Find natural numbers $n$ such that $\frac{3n^{3}-11n}{n + 3}$ is a natural number The question says:

*

*Find natural numbers $n$ so that $\frac{3n^{3}-11n}{n + 3}$ is a natural number.

I will give this information(Actually, they weren't information given by the exercise, they were questions but I already solved them)
$$3n^{3} - 11n = (3n^{2}-9n+16)(n + 3) - 48$$
Natural numbers that divide $48$: $(1,2,3,4,6,8,12,16,24,48)$
$3n^{2}-9n+16$ is a natural number.
My solution:
$$\frac{3n^{3}-11n}{n + 3} = (3n^{2}-9n+16) - \frac{48}{n+3}$$
So that x is a natural number means that:

*

*It's positive


*It's not a non-integer fractional.
The second I proved it easily:
It means that $n + 3$ has to be one divisors of $48$ so
$$n + 3 \in  \{1,2,3,4,6,8,12,16,24,48\}$$
then $$n \in \{3, 5, 9, 13, 21, 45\}$$
The problem is the first condition since I don't know if it's positive or not. I don't know how to prove it.
I don't want to put every $n$ in the equation and test if it's positive because that feels like I'm cheating or something.
EDIT: Oops, sorry I forgot that n must be equal or bigger than 2 but most solutions already knew this. Thank you!
 A: You are almost done with the solution and you got it right.

Note that
$$\begin{align}n(3n^2-11)(n+3)>0, n\in\mathbb Z \end{align}$$
$$\begin{align}&\iff n\in (-\infty ,-4]∪ \left\{-1\right\} ∪ [2,+\infty)&\end{align}$$
This means, you need also negative factors:
$$n+3:=±1,±2,±3, \cdots , ±48$$
Then, remember that
$$\begin{align}n\in (-\infty ,-4]∪ \left\{-1\right\} ∪ [2,+\infty)\end{align}$$
also must be hold.
A: "I don't know if it's positive or not. I don't know how to prove it."
Well, you could try just plugging them in.... if $n = 3$ then $(3n^{2}-9n+16) - \frac{48}{n+3}= 3\cdot 3^2 - 9\cdot 3 +16 -\frac {48}{3+3} = ....$
Okay that's tedious. but....
$\frac {3n^2-11n}{n+3}$ will be positive if the numerator or denominator are both positive of both negative.  If we assume $n \in \mathbb N$ we assume $n>0$ so $n + 3 > 0$.
So we need to find when $3n^3 - 11n > 0$ of when $3n^3 > 11n$.  As we are assume $n > 0$ this will occur if $3n^2 > 11$ or $n^2 > 3\frac 23$ which if we assume $n$ is natural is equivalent to $n \ge 2$.
.....
Alternatively... your solutions set dictates $n \ge 3$.  If $n\ge 3$ then $n+3 \ge 6$ and $\frac {48}{n+3} \le 8$ and $3n^2 - 9n +16-\frac {48}{n+3} \ge 3n^2 - 9n +8$.
And as $n \ge 3$ then $3n^2 = (3n)n \ge 9n$.  So $3n^2-9n\ge 0$ and $3n^2 - 9n +16-\frac {48}{n+3} \ge 3n^2 - 9n +8 > 8 > 0$.
A: Your approach yielded valid candidates for solutions, and there is nothing wrong or "cheating" with testing a small number of candidates to find the true solutions. However, there is a slightly different approach which might interest you.
When looking at $\frac{3n^3-11n}{n+3}$, there is no easy way to see the outcome of division by $n+3$. But you can restate the fraction in terms of a new parameter, $k=n+3$. The requirement $n \in \mathbb N \Rightarrow n>0 \Rightarrow k>3$ $$\frac{3n^3-11n}{n+3}=\frac{3(k-3)^3-11(k-3)}{k}=\frac{3k^3-27k^2+80k-48}{k}$$
Plainly, it must be the case that $k\mid 48$, which is exactly what you found by your approach.
