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?
 A: You already ofund a solution $n=24$. 
For $n<24$, we have
$$ \left\lfloor\frac n2\right\rfloor \left\lfloor\frac n3\right\rfloor\left\lfloor\frac n4\right\rfloor\le \frac n{24}\cdot n^2<n^2.$$
For $n\ge 30$, we have
$$\begin{align}\left\lfloor\frac n2\right\rfloor \left\lfloor\frac n3\right\rfloor\left\lfloor\frac n4\right\rfloor -n^2&\ge\frac{n-1}2\frac{n-2}3\frac{n-3}4-n^2\\&=\frac{n^3-30n^2+11n-6}{24}\\&\ge\frac{11n-6}{24}\\&>0.\end{align}$$
Hence we need at most check $25\le n\le 29$. 
To simplify checks for these cases, note that $\left\lfloor\frac n2\right\rfloor$, $ \left\lfloor\frac n3\right\rfloor$, $\left\lfloor\frac n4\right\rfloor$ are three distinct integers $\ge 6$. As $29^2$ and $5^4$ cannot be written as product of three distinct integers $\>1$, we can exclude $n=29$ and $n=25$. For $3^6$, the only way to write it as product of three distinct integers is as $3\cdot 9\cdot 27$, but $3<6$, so we can exclude $n=27$ as well.
For $n=26$, we see that only one of the three factors is a multiple of $13$, so this can be ruled out as well. Fially, for $n=28$, $\lfloor \frac n3\rfloor=9$, but $28^2$ is not a multiple of $3$. 
A: 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.
A: A lengthy approach
As $[2,3,4]=12,$ let us try with all $12$ in-congruent residues $\pmod{12}$
If $n=12k$
$6k(4k)(3k)=(12k)^2\iff k=2$
If $n=12k+1,$
$6k(4k)(3k)=(12k+1)^2$ which is untenable as LHS is divisible by $6$
If $n=12k+2,(6k+1)(4k)(2k)=(12k+2)^2\iff6k+1=2k^2$ which is even
If $n=12k+3,(6k+1)(4k+1)2k=(12k+3)^2 which is odd unlike the LHS
If $n=12k+4,(12k+4)^2=(6k+2)(4k+1)2k\iff4(3k+1)=4k+1$ which is odd unlike LHS
and so on
