(1)定理:設(shè)x0,x1,x2,...是無(wú)窮實(shí)數(shù)列,xj>0,j>=1,那么,
      (i)對(duì)任意的整數(shù) n>= 1, r>=1有
            <X0,...,Xn-1,Xn,...,Xn+r> = <X0,...,Xn-1,<Xn,...,Xn+r>> 
            =   <X0,...,Xn-1,Xn+1/<Xn+1,...,Xn+r>>.
      特別地有
            <X0,...,Xn-1,Xn,Xn+1> = <X0,...,Xn-1,Xn+1/Xn+1>
      注:用該定理可以求連分?jǐn)?shù)的值

(2)對(duì)于連分?jǐn)?shù)數(shù)數(shù)列 <X0,...Xn> 有遞推關(guān)系:
      Pn = XnPn-1+Pn-2;
      Qn = XnQn-1+Qn-2;
      定義:  P-2 = 0; P-1 = 1; Q-2 = 1; Q-1 = 0;
      所以:  P0 = X0; Q0 = 1; P1 = X1X0+1; Q1 = X1;
      特別地:當(dāng) Xi=1 時(shí), {Pn}, {Qn}為Fbi數(shù)列

(3)pell方程: x^2+ny^2=+-1的解法:
      若n是平方數(shù),則無(wú)解, 否則:
      先求出sqrt(n)的連分?jǐn)?shù)序列<x0,x1..xn> 其中xn = 2*x0;
      對(duì)于 x^2+ny^2=-1
      若n為奇數(shù),則 x=Pn-1, y=Qn-1; n為偶數(shù)時(shí)無(wú)解
      對(duì)于 x^2+ny^2=1
      若n為偶數(shù),則 x=Pn-1, y=Qn-1; n為奇數(shù)時(shí)x=P2n-1, y=Q2n-1
      注:以上說(shuō)的解均為最小正解