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勾股数组研究

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发表于 2023-2-21 21:27 | 显示全部楼层 |阅读模式
本帖最后由 朱明君 于 2023-3-19 13:33 编辑


(x2)2=mn,x4m>n, mn
x<(mn), x=amn=b,  m+n=c,
x>(mn), x=b,  mn=a,  m+n=c,
a2+b2=c2
ab



(x/2)2=mnx4mnmn
x2+(mn)2=(m+n)2

x=mnx3m>nmn
x2[(m2n2)/2]2=[(m2+n2)/2]2

x=m+nx2mn
[m(x+n)]2+(2xn)2=(x2+n2)2

x=m+nx3m>nmn
[x(mn)]2+(2mn)2=(m2+n2)2      


1,(x/2)2=mn,x4
x2+(mn)2=(m+n)2
mn1
x2+(mn)2=(m+n)2
n(x/2)2
(x/2)2=1×Fn11×Fn22××Fnnn, F
(x/2)n(x/2)2/n=m
(x/2)2=1×Fn11×Fn22××Fnnn,1
1n(x/2)()
(x/2)n
x=60
(60/2)2=900=1×22×32×52,
11=1,  21=2,  31=3,  51=5,
22=4,  32=9,  52=25,
2×3=6, 2×5=10,  3×4=12,  3×5=15,  2×9=18,  4×5=20,
n3012345691012151825(13)
(X/2)2/n=m

n=1,m=900n=2m=450n=3m=300n=4m=225
n=5m=180n=6m=150n=9m=100n=10m=90
n=12m=75n=15,m=60n=18m=50n=20m=45
n=25m=36

602+(9001)2=(900+1)2(
602+(4502)2=(450+2)2
602+(3003)2=(300+3)2
602+(2254)2=(225+4)2(
602+(1805)2=(180+5)2
602+(1506)2=(150+6)2
602+(1009)2=(100+9)2(
602+(9010)2=(90+10)2
602+(7512)2=(75+12)2
602+(6015)2=(60+15)2
602+(5018)2=(50+18)2
602+(4520)2=(45+20)2
602+(3625)2=(36+25)2(

(x/2)2=mn()
(4/2)2=4×1(345)(
(6/2)2=9×1(8610)(
(8/2)2=16×1(15817)(
(8/2)2=8×2(6810)                  
(10/2)2=25×1(241026)(   
(12/2)2=36×1(351237)(
(12/2)2=18×2(161220)
(12/2)2=12×3(91215)
(12/2)2=9×4(51213)(
(14/2)2=49×1(481450)


2,x2=mn(X3),m>n, m,n
x2+[(mn)/2]2=[(m+n)/2]2
mn1
x2+[(mn)/2]2=[(m+n)/2]2
nX2
X2=1×Fn11×Fn22××Fnnn,(F)
Xn,(X2)/n=m
X2=1×Fn11×Fn22××Fnnn11n
X()Xn
X=15
X=15,  152=1×32×52,
11=1,  31=3,  51=5,
32=9,  52=25,
n151359(4)
X2/n=m
n=1,  m=225n=3m=75n=5m=45n=9m=25

152+[(2251)/2]2=[(225+1)/2]2(
152+[(753)/2]2=(75+3)/2]2
152+[(455)/2]2=[(45+5)/2]2
152+[(259)/2]2=[(25+9)/2]2(

x2=mn()
32=9×1(345)(
52=25×1(51213)(
72=49×1(72425)(              
92=81×1(94042)(     
92=27×3(91215)  
112=121×1(116061)(      
132=169×1(138485)(         
152=225×1(15112113)(
152=75×3(153639)
152=45×5(152025)
152=25×9(15817)(


3,X
X
X=Fn11×Fn22××Fnnn,(X3Fn)
XL,2n+1
X(L=[(2n1+1)(2n2+1)(2nn+1)1]/2
X(L=[(2n1+12)(2n2+1)(2nn+1)1]/2
X=15,  15=31×51,
[(2×1+1)×(2×1+1)1]/2=4
X=60,  60=22×31×51,
[(2×2+12)×(2×1+1)×(2×1+1)1]/2=13

