Solving an equation with floor function I need some help regarding this question.

Solve the following equation in natural number $x$, where $m,k$ are fixed naturals:
$m\left\lfloor \sqrt{\dfrac{x}{k}}\right\rfloor = x$.

I think the answer is $x = m\left\lfloor\dfrac{m}{k}\right\rfloor$. But I do not know how to prove this claim.
I would be happy if somebody would help me solve this problem.
Thank you,
Isomorphism

Since $x$ must be a multiple of $m$, write $x=my$.
Then the equation becomes $$y=\left\lfloor\sqrt{\frac {my}k}\right\rfloor$$
and  equivalent to
$$y\le \sqrt{\frac {my}k}<y+1,$$
i.e. 
$$y^2\le \frac {my}k <y^2+2y+1$$
or (using $y\ne 0$, though in fact $x=0$ is a trivial solution)
$$y\le \frac {m}k <y+2+\frac1y\le y+3.$$
Therefore $y=\left\lfloor \frac mk\right\rfloor$,  $y=\left\lfloor \frac mk\right\rfloor-1$, and in rare cases $y=\left\lfloor \frac mk\right\rfloor-2$ are the solutions - the latter only if $m\ge 3k$ and $\frac mk-\left\lfloor \frac mk\right\rfloor<\frac 1{\left\lfloor \frac mk\right\rfloor-2}$.
