<s>[INST] 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?
 [/INST] 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
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