4x=m+nx2mn
[m(x+n)]2+(2xn)2=(x2+n2)2
(x+n)m
[m(x+n)]2+(2xn)2=(x2+n2)2

x=m+n()
2=1+1(345)(
3=1+2(51213)(
3=2+1(8610)
4=1+3(72425)(
4=2+2(121620)
4=3+1(15817)(
5=1+4(94041)(
5=2+3(163034)
5=3+2(212029)(
5=4+1(241026)
6=1+5(116061)(
6=2+4(204852)
6=3+3(273645)
6=4+2(322440)
6=5+1(351237)(  


5x=m+nx3m<n<x,xmn
[x(nm)]2+(2mn)2=(m2+n2)2
xmn,
[x(nm)]2+(2mn)2=(m2+n2)2

x=m+n()
3=1+2(345)(
4=1+3(8610)
5=1+4(15817)(
5=2+3(51213)(
6=1+5(241026)
6=2+4(121620)
7=1+6(351237)(
7=2+5(212029)(
7=3+4(72425)(
8=1+7(481450)
8=2+6(322440)
8=3+5(163034)


6
n+1n
32+42=52
102+112+122=132+142
212+222+232+242=252+262+272
362+372+382+392+402=412+422+432+442

n+1=n
2n+11n(2n+1),
1n(2n+3),n(2n+1)+n
1n(2n+1)1n(2n+1)+n
1n(2n+1)+n+11n(2n+3)

n1xnx(2n+1)
n(2n+1)+(x1)=z


7x=(a+b++n)3
(a2+b2++n2)yabn
a2+b2++n2+{(x21)2(x2y)2}2={(x2+1)2(x2y)2}2
x=5,     52+122=132,
5=1+1+1+1+1,12+12+12+12+12+22=32
5=1+1+1+2,12+12+12+22+32=42
5=1+1+3,12+12+32+52=62
5=1+2+2,12+22+22+42=52
5=1+4,12+42+82=92
5=2+3,22+32+62=72
简化公式:
(a2+b2++n2)=x,(a+b++n)3abnx
a2+b2++n2+{(x1)2}2={(x+1)2}2      

8x
x2+(x+1)2+[x(x+1)]2=[x(x+1)+1]2
x=1,12+22+22=32
x=2,22+32+62=72
x=3,32+42+122=132
x=4,42+52+202=212


9x2nx
xn+xn++xn=x(n+1)),       x(xn)=x(n+1)
x=2   2n+2n=2(n+1)
x=3   3n+3n+3n=3(n+1)
x=4   4n+4n+4n+4n=4(n+1)


10,2nn1
(3x+1)/2n=ZxZ
n
x()=2(n+1)×N+2n+{[2(n+1)1]/3}
z()=6N+5
N0
n
x()=2(n+1)×N+[(2n1)/3]
z()=6N+1
nN0


11,x2+yn=z2
[y(n1)y]/2=x[y(n1)+y]/2=z
y2n4
x2+yn=z2


12,
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,
:
a,b,c,d,
(ad)2+(2bc)2=(b2+c2)2
3
1a2b3c
[a(b+c)]2+(2bc)2=(b2+c2)2


13,x=aby=acz=bc(abc)
(xy)2(xz)2=(yz)2,


14,x=b+2(a+c)y=a+2(b+c)z=c+2(a+b+c)abc
x2+y2=z2,


15,xx2+(2x)2+(2x)2=(x+2x)2


16,n2
[(2n1)n]n2+[(2n1)n1]n1=[2×(2n1)n2]n
实例:



17,xn(2n)x=(2x)n2×(2n)x=2nx+1
(2n)x(2n)x=2(nx1)
(2x)n(2x)n=2(nx1)
(2n)x(2x)n=2(nx1)



18an+bn=z,  az=x,  bz=y,
abn
xn+yn=zn+1


19,a=b=2,   c=2(n+1)÷2,  nan+bn=c2,
实例:
2121=22
2323=42
2525=82
2727=162
......


