1. ## Solve the equation

$\displaystyle \lfloor x \rfloor =x^2(x^2-2)$ where $\displaystyle \lfloor x \rfloor$ denotes the floor function.

2. Originally Posted by james_bond
$\displaystyle \lfloor x \rfloor =x^2(x^2-2)$ where $\displaystyle \lfloor x \rfloor$ denotes the floor function.
With the floor on the left, the right side must be integer valued which, in turn means $\displaystyle x^2$ must be integer valued. x is either an integer or the square root of an integer.

So just try some integers. If x= 0, this is 0= 0(0-2) which is certainly true!
If x= 1, 1= 1(1-2)= 0-1 which is NOT true. If x= -1, -1= 1(1-2)= -1 which is true!

If $\displaystyle x= \sqrt{2}$ this is 1= 2(2-1) which is NOT true.

For x> 1, the right hand side will be negative so there can be no further negative roots.

If $\displaystyle x= \sqrt{3}$ this is 1= 3(3-1) which is NOT true.

If x= 2, 2= 4(4-2)= 8 which is NOT true.

If $\displaystyle x= \sqrt{5}$ this is 2= 5(5-1)= 20 and it should be clear that if you the difference between right and left side just getting larger and larger- the right side is growing as $\displaystyle x^4$ while the right side is only growing as x.

The roots to this equation are x= -1 and x= 0.

3. Unfortunately $\displaystyle \sqrt{1+\sqrt{2}}$ is also a solution