Find all $n$ $\in$ $\Bbb Z^+$ such that: $\lfloor\frac{n}{2}\rfloor \cdot \lfloor \frac{n}{3} \rfloor \cdot \lfloor \frac{n}{4} \rfloor = n^2$ Find all the numbers $n$ $\in$ $\Bbb Z^+$ such that:
$$\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor = n^2$$
I never worked before with floor function so i'm not completely sure how to solve this. I think (only because $n$ $\in$ $\Bbb Z^+$), i can just multiply (skipping the floor function) and get the answer of $n=24$, but this is floor function so i don't know if there are more solutions. 
Any hints?

We have
$$
\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 =0
$$
Note that:
$$
x > \left\lfloor x \right\rfloor
$$
Thus:
$$
\frac{n}{2}\cdot \frac{n}{3}\cdot \frac{n}{4}  - n^2 >\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2  = 0
$$
Thus:
$$
\frac{n^3}{24}-n^2= n^2\left( \frac{n}{24}-1\right) >0
$$
Hence we are looking for a number larger than 24.
At the same time:
$$
\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 > \left( \frac{n}{2}-1\right) \cdot\left( \frac{n}{3}-1\right) \cdot\left( \frac{n}{4}-1\right) -n^2
$$
Therefore:
$$
0>\left( \frac{n}{2}-1\right) \cdot\left( \frac{n}{3}-1\right) \cdot\left( \frac{n}{4}-1\right) -n^2
$$
This function has only one real root around 32.216 I'm lazy, sorry about that.
Hence we are looking for $n$ in the interval $[24,32]$. I'll check it in Python:
from math import floor
for n in range(24, 33):
    if (floor(n/2)*floor(n/3)*floor(n/4)-n*n) == 0:
        print(n)

And the only answer is 24.
