Find all $n$ $\in$ $\Bbb Z^+$ such that: $\lfloor\frac{n}{2}\rfloor \cdot \lfloor \frac{n}{3} \rfloor \cdot \lfloor \frac{n}{4} \rfloor = n^2$

Question

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?

Answer 1

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 $3$.

Answer 2

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.

Answer 3

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

Answer 4

3 solutions; 24, 25, 26 $${n}^{\mathrm{2}} =\lfloor\frac{{n}}{\mathrm{2}}\rfloor\lfloor\frac{{n}}{\mathrm{3}}\rfloor\lfloor\frac{{n}}{\mathrm{4}}\rfloor \\ $$ $${n}=\mathrm{12}{m}+{k} \\ $$ $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}{\boldsymbol{{k}}}&\hline{\boldsymbol{{n}}^{\mathrm{2}} =\lfloor\frac{\boldsymbol{{n}}}{\mathrm{2}}\rfloor\lfloor\frac{\boldsymbol{{n}}}{\mathrm{3}}\rfloor\lfloor\frac{\boldsymbol{{n}}}{\mathrm{4}}\rfloor}&\hline{\boldsymbol{{m}}}&\hline{\boldsymbol{{n}}}\\{\mathrm{0}}&\hline{\left(\mathrm{12}{m}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}\right)\left(\mathrm{4}{m}\right)\left(\mathrm{3}{m}\right)}&\hline{\mathrm{2}}&\hline{\mathrm{24}}\\{\mathrm{1}}&\hline{\left(\mathrm{12}{m}+\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}\right)\left(\mathrm{4}{m}\right)\left(\mathrm{3}{m}\right)}&\hline{\varnothing}&\hline{\varnothing}\\{\mathrm{2}}&\hline{\left(\mathrm{12}{m}+\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{1}\right)\left(\mathrm{4}{m}\right)\left(\mathrm{3}{m}\right)}&\hline{-\frac{\mathrm{1}}{\mathrm{6}}}&\hline{\mathrm{0}}\\{\mathrm{3}}&\hline{\left(\mathrm{12}{m}+\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{1}\right)\left(\mathrm{4}{m}+\mathrm{1}\right)\left(\mathrm{3}{m}\right)}&\hline{-\frac{\mathrm{1}}{\mathrm{4}}}&\hline{\mathrm{0}}\\{\mathrm{4}}&\hline{\left(\mathrm{12}{m}+\mathrm{4}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{2}\right)\left(\mathrm{4}{m}+\mathrm{1}\right)\left(\mathrm{3}{m}+\mathrm{1}\right)}&\hline{\frac{\mathrm{7}}{\mathrm{4}}}&\hline{\mathrm{25}}\\{\mathrm{5}}&\hline{\left(\mathrm{12}{m}+\mathrm{5}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{2}\right)\left(\mathrm{4}{m}+\mathrm{1}\right)\left(\mathrm{3}{m}+\mathrm{1}\right)}&\hline{\varnothing}&\hline{\varnothing}\\{\mathrm{6}}&\hline{\left(\mathrm{12}{m}+\mathrm{6}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{3}\right)\left(\mathrm{4}{m}+\mathrm{2}\right)\left(\mathrm{3}{m}+\mathrm{1}\right)}&\hline{\frac{\mathrm{5}}{\mathrm{3}}}&\hline{\mathrm{26}}\\{\mathrm{7}}&\hline{\left(\mathrm{12}{m}+\mathrm{7}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{3}\right)\left(\mathrm{4}{m}+\mathrm{2}\right)\left(\mathrm{3}{m}+\mathrm{1}\right)}&\hline{\varnothing}&\hline{\varnothing}\\{\mathrm{8}}&\hline{\left(\mathrm{12}{m}+\mathrm{8}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{4}\right)\left(\mathrm{4}{m}+\mathrm{2}\right)\left(\mathrm{3}{m}+\mathrm{2}\right)}&\hline{\frac{\mathrm{3}}{\mathrm{2}}}&\hline{\mathrm{26}}\\{\mathrm{9}}&\hline{\left(\mathrm{12}{m}+\mathrm{9}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{4}\right)\left(\mathrm{4}{m}+\mathrm{3}\right)\left(\mathrm{3}{m}+\mathrm{2}\right)}&\hline{-\frac{\mathrm{3}}{\mathrm{4}}}&\hline{\mathrm{0}}\\{\mathrm{10}}&\hline{\left(\mathrm{12}{m}+\mathrm{10}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{5}\right)\left(\mathrm{4}{m}+\mathrm{3}\right)\left(\mathrm{3}{m}+\mathrm{2}\right)}&\hline{-\frac{\mathrm{5}}{\mathrm{6}}}&\hline{\mathrm{0}}\\{\mathrm{11}}&\hline{\left(\mathrm{12}{m}+\mathrm{11}\right)^{\mathrm{2}} =\left(\mathrm{6}{m}+\mathrm{5}\right)\left(\mathrm{4}{m}+\mathrm{3}\right)\left(\mathrm{3}{m}+\mathrm{2}\right)}&\hline{\varnothing}&\hline{\varnothing}\\\hline\end{array} \\ $$ $${n}=\left\{\mathrm{24},\mathrm{25},\mathrm{26}\right\} \\ $$