Prove that if $p$ is a prime, $a$ is an integer, and $p$ divides $a^2$, then $p$ divides $a$. I think I need to use the Fundamental Theorem of Arithmetic but that only applies to $a > 1$ so I think I need to do it by cases.
Case 1) $a = 0$.
Case 2) $a = 1$.
Case 3) $a > 1$.
Case 4) $a < 0$.
Cases 1 and 2 are easy.  I think Case 4 will follow from Case 3 but I struggle with Case 3.
Case 3) $a > 1$. Then by Fundamental Theorem of Arithmetic, $a$ is equal to a unique product of primes. Then, $a^2$ is a product of those primes squared. Then, because $p$ divides $a^2$, $a^2 = pk$ for some $k$ (an integer). So that product of primes squared equals $pk$. Now how do I argue that one of those primes must be $p$?  Once we see that one of those primes is $p$ (i.e. $p$ is a prime factor of $a$), we can say that $p$ divides $a$.
So my questions are:
How do I argue that one of those primes must be $p$?
Can the proof be done without resorting to four cases?
 A: If $p$ does not divide $a$ then $\gcd(p,a)=1.$ By Euclidean Algorithm (or Bezout) there are integers $x$ and $y$ so that $px+ay = 1$.  Multiply by $a$
to get $apx+a^2y = a.$ Since $p$ divides the left side, it must divide the right, contradiction.
A: Assume that $p \nmid a$. Then, $a=p \cdot q +r$ for some $r >0$,  s.t. $r \neq 0 (modp) $.
Since,  ${p \mid a^2 } \Rightarrow {a^2 \equiv 0 (mod p)}$
So, $a^2=(p \cdot q +r)^2=p^2 q^2+2pqr+r^2 \equiv r^2 (modp)\neq 0 (modp)  \Rightarrow p \nmid a^2$ Contradiction.
A: Using prime factorization of $a = p_1^{n_1}p_2^{n_2}\cdots p_m^{n_m}$ where $p_j$'s are distinct primes, and $n_j$'s are natural numbers. Thus $p\mid a^2 = p_1^{2n_1}p_2^{2n_2}\cdots p_m^{2n_m}$. Since $p$ is a prime $p = p_k$, for some $1 \le k \le m$, and this implies $p \mid p_1^{n_1}p_2^{n_2}\cdots p_m^{n_m} = a$
A: Here is my understanding for what it's worth.
If $a$ is some integer then using the Fundamental Theorem of Arithmetic like you mentioned $a$ can be written as a product of prime numbers. Let's say $a=bcd$ where $b$ , $c$ and $d$ are prime numbers. From this we can see that $a^2=bbccdd$. If $p$ divides evenly into $a^2$ then $p$ must be equal to either one of $b$ , $c$ or $d$. It can't be a product of any of them as it is prime itself. Let's say $p=c$. We can see that $c$ is also a factor of $a$. Whichever factor of $a^2$ that equals $p$ also exists in $a$.    
A: This is a special case
of my question
(which I answered)
here:
If $n \mid a^2 $, what is the largest $m$ for which $m \mid a$?
In particular,
I show that if
$p^m | a^2$,
then
$p^{\lceil m/2 \rceil} | a
$.
This question is
the case $m=1$.
A: Let the prime factorisation of a be as follows :
 $a = p_1p_2 \dots p_n$, where $p_1,p_2, \dots p_n$ are primes, not necessarily distinct. 
Therefore, $a_2 = (p_1p_2 \dots p_n)(p_1p_2 . . . p_n) = p_{1^2}p_{2^2} \dots p_{n^2}$.
Now, we are given that $p$ divides $a{^2}$. Therefore, from the Fundamental Theorem of Arithmetic, it follows that $p$ is one of the prime factors of $a{^2}$. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of $a{^2}$ are $p_1p_2 \dots p_n$. So $p$ is one of $p_1, p_2,\dots, p_n$.
Now, since $a = p_1p_2 \dots p_n$, $p$ divides $a$.
