■整数であるか? (その85)

フィボナッチ数列

f(x)=(x)/(1-x-x^2)=(1/√5)/(1-αx)-(1/√5)/(1-βx)

α=(1+√5)/2、β=(1-√5)/2,αβ=-1,α^2+β^2=3、β=-1/α

an=1/√5・{α^n-β^n}

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F2n+1=Fn^2+Fn+1^2l

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a2n+1=1/√5・{α^2n+1-β^2n+1}

a2n-1=1/√5・{α^2n-1-β^2n-1}

(a2n+1)^2=1/5・{α^4n+2-2(αβ)^2n+1+β^4n+2}=1/5・{α^4n+2+2+β^4n+2}

(a2n-1)^2=1/5・{α^4n-2-2(αβ)^2n-1+β^4n-2}=1/5・{α^4n-2+2+β^4n-2}

(a^2+b^2+1)=1/5・{α^4n+2+2+β^4n+2+α^4n-2+2+β^4n-2+5}

(a^2+b^2+1)=1/5・{α^4n+2+β^4n+2+α^4n-2+β^4n-2+9}

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a2n+1・a2n-1=1/5・{α^2n+1-β^2n+1}{α^2n-1-β^2n-1}

a2n+1・a2n-1=1/5・{α^4n+β^4n-α^2n+1β^2n-1-α^2n-1β^2n+1}

a2n+1・a2n-1=1/5・{α^4n+β^4n-α^2(αβ)^2n-1-(αβ)^2n-1β^2}

a2n+1・a2n-1=1/5・{α^4n+β^4n+α^2+β^2}

a2n+1・a2n-1=1/5・{α^4n+β^4n+3}

3a2n+1・a2n-1=1/5・{3α^4n+3β^4n+9}

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5(a^2+b^2+1)-15ab

={α^4n+2+β^4n+2+α^4n-2+β^4n-2+9-3α^4n-3β^4n-9}

={α^4n(α^2+α^-2)+β^4n(β^2+β^-2)-3α^4n-3β^4n}

={α^4n(α^2+α^-2-3)+β^4n(β^2+β^-2-3)}

={α^4n(α^2+β^2-3)+β^4n(α^2+β^2-3)}

=0

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a=F2k-1,b=F2k+1のとき、

(a^2+b^2+1)/ab=3

が成り立つ

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