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用连分数求PELL方程

Posted by ait2008 at 2009-04-21 11:52:31 on Problem 1320
1+2+...+(a-1) = (a+1)+(a+2)+...+b==> 8*a^2     = 4*b^2 +4*b +1-1==> 2*(2*a)^2 = (2*b+1)^2 -1==> (2*b+1)^2 - 8*(a)^2 = 1令  x = 2*b+1, y= a则  x^2 - 8*y^2 = 1 最小正解可以1.直接试为 3,1
2.连分数 sqrt(8)=<2,1,4> ,周期为2,及<2,1>=3/1 得3,1所有正解为 x+y*sqrt(8)=(3+sqrt(8))^j (j=1,2,3……)
x[n+1]+y[n+1]*sqrt(8)=(x[n]+y[n]*sqrt(8))*(3+sqrt(8))=3*x[n]+8*y[n]+(3*y[n]+x[n])*sqrt(8)可得递推式。

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