determine with proof, all the positive integers $n$ for which $n$ is not the square of any integer and $[\sqrt{n}]^3$ divides $n^2$ determine with proof, all the positive integers $n$ for which:

*

*$n$ is not the square of any integer and,

*$[\sqrt{n}]^3$ divides $n^2$. where $[x]$ is the greatest integer less than or equal to $x$.

My Approach:
Well I tried a few test cases and I concluded that no prime number greater than 4 holds true to this statement. Except that, I couldn't recognize a pattern or make a solution on my own so please help me out.
 A: Hint: $\,[\sqrt n]\! =\! k < \sqrt n < k\!+\!1 \!\iff\! k^2 < n < (k\!+\!1)^2 \!\!\iff\! 0 < \overbrace{n\!-\!k^2}^{\color{#0a0}{\textstyle j}} \color{#0a0}{< 2k+1}$
Thus $\, k^3\mid (k^2\!+\!j)^2 \Rightarrow  k^2\mid j^2 \Rightarrow k\mid j\, $ via Rational Root Test, so cancelling $\,k^2$
yields $\ k\mid (k\!+\!j/k)^2 \Rightarrow \,\bbox[5px,border:1px solid #c00]{\color{#c00}k\mid (j/k)^2\,\color{#c00}{\mid\, 4}}\ $ by $\,\color{#0a0}{j/k \le 2}$
A: Consider $k$ such that $k^2<n<(k+1)^2$. Then $\lfloor n\rfloor=k$ and $n=k^2+i$, $1\leq i\leq 2k$. We get the following:
$$k^3|(k^2+i)^2\Leftrightarrow k^3|k^4+i(2k^2+i)\Leftrightarrow k^3|i(2k^2+i)$$
$\text{If }k>1$
Consider a prime $p$ that divides $k$ and let $v_p(k)$ the exponent of $p$ in the factorisation of $k$. We must have $$v_p(k^3)\leq v_p(i(2k^2+i))\Leftrightarrow3\cdot v_p(k)\leq v_p(i)+v_p(2k^2+i)$$ Suppose $v_p(i)\leq v_p(k)$. Then, obivously, $v_p(2k^2+i)=v_p(i)$ so $v_p(i)+v_p(2k^2+i)\leq 2\cdot v_p(k)$, which would lead to $3\cdot v_p(k)\leq2\cdot v_p(k)$ which is a contradiction ($p$ divides $k$ so $v_p(k)\geq 1$).
Thus, for any $p $that divides $k$, $v_p(i)>v_p(k)$ so $k|i$ and $k\neq i$. However, $i\leq 2k$, so this implies $i=2k$ (so $n=k^2+2k$) which gives us $$k^3|2k(2k^2+2k)\Leftrightarrow k^3|4k^3+4k^2\Leftrightarrow k^3|4k^2 \Leftrightarrow k|4$$
Thus, just check cases $k=2,4$, which give $n=8,24$
$\text{If }k=1$
Then this gives us $n=2,3$ and again, just check those cases.
