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

Question

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.

the greatest integer function would evaluate $0$ right? then how would it be divisible — Patrick Schick, Dec 01 '20 at 11:52
@ShlokJain, what cosmo5 said. I agree, though, that no prime $n\ge5$ will work. — Barry Cipra, Dec 01 '20 at 12:05

Bill Dubuque · Answer 1 · 2020-12-01T14:38:47.093

1

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 $\,\color{#0a0}{j/k \le 2}$

edited Dec 01 '20 at 14:38

answered Dec 01 '20 at 14:10

Bill Dubuque

272,048

Can You please explain the last line k∣(j/k)2∣4 by j/k≤2. How did you arrive k | 4? – Shashikant Aug 30 '21 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? – Shashikant Aug 30 '21 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}\ \ $ – Bill Dubuque Aug 30 '21 at 12:55
But in case of when j/k = 1, k | 4 will not hold true? – Shashikant Aug 30 '21 at 14:31
@Shashikant Why? – Bill Dubuque Aug 30 '21 at 14:43
I am not able to get this because I think if j/k = 1 and j = 1 and k | 4 then it implies that k = 1. – Shashikant Aug 30 '21 at 17:39
In that case (when k =1) n = 2, 3 will be the solution. Right? – Shashikant Aug 30 '21 at 17:47
@Sha Correct for $k=1$. There are only $6$ cases to check $,k=1,2.4,$ and $,j = k,,2k\ \ $ – Bill Dubuque Aug 30 '21 at 18:16

score 0 · Answer 2 · answered Dec 01 '20 at 12:04

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, 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.

@ShlokJain It's a bit easier to use the Rational Root Test - see my answer. — Bill Dubuque, Dec 01 '20 at 14:17

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

2 Answers2

$\text{If }k>1$

$\text{If }k=1$