■ペル恒等式(その14)

 ブラーマグプタの恒等式

  (x1^2−Ny1^2)(x2^2−Ny2^2)=(x1x2+Ny1y2)^2−N(x1y2+x2y1)^2

と,Speckmann-Ricaldeの恒等式

  (k^2m±1)^2−(k^2m^2±2m)k^2=1

の関係がわかりにくい.

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 (x1^2−Ny1^2)(x2^2−Ny2^2)=1のとき,

 (x1^2−Ny1^2)=±1,(x2^2−Ny2^2)=±1

  N=(x1^2±1)/y1^2=(x2^2±1)/y2^2

=(x1^2±1)^1/2(x2^2±1)^1/2/y1y2

  x1^2y2^2±y2^2=x2^2y1^2±y1^2

  x1^2y2^2−x2^2y1^2=±(y1^2−y2^2)

であるが,

  (x2^2−Ny2^2)=ε^n(x1^2−Ny1^2)

も成り立つ.

 一方,

  (k^2m+1)^2−(k^2m^2+2m)k^2=1

k^2m^2+2m=Nとおくと,

k^2=(N−2m)/m^2

k^2m+1=(N−2m)/m+1=N/m−1

(N/m−1)^2−N(N−2m)/m^2=1

  (k^2m−1)^2−(k^2m^2−2m)k^2=1

k^2m^2−2m=Nとおくと,

k^2=(N+2m)/m^2

k^2m−1=(N+2m)/m−11=N/m+1

(N/m+1)^2−N(N+2m)/m^2=1

(N/m±1)^2−N(N±2m)/m^2=1

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[まとめ]両者は異なる恒等式に思えるのだが・・・

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