Generalization of "If $p$ is an integer and $p^2$ is even then $p$ is even" The statement in the title is true.
I am just wondering if a general statement of the the above statement which I state below is true.

Proposition: If $p$ is an integer and $n$ divides $p^2$ then $n$ divides $p$. 

 A: Counterexample:
$9$ divides $36$, but $9$ does not divide $6$.
A: There is a generalization: if $p$ is an integer and $n$ is prime, then if $n$ divides $p^2$, we have that $n$ divides $p$. This is obvious when you examine the prime factorization of $p$. 
A: Your generalisation is true iff $n$ is a so-called square-free integer. Square-free means that there is no square number (other than $1$) which divides $n$. Alternatively, any prime in the prime factorisation of $n$ appears only once.
Your original theorem uses $n=2$, which is a prime and therefore square-free.
A: Suppose $p$ factors into distinct primes $q_i$ as
$p = \displaystyle \prod_1^k q_i^{\alpha_i}, \; \alpha_i \ge 1; \tag 1$
then
$p^2 = \displaystyle \prod_1^k q_i^{2\alpha_i}; \tag 2$
now choose $\beta_i$, $1 \le i \le k$, such that
$\alpha_i < \beta_i \le 2\alpha_i; \tag 3$
then if
$n = \displaystyle \prod_1^k q_i^{\beta_i}, \tag 4$
we clearly have $n \mid p^2$; but $n \not \mid p$ since the exponents of its prime factors $q_i$ are all greater than those occurring in (1).
A: Well, the statement isn't true for all n that divide $p^2$ for example, $4|36$ which is $6^2$ but 4 doesn't divide 6.
In general if we write our integer $$m = p_1^{a_ 1}p_2^{a_2}...p_k^{a_k}$$ as the product of primes, then we know that $$m^2= p_1^{2a_ 1}p_2^{2a_2}...p_k^{2a_k}$$
Thus if an integer $n$ divides $m^2$ it will only imply that $n$ divides $m$ if each prime dividing $n$ is also a prime that divides $m^2$ and the powers of any of those primes cannot be larger than $a_i$ for $1\leq i \leq k$.
I hope that helps!
