Let $n$ be a positive integer. Prove that if $18|n^3$, then $18|n^2$.

Question

Let $n$ be a positive integer. Prove that if $18|n^3$, then $18|n^2$.

I realize that this does not also apply to $18|n$, since it works for $n=6$. I think that in order to solve this, the fundamental theorem of arithmetic needs to be used.

Answer 1

Hint: Consider the prime factorization. If $p\mid n^2$, then $p\mid n$. In this case, you have $n^3$...

Answer 2

Suppose $18$ divides $n^3$. This means that there is some integer $k$ such that $18k=n^3$. Notice that $18=2\cdot 3^2$. So $2$ divides $n^3$ and $3^2$ divides $n^3$.

If $2$ divides $n^3$, what can you say about $n$ itself?

If $3$ divides $n^3$, again what can be said about $n$?

Consider prime factorizations.

I hope this helps!