If x is and integer [x]=x and the equation
x^{3} + 2x^{2} = x^{3} + 2x^{2}is always true.
If x is not an integer, let x=y+f where y is an integer and 0<f<1.
The equation becomes y^{3} + 2(y+f)^{2} = (y+f)^{3} + 2y^{2}.
y^3 + 2y^2 + 4yf + 2f^2 = y^3 + 3y^2f + 3yf^2 + f^3 + 2y^2
4yf + 2f^2 = 3y^2f + 3yf^2 + f^3
4y + 2f = 3y^2 + 3yf + f^2
3y^2 + 3yf + f^2  4y 2f = 0
Which is a general conic. The discriminant test shows it is an ellipse, so there are not too many possibilities for y.
If y<0, f is nonreal.
If y=0, f=0 or 2. Not allowed.
If y=1, f=ϕ or ϕ1. The second is a possibility around 0.618
If y=2, f=2. Not allowed.
If y>2, f is nonreal.
So the only noninteger x = y+f = ϕ.

Posted by Jer
on 20240416 13:37:17 