整数平方根的连分数表示

elim

设\(\small\,(n\in\mathbb{N}^+)\wedge(\sqrt{n}\not\in\mathbb{N}),\;F(n)=\{(m,k)\in\mathbb{Z}\times\mathbb{N}:\,k\mid n-m^2\in(0,n)\}\)
令\(\,{\small\lambda(m,k)=}\lfloor\frac{\sqrt{n}+m}{k}\rfloor,\;m_1=k\cdot\lambda(m,k)-m,\,k_1=\large\frac{n-m_1^2}{k}.\;\)则有
\({\small\sqrt{n}-m_1=}\big(\frac{\sqrt{n}+m}{k}{\small-\lambda(m,k)}\big)k>0,\;\sqrt{n}+m_1>\sqrt{n}-m>0\)
\(\therefore\;k\mid n-m^2+2mk\lambda-(k\lambda)^2=n-m_1^2>0,\;(m_1,k_1)\in F(n)\)
\(\therefore\;\frac{k}{\sqrt{n}+m}=\frac{k}{k\lambda+\sqrt{n}+m-k\lambda}=\frac{k}{k\lambda+\sqrt{n}-m_1}=\frac{1}{\lambda+\large\frac{k_1}{\sqrt{n}+m_1}}\)
令\(\,\psi^{\langle k+1\rangle}=\psi(\psi^{\langle k\rangle}),\;(m_j,k_j)=\psi^{\langle j\rangle}(m,k),\;\lambda_j=\lambda(m_{j-1},k_{j-1})\)
定义\(\,[a,b]=a+\large\frac{1}{[b]},\,\)则有\(\small\,1\le m,\,L\in\mathbb{N}\,\)使\(\small(m_{u+L},k_{u+L})=(m_u,k_u)\)
\(\,\frac{k}{\sqrt{n}+m}=[0,\lambda_1+\frac{k_1}{\sqrt{n}+m_1}]=[0,\lambda_1,\ldots,\lambda_{u}+\frac{k_u}{\sqrt{n}+m_u}]\)
\(\qquad\cdots=[0,\lambda_1,\ldots,\lambda_{u},\ldots,\lambda_{u+L}+\frac{k_{u+L}}{\sqrt{n}+m_{u+L}}]\)
\(\qquad\cdots=[0,\lambda_1,\ldots,\lambda_{u},\dot{\lambda}_{u+1},\ldots,\dot{\lambda}_{u+L}]\)
\(\qquad\cdots=[0,\lambda_1,\ldots,\lambda_{u},\overline{{\lambda}_{u+1},\ldots,\lambda}_{u+L}]\)
这是因为\(F(n)\)是有限集.
取\(\,m,k\in\mathbb{N}\,\)使\(\small\,m^2< n< (m+1)^2,\,k=n-m^2,\,\)则\(\small\,(m,k)\in F(n).\)
可见\(\,\sqrt{n}=[m;\lambda_1,\ldots,\lambda_u,\overline{\lambda_{u+1},\ldots,\lambda}_{u+L}]\,\)是循环连分数.

luyuanhong

楼上 elim 的帖子很好！已收藏。

elim

命题 若\(\,a_i\in\mathbb{N}^+,\;\alpha=[\overline{a_1,\ldots,a}_L]\;\small(L>1)\,\)
\(\qquad\)则\(\,\alpha\,\)是整系数二次方程的根.
证明：此时\(\,\alpha=[a_1,\ldots,a_L+\alpha].\) 当\(\,L=2\,\)时
\(\,\alpha=a_1+\large\frac{1}{a_2+\alpha}\;\)即\(\,\alpha^2+(a_2-a_1)\alpha-a_1a_2-1=0\)
当\(\,L>2\,\)时\(\,\alpha=[a_1,\ldots,a_L+\alpha]=[a_1,\ldots,a_{L-1}+\frac{1}{a_L+\alpha}]\)
所以据归纳法原理, 命题对一切\(L>1\)成立.

推论1 循环连分数的值具有一般形式\(\,\frac{u\pm\sqrt{|v|}}{w}\;\small(u,v,w\in\mathbb{Z})\)

推论2 若\(\,k,\,n\in\mathbb{N}^+,\, k>2,\,\sqrt[k]{n}\not\in\mathbb{N},\,\)则
\(\qquad\sqrt[k]{n}\,\)的连分数表示是无穷不循环的. 这个结果出乎意料.

elim

\(\sqrt[3]{3}\) 的连分数表示:

elim

设 \(n\in M=\{ m\in\mathbb{N}:\,\sqrt{m}\not\in\mathbb{N}\},\) 定义集合
\(E(n)=\{(m,k)\in\mathbb{Z}^2:\,0< k\mid n-m^2>0\}\). 令
\(\small\,\lambda(m,k)=\big\lfloor{\small\dfrac{\sqrt{n}+m}{k}}\big\rfloor,\;\eta(m,k)=\small\dfrac{n-(k\lambda(m,k)-m)^2}{k}\)
\(\psi(m,k)=(k\lambda(m,k) -m, \eta(m,k))\).
试证 \(\psi(E(n))\subset E(n)\). 讨论\(\,\psi\,\)是否是满射, 是否是单射.