# 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.

• What about the primes $n=2$ and $3$? Commented Dec 1, 2020 at 11:47
• the greatest integer function would evaluate $0$ right? then how would it be divisible Commented Dec 1, 2020 at 11:52
• $[\sqrt{2}]=1$, $[\sqrt{3}]=1$. Commented Dec 1, 2020 at 11:59
• @ShlokJain, what cosmo5 said. I agree, though, that no prime $n\ge5$ will work. Commented Dec 1, 2020 at 12:05
• okay I will edit that into the question Commented Dec 1, 2020 at 12:10

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}$$

• Can You please explain the last line k∣(j/k)2∣4 by j/k≤2. How did you arrive k | 4? Commented Aug 30, 2021 at 12:39
• I think I got it how it is. Since 0 < j < 2k + 1 , therefore we can write 0 < j/k <= 2. Is my understanding correct? Commented Aug 30, 2021 at 12:53
• @Shashikant Yes, $\,\color{#0a0}{j/k}\,$ is $\color{#c00}1\,$ or $\,\color{#c00}2\,$ so its square $\color{#c00}{{\rm divides}\ 4}\ \$ Commented Aug 30, 2021 at 12:55
• But in case of when j/k = 1, k | 4 will not hold true? Commented Aug 30, 2021 at 14:31
• @Shashikant Why? Commented Aug 30, 2021 at 14:43

Consider $$k$$ such that $$k^2. Then $$\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.

• don't you think $n = 8$ would also work Commented Dec 1, 2020 at 12:12
• @ShlokJain It's a bit easier to use the Rational Root Test - see my answer. Commented Dec 1, 2020 at 14:17