■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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