Show that if $p\mid m^2$ then $p \mid m$ and hence $p^2 \mid m^2$
I don't understand what is being asked of me. I thought this question was asking if $p$ divides $m^2$, then $p$ divides $m$, and hence, $p^2$ divides $m^2$.
$4$ divides $2^2$, but does not divide $2$. So Im trying to figure out what's being asked of me.
IF it's any help, this is for an abstract algebra class
If the statement were corrected to include that $p$ was a prime, then it's true. [And you might try to prove it; if you have unique factorization, then you get it very quickly. But you might also try to use the definition of a prime].