20,m(m+1)=x,m(4x+1)2=[m+(m+1)]2,   


21,n0(2n)n+2+(2n)n+2=(2×(2n))n+1
实例
n=0,   12+12=21,
n=1,   23+23=42,
n=2,   44+44=83,
n=3,   85+85=164,
n=4,   166+166=325,


22
项数,     n=1,2,3,4,5,6,7,8,9,10,11,....
兔子数,F=0,1,1,2,3,5,8,13,21,34,35,....
a2n3n1
n13,(+1)
n22,(+1)
n31,(+1)
((an1)n1)n2+((an1)n21)n1=(a(an1)n32)n
注:在实际操作运算中,我们要将公式中的n(项数)换成对应的兔子数。
实例:


23,a=b=2n,c=2(2n),n0abc(a+b+c)=2n2,


24,n2(2n)n2(2n2)n=(2n1)n1


25,122344566788...
n3=x()n3
a(n)=nx


26,x3+3xy+n=y3
y3,x=y2,n=3xy+8,x3+3xy+n=y3


27,(m2+n2)=Cm>nmnC
{C(m2n2)}2+(2mnC)2=C4

28,abCa<b<C
(b2a2)2+(2ab)2=C4
(aC)2+(bC)2=(CC)2=C4


29,a3n2
((an5)4n)n1+((an5)2n)2n2+((an5)2n1)2n1+((an5)2n2)2n+
\left( \left( a^n-5\right)^n\right)^{4n-4}+\left( \left( a^n-5\right)^{n-1}\right)^{4n}=\left( a\left( a^n-5\right)^{4n-4}\right)^n
实例



30,设n为大于等于2的正整数,则
\left( 4^{\left( n+1\right)\times\left( n+2\right)\times n}\right)^{n-1}+\left( 4^{\left( n+1\right)\times\left( n+2\right)\times\left( n-1\right)}\right)^n+
\left( 4^{\left( n\times\left( n+1\right)\times\left( n-1\right)+\left( n\times\left( n-1\right)\right)\right)}\right)^{n+1}+\left( 4^{\left( n\times\left( n+1\right)\times\left( n-1\right)\right)}\right)^{n+2}=\left( 4^{\left( n\times\left( n+1\right)-1\right)}\right)^{\left( n\times\left( n+1\right)-1\right)}
实例


31,设n为≥1的正整数,则(n+(n+1))^2-(n(n+1))×2^2=1,
设xdy均为正整数,则x^2-d^2y^2=1,无正整数解


32,




33,
求数例3,7,11,17,23,31,39,49…。正整数解的通项公式
设n为大于等于2的正整数,
a\left( n\right)=\frac{n\left( n+1\right)+\left( n-x\right)}{2}{,}其中n为偶数,则x为2{,}\ \ \ \ \ n为奇数,则x为1{,}
求数例7,13,23,33,47,61,79,97…。正整数解的通项公式
设n为大于等于2的正整数,
a(n)=n(n+1)+(n-x), 其中n为偶数,则x为1,其中n为奇数,则x为2,




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发表于 2023-2-22 17:05 | 显示全部楼层
试试,配个通项公式。谢谢!

\big(4^{024}\big)^{1}+\big(4^{012}\big)^{2}+\big(4^{008}\big)^{03}+\big(4^{006}\big)^{04}=\big(4^{05}\big)^{05}

\big(4^{060}\big)^{2}+\big(4^{040}\big)^{3}+\big(4^{030}\big)^{04}+\big(4^{024}\big)^{05}=\big(4^{11}\big)^{11}

\big(4^{120}\big)^{3}+\big(4^{090}\big)^{4}+\big(4^{072}\big)^{05}+\big(4^{060}\big)^{06}=\big(4^{19}\big)^{19}

\big(4^{210}\big)^{4}+\big(4^{168}\big)^{5}+\big(4^{140}\big)^{06}+\big(4^{120}\big)^{07}=\big(4^{29}\big)^{29}

\big(4^{336}\big)^{5}+\big(4^{280}\big)^{6}+\big(4^{240}\big)^{07}+\big(4^{210}\big)^{08}=\big(4^{41}\big)^{41}

\big(4^{504}\big)^{6}+\big(4^{432}\big)^{7}+\big(4^{378}\big)^{08}+\big(4^{336}\big)^{09}=\big(4^{55}\big)^{55}
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发表于 2023-2-24 11:15 | 显示全部楼层
试试,配个通项公式。谢谢!

