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By applying long division, we can find this

$3n^{3}-11n = (n+3)(3n^{2}-9n+16) - 48$

Which is

$\frac{3n^{3}-11n}{n+3} = (3n^{2}-9n+16) - \frac{48}{n+3}$

The question: Find all $n > 2$ natural numbers so that $\frac{3n^{3}-11n}{n+3}$ is a natural number.

The book solved it this way but I think it's wrong:

$3n^{2}-9n+16 > 0$

That means $n+3$ has to be a divisor of $48$

$n + 3 = (1,2,3,4,6,8,12,16,24,48)$

So $n = (3,5,9,13,21,45)$

But I think this way is wrong because this is only absolutely correct if $-\frac{48}{n+3} > 0$ but we know that it's $< 0$ since $n > 2$ so we have no way of knowing if it's truly a natural number and not a negative one.

I think this way is only correct in that its result will be non-decimal which is half the correct answer but that's it.

What do you think?

EDIT: Okay. So I found the correct answer, thank you all. And yes this was a duplicate but I really didn't notice. Sorry for that!

Bill Dubuque
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TechnoKnight
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  • Of that set of solutions, if any of them is plugged in does the whole quantity become negative? – abiessu Jan 23 '22 at 16:20
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    Note that for $n > 2$, $( 3n^3 - 11n) / (n+3)$ is a positive number, so you don't have to check for it in another form. – Calvin Lin Jan 23 '22 at 16:28
  • @abiessu I think if we test for every solution, we would find them positive. However, I really don't like it. I would rather have a dynamic math proof ensuring that it's positive without testing. – TechnoKnight Jan 23 '22 at 16:31
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    OP, this question is a duplicate. You asked this before – Mike Jan 23 '22 at 16:33
  • By the first dupe $, n+3\mid f(n) = 3n^3-11n\iff n+3\mid f(-3)=-48\iff n,$ is $3$ less than a divisor of $48,,$ so simply test which of those satisfy the given constraints on $,n.\ \ $ – Bill Dubuque Jan 23 '22 at 17:01

1 Answers1

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The original quantity is positive when $3n^2-9n+16\gt 0$ and $3n^2-9n+16\gt\frac{48}{n+3}$. The first condition is satisfied for every $n\in \Bbb N$. The second is satisfied for $n\ge 3$. Then only divisors of $48$ need to be considered in order for the original expression to be an integer.

abiessu
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