>>819

x^n+y^n=(x+n^{1/(n-1)})^n…(3)
n=3、x=(3/8)(3^(1/6))+(1/8)(3^(1/2))+(3/8)(3^(5/6))、y=2ｘとおく、
左辺
x^3+y^3=9x^3=(4617/512)(3^(1/6))+(3915/512)(3^(1/2))+(2673/512)(3^(5/6))
右辺
(x+3^{1/(3-1)})^3=(x+3^(1/2))^3=(4617/512)(3^(1/6))+(3915/512)(3^(1/2))+(2673/512)(3^(5/6))

よってこのx、yはx^n+y^n=(x+n^{1/(n-1)})^n…(3)を満たす
x:y=1:2

x^n+y^n=(x+n^{1/(n-1)})^n…(3)
n=3、x=(1/27)(3^(1/2))+(1/27)(3^(1/2))(2^(2/3))(7^(1/3))+(2/27)(3^(1/2))(2^(1/3))(7^(2/3))、y=3xと置く
左辺
x^3+y^3=28x^3=(3052/729)(3^(1/2))+(532/729)(3^(1/2))(2^(2/3))(7^(1/3))+(560/729)(3^(1/2))(2^(1/3))(7^(2/3))
右辺
(x+3^{1/(3-1)})^3=(x+3^(1/2))^3=3052/(243 sqrt(3)) + (532 2^(2/3) 7^(1/3))/(243 sqrt(3)) + (560 2^(1/3) 7^(2/3))/(243 sqrt(3))

よってこのx、yはx^n+y^n=(x+n^{1/(n-1)})^n…(3)を満たす
x:y=1:3


n=3、x=(1/64)(3^(1/2))+(1/64)(3^(1/2))(65^(1/3))+(1/64)(3^(1/2))(65^(2/3))、y=4xと置く
左辺
x^3+y^3=65x^3=(912795/262144)(3^(1/2))+(76635/262144)(3^(1/2))(65^(1/3))+(39195/262144)(3^(1/2))(65^(2/3))
右辺
(x+3^{1/(3-1)})^3=(x+3^(1/2))^3=(912795/262144)(3^(1/2))+(76635/262144)(3^(1/2))(65^(1/3))+(39195/262144)(3^(1/2))(65^(2/3))

よってこのx、yはx^n+y^n=(x+n^{1/(n-1)})^n…(3)を満たす
x:y=1:4

n=3、x=(8/27)(3^(1/2))+(4/27)(3^(1/2))(35^(1/3))+(2/27)(3^(1/2))(35^(2/3))、y=3x/2と置く
左辺
x^3+y^3=65x^3=(912795/262144)(3^(1/2))+(76635/262144)(3^(1/2))(65^(1/3))+(39195/262144)(3^(1/2))(65^(2/3))
右辺
(x+3^{1/(3-1)})^3=(x+3^(1/2))^3=(912795/262144)(3^(1/2))+(76635/262144)(3^(1/2))(65^(1/3))+(39195/262144)(3^(1/2))(65^(2/3))

よってこのx、yはx^n+y^n=(x+n^{1/(n-1)})^n…(3)を満たす
x:y=1:4

このように、(3)のx、yが整数比となる(3)の解はいくらでもあります。
その中に、x,y,zが整数比になるものがない、という証拠が、ありません。