For which $n\in\mathbb{N}$ is $(n+2)\mid(n^3+14)$ true?

Question

Tried substitution $n+2=p$, but it only gets more complicated.

For $ n=1 $ this is true, but how can I find other $ n$'s?

Answer 1

Since $$n^3+14=n^3+8+6=(n+2)(n^2-2n+4)+6$$ it follows that $n+2$ divides $n^3+14$ iff $n+2$ divides $6$.

Answer 2

Hint:

Mod $n+2$, $n\equiv-2$ so $n^3\equiv-8$ so $n^3+14\equiv6$.

Now find $n$ such that $6\equiv0\bmod n+2$.

Answer 3

Hint: $$n^3+8=(n+2)((n+2)^2-6n)$$

Answer 4

$n^3+2=(n+2)(n^2-2n+4)=6n+8\mid 6$?

Answer 5

I think this is only possible for n=1.

if we factorise the dividend with the divisor then the remainder is the number 6. The prime factorisation of 6 is 2x3.

(n+2) is certainly more than 2.

So n+2 = 3

Is the only case where we can factorise the dividend with the divisor perfectly.

And that means that n = 1