If $f(x)=\big\lfloor x\lfloor x\rfloor\big\rfloor$ for all $x\geq 0$, then for an integer $n$, solve for $x\geq 0$ such that $f(x)=n$. 
Let $f(x) = \big\lfloor x \lfloor x \rfloor \big\rfloor$ for $x \ge 0.$
(a) Find all $x \ge 0$ such that $f(x) = 1.$
(b) Find all $x \ge 0$ such that $f(x) = 3.$
(c) Find all $x \ge 0$ such that $f(x) = 5.$
(d) Find the number of possible values of $f(x)$ for $0 \le x \le 10.$

Attempt.
I've solved part (a), but I'm stuck on how to solved Part (b), (c), and (d).
The answer I got for (a) is $1 \leq x < 2$.  If $0\le x<1$, then $\lfloor x\rfloor=0$, so $f(x)=0\ne 1$.  If $x\ge 2$, then $\lfloor x\rfloor=2$ and $x\lfloor x\rfloor \ge 4$, so $f(x)\ge 4$.  Hence, when $f(x)=1$ we must have $1\le x<2$.  This means $\lfloor x\rfloor=1$ so $1\le x\lfloor x\rfloor<2$ and $f(x)=1$.
Could someone help me out with the other parts of the question? Thanks! (Also, there is a hint to divide into cases based on the value of $\lfloor x \rfloor.$
but I don't exactly understand.)
 A: If $x$ were allowed to be negative this would be a real pain but if $x \ge 0$ then $[x] \ge 0$
If $[x] = n$ then $n \le x < n+1$ and $n^2 \le nx \le n^2 + n$ with the second equality holding only if $n= 0$....
So if $0 \le x < 1$ then $[x[x]] = 0$.
If $x \ge 1$ then $[x] = n \ge 1$ and $n \le x < n+1$ so $1\le n^2 \le nx = x[x] < n^2 + n$ so $n^2 \le [nx]=[x[x]] < [nx]+1 \le  n^2+n$.
So if $f(x) = 0$ then $x\in[0,1)$
And if $f(x) = k$ then there is an $n\in \mathbb N$ so that $n^2 \le k < n^2 + n$ and $x \in [n,n+1)$ (but there is only one such $n$.)
But if $k$ is such that there is no such $n$ (which will happen frequently if there is an $m$ so that $m^2 + m \le k < (m+1)^2$) there will be no solutions.
So $f(x) = 1$ means that $1^2 \le f(x) < 1^2 + 1$ so $1\le x < 2$.  $f(x) =3$ means $n^2 \le 3< n^2 +n$ which is impossible.
$f(x) = 5$ means that $2^2 \le f(x) < 2^2 + 2$ so $2\le x < 3$.
And $0 =0$ and that occurs if $x\in [0,1)$.
And $1 \le 1 < 2$ and so if $ x\in [1,2)$ then $f(x)=1$..
And $2^2 \le 4$ and $5 < 2^2+2$ so $f(x) =4,5$ are possible if $x\in [2,3)$.  If $x < 2.5$ then $x[x]=2x < 5$ and $f(x) =4$.  If $4x \geq 2.5,$ then $f(x) =5$.
$3^2 \le 9,10,11 < 3^2 + 3$ so $f(x) = 9,10,11$ are possible.  if $x \in [3,4)$.  If so then $[x] = 3$ and $[x[x]] = 9$ if $x < 3\frac 13$ and $f(x) =10$ if $3\frac 13\le x < 3\frac 23$ and if $x \ge 3\frac 23$ then $f(x)=11$.  If $x \ge 4$ then $f(x) \ge 16$.
