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?

  • 6
    $\begingroup$ It seems to me that the solution of $(\frac {n}{2} -1)(\frac {n}{3}-1)(\frac {n}{4} -1) = n^2$ should provide an upper bound for possible solutions, and this should give a manageable number of cases to check. $\endgroup$ Feb 19, 2019 at 16:29
  • 1
    $\begingroup$ Yep, that's a good starting point. For all $n \ge 33$, we have $(\tfrac{n}{2}-1)(\tfrac{n}{3}-1)(\tfrac{n}{4}-1) > n^2$. Thus, we only need to check $1 \le n \le 32$. $\endgroup$
    – JimmyK4542
    Feb 19, 2019 at 16:31
  • 2
    $\begingroup$ Using the fact that $x-1\leq [x]<x$, we have $$ \frac{(n-2)(n-3)(n-4)}{24}\leq n^2<\frac{n^3}{24}$$ which implies $24\leq n$. $\endgroup$
    – Qurultay
    Feb 19, 2019 at 16:33
  • $\begingroup$ And for $n \le 23$ we have $(n/2)(n/3)(n/4) < n^2$. So we don't need to check $n \le 23$. It suffices to check $24 \le n \le 32$. $\endgroup$ Feb 19, 2019 at 16:33

3 Answers 3


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 $3$.

  • $\begingroup$ (+1) I took too long working on my answer, that I did not see yours, which is essentially the same. $\endgroup$
    – robjohn
    Feb 19, 2019 at 19:36

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: $$ \frac{n}{2}\cdot \frac{n}{3}\cdot \frac{n}{4} - n^2 >\left\lfloor\frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor - n^2 = 0 $$


$$ \frac{n^3}{24}-n^2= n^2\left( \frac{n}{24}-1\right) >0 $$

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:

And the only answer is 24.


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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