- 267 名前：１３２人目の素数さん mailto:sage [2012/07/06(金) 06:57:10.23 ]
- >>256,258の詳細
i=√-1=isin(2π/4)+cos(2π/4)
ω=isin(2π/3)+cos(2π/3)
α=isin(2π/12)+cos(2π/12)とおくと
x^12-1 = (x-α)(x-α^2)(x-α^3)(x-α^4)(x-α^5)(x-α^6)*……
……(x-α^7)(x-α^8)(x-α^9)(x-α^10)(x-α^11)(x-α^12)
ここでα^12=1, α^6=-1, α^4=ω, α^3=i,
α=α^13=(α^6)(α^3)(α^4)=-iω
x-α^12 = x-1
x-α^6 = x+1
(x-α^3)(x-α^9) = x^2+1
(x-α^4)(x-α^8) = x^2+x+1
(x-α^2)(x-α^10) = x^2-x+1
(x-α)(x-α^11) = (x+iω)(x-iω^2) = x^2-(√3)x+1
(x-α^5)(x-α^7) = (x-iω)(x+iω^2) = x^2+(√3)x+1
補足：ω-ω^2={-1+(√3)i}/2 - {-1-(√3)i}/2 = (√3)i
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