13/03/21 21:10:29.61
>>34-36
√(x+1) = y とおいて
x = y^2 -1 = (y-1)(y+1),
0 = (右辺)^2 - (左辺)^2
= x^3 - (x-1)(y+1)^2
= {x(y-1)^2 - (x-1)}(y+1)^2
= {(x-y)(y+1)}^2
= {(y^2 -y-1)(y+1)}^2,
y = -1, φ.
でもいいが
13/03/21 21:10:29.61
>>34-36
√(x+1) = y とおいて
x = y^2 -1 = (y-1)(y+1),
0 = (右辺)^2 - (左辺)^2
= x^3 - (x-1)(y+1)^2
= {x(y-1)^2 - (x-1)}(y+1)^2
= {(x-y)(y+1)}^2
= {(y^2 -y-1)(y+1)}^2,
y = -1, φ.
でもいいが