9-4√5 が出て来るτ = √-t ( t > 0 ) k( τ ) = 1/√5 k’( τ ) = 2/√5 として K’( k ) = K( k’ ) = √tK( k ) k( 2τ ) = ( 1-k’( τ ) )/( 1+k’( τ ) ) = ( √5-2 )^2 = 9-4√5 = ( 2k( √-5 ) )k’( √-5 ) )^2 ⇒ t = 5 ???