TITLE: A short or elegant proof for if $p \mid n^2$ then $p \mid n$ when $p$ is prime?
QUESTION [5 upvotes]: Let $n, p \in \mathbb{Z}^{+}$ such that $p$ is prime. Prove $p \mid n^2 \Rightarrow p \mid n$. 
What is a short or elegant proof to this? Some ideas are given at the question
Prove that $\sqrt 5$ is irrational, but I would like to find a proof that does not need to rely upon the Unique Factorization Theorem or Euclid's Lemma, if possible. (In my particular math course,  we haven't reached these results; they are yet to be proven and are inaccessible.) 
I tried to start this proof by assuming $$\tag 1  p\mid ab\implies p\mid a \text{ or } p\mid b$$ The statement seems obviously true, however, my course requires a pedantic proof of that statement, and I am lost on finding one. 
Euclid's Lemma (see snapshot at end) is given in a chapter of the textbook my class has not yet reached, so we may not assume this without proving it first. (But we may assume everything up to and including chapter 8 in this textbook(table of contents)). So I am trying to find a short proof for (1) $$p\mid ab\implies p\mid a \text{ or } p\mid b$$ that does not require Euclid's or Bezout's lemmas, if possible. If it is possible to show (1), then it seems proving $p | n^2 \Rightarrow p | n$ should be straightforward. 
I was sketching out the following proof by $4$ cases. 
Let $a, b, p, m, r_1, r_2 \in \mathbb{Z}$ where $p$ is prime. Prove:
$$p\mid ab\implies p\mid a \text{ or } p\mid b$$
Assume $p\mid ab$. Then $ab = pm$ for some $m \in \mathbb{Z}^{+}$. 
(Note: Cases 1. and 2. are totally wrong, and so I may ask about this proof, $p\mid ab\implies p\mid a \text{ or } p\mid b$, in a new question.)

$p \not \mid a \implies \cdots \implies \cdots \implies p \mid b$
$p \not \mid b \implies \cdots \implies \cdots \implies p \mid a$
$p \mid a$ and $p \mid b$. Done.
$p \not \mid a$ and $p \not \mid b$. We must somehow show this leads to a contradiction. Not sure how to do this. 


Euclid's Lemma

REPLY [4 votes]: You can hide the usage of the linear representation lemma, if you want. However, division with remainder is a must with this definition. Otherwise, you'll prove a statement that is too general to be true.
Now, let's prove that if $(a,b)=1$ and $a\mid bc$, then $a\mid c$ by induction in $a$.
$a=1$ is straightforward ($1$ divides everything). Assume that $A\ge 2$ and we know the result for $a<A$. Let $(A,b)=1$ and $A\mid bc$. Dividing with remainder, we can assume without loss of generality that $b<A$ (this is nothing but one step in the Euclidean algorithm, so, as I said, we still have it in disguise). Also, since $(A,b)=1$ and $A\ge 2$, the case $b=0$ is impossible. Write $Ak=bc$. Since $(b,A)=1$, $b\mid Ak$, $b<A$, we have $b\mid k$ by the induction assumption. Hence $k=mb$ and $Am=c$, i.e., $A\mid c$.
This uses the absence of divisors of $0$, division with remainder, and the induction axiom (which can be relaxed to the condition that no infinite sequence with strictly decreasing norms exists). Removing any of those makes the proof impossible because you can create a ring in which the statement is false then.