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The Floor Equation (Posted on 2024-04-16) Difficulty: 3 of 5
Solve the floor equation:

[x]3 + 2x2 = x3 + 2[x]2

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Full solution Comment 2 of 2 |
If x is and integer [x]=x and the equation x3 + 2x2 = x3 + 2x2is always true.

If x is not an integer, let x=y+f where y is an integer and 0<f<1.
The equation becomes y3 + 2(y+f)2 = (y+f)3 + 2y2.

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 non-real.
If y=0, f=0 or 2.  Not allowed.
If y=1, f=-ϕ or ϕ-1.  The second is a possibility around 0.618
If y=2, f=-2.  Not allowed.
If y>2, f is non-real.

So the only non-integer x = y+f = ϕ.

  Posted by Jer on 2024-04-16 13:37:17
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