(4^{020})^{1}+(4^{010})^{2}+(4^{005})^{04}+(4^{004})^{05}=(4^{07})^{03}
(4^{045})^{2}+(4^{030})^{3}+(4^{018})^{05}+(4^{015})^{06}=(4^{13})^{07}
(4^{084})^{3}+(4^{063})^{4}+(4^{042})^{06}+(4^{036})^{07}=(4^{23})^{11}
(4^{140})^{4}+(4^{112})^{5}+(4^{080})^{07}+(4^{070})^{08}=(4^{33})^{17}
(4^{216})^{5}+(4^{180})^{6}+(4^{135})^{08}+(4^{120})^{09}=(4^{47})^{23}
(4^{315})^{6}+(4^{270})^{7}+(4^{210})^{09}+(4^{189})^{10}=(4^{61})^{31}
(4^{440})^{7}+(4^{385})^{8}+(4^{308})^{10}+(4^{280})^{11}=(4^{79})^{39}
(4^{594})^{8}+(4^{528})^{9}+(4^{432})^{11}+(4^{396})^{12}=(4^{97})^{49}
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发表于 2023-2-25 02:08 | 显示全部楼层
无根之水,
无本之木,
胡编乱造,
扰乱军心!
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发表于 2023-2-25 09:18 | 显示全部楼层
(4^{020})^{1}+(4^{010})^{2}+(4^{005})^{04}+(4^{004})^{05}=(4^{07})^{03}
(4^{045})^{2}+(4^{030})^{3}+(4^{018})^{05}+(4^{015})^{06}=(4^{13})^{07}
(4^{084})^{3}+(4^{063})^{4}+(4^{042})^{06}+(4^{036})^{07}=(4^{23})^{11}
(4^{140})^{4}+(4^{112})^{5}+(4^{080})^{07}+(4^{070})^{08}=(4^{33})^{17}
(4^{216})^{5}+(4^{180})^{6}+(4^{135})^{08}+(4^{120})^{09}=(4^{47})^{23}
(4^{315})^{6}+(4^{270})^{7}+(4^{210})^{09}+(4^{189})^{10}=(4^{61})^{31}
(4^{440})^{7}+(4^{385})^{8}+(4^{308})^{10}+(4^{280})^{11}=(4^{79})^{39}
(4^{594})^{8}+(4^{528})^{9}+(4^{432})^{11}+(4^{396})^{12}=(4^{97})^{49}

\frac{\big(4^{(n^2-1)(n+2)/2}\big)^{n-2}+\big(4^{(n^2-4)(n+1)/2}\big)^{n-1}+\big(4^{(n^2-4)(n-1)/2}\big)^{n+1}+\big(4^{(n^2-1)(n-2)/2}\big)^{n+2}}{\big(4^{(2n^2-5-\cos(n\pi))/2}\big)^{(2n^2-5+\cos(n\pi))/4}}=1
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发表于 2023-2-28 15:51 | 显示全部楼层
求数例3,7,11,17,23,31,39,49…。的通项公式

a(n)=[\frac{n^2+2n}{2}]-1


求数例7,13,23,33,47,61,79,97…。的通项公式

a(n)=2*[\frac{n^2+2n+2}{2}]-3


6楼的通项公式不是很好吗?

点评

6楼的公式不是正整数解的公式  发表于 2023-3-17 05:34
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 楼主| 发表于 2023-2-28 21:42 | 显示全部楼层
Treenewbee老师的公式
已知x^a+y^b=z^c{,}则(xy^c)^a+y^{ac+b}=(zy^a)^c
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发表于 2023-3-1 17:18 | 显示全部楼层
朱明君 发表于 2023-2-28 21:45
x^2-dy^2=1,设n为≥1的正整数,则(n+(n+1))^2-(n(n+1))×2^2=1,

若x,y,d为正整数,\ \ x^2-dy^2=1, \ \ 则d=1, 4, 9, 16, 25, 36,.....无解。
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 楼主| 发表于 2023-3-16 22:28 | 显示全部楼层
本帖最后由 朱明君 于 2023-3-18 13:51 编辑

求数例3,7,11,17,23,31,39,49…。正整数解的通项公式
设n为大于等于2的正整数,
a\left( n\right)=\frac{n\left( n+1\right)+\left( n-x\right)}{2}{,}其中n为偶数,则x为2{,}\ \ \ \ \ n为奇数,则x为1{,}

求数例7,13,23,33,47,61,79,97…。正整数解的通项公式
设n为大于等于2的正整数,
a(n)=n(n+1)+(n-x), 其中n为偶数,则x为1,其中n为奇数,则x为2,


