■奇数の2乗−1の整除性(その8)

以下の式は利用できないだろうか?

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F2n+1=(Fn+1)^2+(Fn)^2

Fn+2Fn-1=(Fn+1)^2-(Fn)^2

Fn+1Fn-1-(Fn)^2=(-1)^n

Fn=FmFn+1-m+Fm-1Fn-m

Ln+m+(-1)^mLn-m=LmLn

L2n+2(-1)^2=(Ln)^2

Ln-1+Ln+1=5Fn

Fn-1+Fn+1=Ln

Fn+2-Fn-2=Ln

Fn+Ln=2Fn+1

F2n=FnLn

Fn+1Ln+1-FnLn=F2n+1

Fn+m+(-1)^nFn-m=LmFn

Fn+m-(-1)^nFn-m=FmLn

LmFn+LnFm=2Fn+2

LmFn-LnFm=(-1)^m2Fn-2

Lm+n-(-1)^mLn-m=5FmFn

(Ln)^2-2L2n=-5(Fn)^2

L2n-2(-1)^2=5(Fn)^2

5(Fn)^2-(Ln)^2=4(-1)^n+1

3Fn+Ln=2Fn+2

5Fn+3Ln=2Ln+2

Ln=Fn+2+2Fn-1

Ln=L1Fn+L0Fn-1

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