Given that $18|n^3$, can we conclude $18|n^2$?

Question

Given that $18|n^3$, can we conclude $18|n^2$?

I have been working on this a lot, and so far I have determined from the prime factorization of $18$, that $2|n$ and $3|n$. Therefore we also know $6|n$. I am not sure how to use this however to show that $18|n^2$.

Could we say that since $6|n$, $36|n^2$, or is this an inaccurate claim?

Answer 1

Write $n=p_1p_2\cdots p_k$, where the $p_i$ are prime (there might be some repetitions). Then $n^3=p_1^3p_2^3\cdots p_k^3$. If $18=3^2\cdot 2$ divides $n^3$, then at least one of the $p_i$ is equal to $3$ and one is equal to $2$, so reordering we may assume $p_1=2$ and $p_2=3$. But then $n^2=p_1^2p_2^2\cdots p_k^2=2\cdot 18 p_3^2\cdots p_k^2$ is divisible by $18$.

You can generalize this to prove the following fact: Suppose that $m$ and $n$ and $k$ such that:

1. the powers of the distinct primes appearing in the prime decomposition of $m$ are all at most $k$;
2. $m$ divides $n^l$ for some $l\geq k$.

Then $m$ divides $n^k$.

Answer 2

If $18 \mid n^3$, then we must have $2 \mid n$ and $3 \mid n$. S0 $6 \mid n$. That is to say, $n = 6d$ for some integer $d$. Since $18 \mid (6d)^3$, for all $d$, we can say that $18 \mid n^3$ if and only if $n = 6d$ for some integer $d$.

So, if $18 \mid n^3$, then $n=6d$ for some integer $d$. It's easy now to check that $18 \mid n^2$ too.