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\frac{\square}{\square}\sqrt{\square}\square_{\baguet}^{\baguet}\overarc{\square}\ \dot{\baguet}\left(\square\right)\binom{\square}{\square}\begin{cases}\square\\\square\end{cases}\ \begin{bmatrix}\square&\square\\\square&\square\end{bmatrix}\to\Rightarrow\mapsto\alpha\ \theta\ \pi\times\div\pm\because\angle\ \infty
\frac{\square}{\square}\sqrt{\square}\sqrt[\baguet]{\square}\square_{\baguet}\square^{\baguet}\square_{\baguet}^{\baguet}\sum_{\baguet}^{\baguet}\prod_{\baguet}^{\baguet}\coprod_{\baguet}^{\baguet}\int_{\baguet}^{\baguet}\lim_{\baguet}\lim_{\baguet}^{\baguet}\bigcup_{\baguet}^{\baguet}\bigcap_{\baguet}^{\baguet}\bigwedge_{\baguet}^{\baguet}\bigvee_{\baguet}^{\baguet}
\underline{\square}\overline{\square}\overrightarrow{\square}\overleftarrow{\square}\overleftrightarrow{\square}\underrightarrow{\square}\underleftarrow{\square}\underleftrightarrow{\square}\dot{\baguet}\hat{\baguet}\vec{\baguet}\tilde{\baguet}
\left(\square\right)\left[\square\right]\left\{\square\right\}\left|\square\right|\left\langle\square\right\rangle\left\lVert\square\right\rVert\left\lfloor\square\right\rfloor\left\lceil\square\right\rceil\binom{\square}{\square}\boxed{\square}
\begin{cases}\square\\\square\end{cases}\begin{matrix}\square&\square\\\square&\square\end{matrix}\begin{pmatrix}\square&\square\\\square&\square\end{pmatrix}\begin{bmatrix}\square&\square\\\square&\square\end{bmatrix}\begin{Bmatrix}\square&\square\\\square&\square\end{Bmatrix}\begin{vmatrix}\square&\square\\\square&\square\end{vmatrix}\begin{Vmatrix}\square&\square\\\square&\square\end{Vmatrix}\begin{array}{l|l}\square&\square\\\hline\square&\square\end{array}
\to\gets\leftrightarrow\nearrow\searrow\downarrow\uparrow\updownarrow\swarrow\nwarrow\Leftarrow\Rightarrow\Leftrightarrow\rightharpoonup\rightharpoondown\impliedby\implies\Longleftrightarrow\leftharpoonup\leftharpoondown\longleftarrow\longrightarrow\longleftrightarrow\Uparrow\Downarrow\Updownarrow\hookleftarrow\hookrightarrow\mapsto
\alpha\beta\gamma\Gamma\delta\Delta\epsilon\varepsilon\zeta\eta\theta\Theta\iota\kappa\varkappa\lambda\Lambda\mu\nu\xi\Xi\pi\Pi\varpi\rho\varrho\sigma\Sigma\tau\upsilon\Upsilon\phi\Phi\varphi\chi\psi\Psi\omega\Omega\digamma\vartheta\varsigma\mathbb{C}\mathbb{H}\mathbb{N}\mathbb{P}\mathbb{Q}\mathbb{R}\mathbb{Z}\Re\Im\aleph\partial\nabla
\times\cdot\ast\div\pm\mp\circ\backslash\oplus\ominus\otimes\odot\bullet\varnothing\neq\equiv\not\equiv\sim\approx\simeq\cong\geq\leq\ll\gg\succ\prec\in\ni\cup\cap\subset\supset\not\subset\not\supset\notin\not\ni\subseteq\supseteq\nsubseteq\nsupseteq\sqsubset\sqsupset\sqsubseteq\sqsupseteq\sqcap\sqcup\wedge\vee\neg\forall\exists\nexists\uplus\bigsqcup\bigodot\bigotimes\bigoplus\biguplus\bigcap\bigcup\bigvee\bigwedge
\because\therefore\angle\parallel\perp\top\nparallel\measuredangle\sphericalangle\diamond\diamondsuit\doteq\propto\infty\bowtie\square\smile\frown\bigtriangledown\triangle\triangleleft\triangleright\bigcirc \wr\amalg\models\preceq\mid\nmid\vdash\dashv\nless\ngtr\ldots\cdots\vdots\ddots\surd\ell\flat\sharp\natural\wp\clubsuit\heartsuit\spadesuit\oint\lfloor\rfloor\lceil\rceil\lbrace\rbrace\lbrack\rbrack\vert\hbar\aleph\dagger\ddagger

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