■整数であるか? (その98)
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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