■ペル恒等式(その18)

  1={(x^2+Ny^2)/(x^2−Ny^2)}^2−N(2xy/(x^2−Ny^2))^2

  (k^2m+1)^2−(k^2m^2+2m)k^2=1

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N←→k^2m^2+2m

k^2m+1←→(x^2+Ny^2)/(x^2−Ny^2)

←→2Ny^2/(x^2−Ny^2)+1

k^2←→{2xy/(x^2−Ny^2)}^2

とする.

k^2←→{2xy/(x^2−Ny^2)}^2

k^2m←→2Ny^2/(x^2−Ny^2)

m←→N(x^2−Ny^2)/2x^2

とすると

k^2m^2+2m←→N

が成り立つ.

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 逆変換も欲しいところである.そこで,

k^2m←→2Ny^2/(x^2−Ny^2)

m←→N(x^2−Ny^2)/2x^2

を用いる.

k^2m(x^2−Ny^2)=2Ny^2

k^2mx^2−N(k^2m+2)y^2=0

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

Nx^2−N^2y^2=2mx^2

(N−2m)x^2−N^2y^2=0

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

を代入すると

(N−2m)x^2=N^2y^2=Nk^2mx^2/(k^2m+2)

しかし,これではx^2=y^2=0になってしまう.

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