Q: $p$ is a prime, $n$ is an integer. $p^5$ divides $n^2$ then which of the following is correct I have a question from an old math exam and I am unable to figure out the answer. 
The question reads:
Suppose that $p$ is a prime, and $n$ is an integer. Say that $p^5$ divides $n^2$, then which of the following is true?


*

*$p^{11}$ divides $n^3$

*$p^{10}$ divides $n^3$

*$p^6$ divides $n^2$

*$p^5$ divides $n$

*$p^4$ divides $n$
I know that n can be listed as a product of primes 
$$ n = p_1^{\alpha_1}, \ldots , p_i^{\alpha_i}$$
which means that
$$n^2 = p_1^{2\alpha_1}, \ldots , p_i^{2\alpha_i}$$
Furthermore, $$p^5 \in \Bigl\{p_1^{2\alpha_1}, \ldots , p_i^{2\alpha_i} \Bigr\}$$
My instinct tells me that the answer is number $4$, but when I list it out as above I get more confused. 

A: IF $p^5$ divides $n^2$ then $p^3$ must divide $n$. [Can you see why].
1.and 2. and 4. and 5. are not true take $n=p^3$.


*is true see above


A: If $p^5\mid n^2$, then $n=pm$, for some $m\in\mathbb N$. Therefore, $n^2=p^2m^2$. Since $p^5\mid n^2$, this implies that $p^3\mid m^2$. So, $m=pq$, for some $q\in\mathbb N$, and therefore $m^2=p^2q^2$. But then, since $p^3\mid m^2$, $p\mid q^2$. So, $q=pr$, for some $r\in\mathbb N$. But$$q=pr\implies q^2=p^2r^2\implies m^2=p^4r^2\implies n^2=p^6r^2.$$So, the third option holds.

A: Take $n=8$ then, with $p=2$, we have that $p^3\mid n$, $p^6\mid n^2$ and $p^9\mid n^3$. So we have that $3$ is true but $1,2,4$ and $5$ are necessarily false.
