*** edited to show correct argument of floor function in solution.
F((x+1)
^{3}) = x
^{3}, where F(x) = x  floor(x)
0 <= F(x) < 1, so 0 <= x < 1
F((x+1)^{3}) = x^{3 }+ 3x^{2 }+ 3x + 1  floor((x+1)^{3}) = x^{3}
^{}
floor((x+1)^{3}) =3x^{2 }+ 3x + 1 = 3(x^{2 }+ x) + 1, an integer >=0 and < 7 as x < 1
x^{2 }+ x = 5/3, x = 1/2+sqrt(23/3)/2
x^{2 }+ x = 4/3, x = 1/2+sqrt(19/3)/2
x^{2 }+ x = 1, x = 1/2+sqrt(5)/2
x^{2 }+ x = 2/3, x = 1/2+sqrt(11/3)/2
x^{2 }+ x = 1/3, x = 1/2+sqrt(7/3)/2
x^{2 }+ x = 0, x = 0
Edited on September 27, 2016, 6:29 pm
Edited on September 27, 2016, 6:30 pm

Posted by ken
on 20160927 13:43:49 