^1    (2是最小的)
Abs[EulerGamma-4/7]<(1/10)^2     (7是最小的)
Abs[EulerGamma-15/26]<(1/10)^3   (26是最小的)
Abs[EulerGamma-71/123]<(1/10)^4    (123是最小的)
Abs[EulerGamma-157/272]<(1/10)^5
Abs[EulerGamma-228/395]<(1/10)^6
Abs[EulerGamma-2579/4468]<(1/10)^7
Abs[EulerGamma-3035/5258]<(1/10)^8
Abs[EulerGamma-15403/26685]<(1/10)^9
Abs[EulerGamma-33841/58628]<(1/10)^10

得到一串数(OEIS没有):  1,4,15,71,157,228,2579,3035,......

蔡家雄

设正整数 a, b 满足 ab+9 可以整除 a^2+b^2，求 a, b 各几何？

Treenewbee

蔡家雄 发表于 2023-12-13 14:21
设正整数 a, b 满足 ab+9 可以整除 a^2+b^2，求 a, b 各几何？

\[ a< b<10000\]的解：
{{6,24},{1,32},{2,44},{1,73},{9,81},{4,82},{24,90},{7,176},{12,192},{90,336},{15,375},{8,521},{18,648},{81,720},{32,799},{44,878},{21,1029},{336,1254},{24,1536},{82,1636},{27,2187},{30,3000},{192,3060},{33,3993},{176,4393},{1254,4680},{73,4744},{36,5184},{720,6399},{39,6591},{42,8232},{375,9360}}

蔡家雄

这是我以前的解答

http://www.mathchina.com/bbs/for ... 1&fromuid=45368

蔡家雄

设正整数 a, b 满足 ab+9 可以整除 a^2+b^2，

好像当 a, b 是3的倍数时，(a^2+b^2)/(ab+9)=完全平方数，如果是真的，可以证明吗？

及 Treenewbee 的计算，a < b <10000，

{{6,24},{1,32},{2,44},{1,73},{9,81},{4,82},{24,90},{7,176},{12,192},{90,336},{15,375},{8,521},{18,648},{81,720},{32,799},{44,878},{21,1029},{336,1254},{24,1536},{82,1636},{27,2187},{30,3000},{192,3060},{33,3993},{176,4393},{1254,4680},{73,4744},{36,5184},{720,6399},{39,6591},{42,8232},{375,9360}}

Treenewbee

\[a=3 n;b=3 n^3;c=\frac{a^2+b^2}{a b+9}=n^2\]