20/06/22 10:35:54.84 HOq0vlXr.net
>>572
n=3 の場合
y^3 = z^3 - x^3 + 1
= (z-x)(zz+xz+xx) + 1
= (z-x){(2x+z)^3 - (z-x)^3}/9x + 1,
ここで 9x = (z-x)^4 とすれば
y^3 = {(2x+z)/(z-x)}^3,
y = (2x+z)/(z-x)
= 3x/(z-x) +1
= (1/3)(z-x)^3 + 1,
z = x + (z-x),
これより
x = 9t^4,
y = 9t^3 + 1,
z = 9t^4 + 3t,