ペル方程式x^2-(a^2-1)y^2=1の性質を調べよう。

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{a+(a^2-1)^1/2}{a-(a^2-1)^1/2}=1

{a+(a^2-1)^1/2}^n{a-(a^2-1)^1/2}^n=1

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{a+(a^2-1)^1/2}^n=xn(a)+yn(a)(a^2-1)^1/2

{a-(a^2-1)^1/2}^n=xn(a)-yn(a)(a^2-1)^1/2

xn(a)^2-(a^2-1)yn(a)^2=1

xn(1)=1,yn(1)=nと約束しておく

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x0(a)=1,x1(a)=a,x2(a)=2a^2-1

y0(a)=0,y1(a)=1,y2(a)=2a

i＜j,a＞1のときxi(a)＜xj(a), yi(a)＜yj(a)

xn(a)とyn(a)は互いに素

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xn+1(a)=axn(a)+(a^2-1)yn(a)

yn+1(a)=xn(a)+ayn(a)

xn+1(a)=2axn(a)-xn-1(a)

yn+1(a)=2ayn(a)-yn-1(a)

ym+n(a)=xn(a)ym(a)+xm(a)yn(a)

yn(a)|yt(a)←→n|t

yk+u(a)=-yk+u-1(a) (mod yk+yk+1)

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yn(a)=n (mod a-1)

yn(a)^2|y(nyn(a))(a)

yn(a)^2|yt(a)→yn(a)|t

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yn+2k+1(a)=yn(a) (mod yk(a)+yk+1(a))

y2n+1(a)=(yn+1(a)-yn(a))(yn+1(a)+yn(a))

i,j＜2k+1のとき、i≠jならばyi(a)≠yj(a) (mod yk(a)+yk+1(a))

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