A short or elegant proof for if $p \mid n^2$ then $p \mid n$ when $p$ is prime?

The defining property of a prime number is that $$\tag 1 p\mid ab\implies p\mid a \text{ or } p\mid b$$

Thus, if $p\mid a^2$ then $p\mid a$.

ADD I am guessing you define a number $p$ to be prime if the only divisors of $p$ are $1$ and $p$ itself. Euclid's lemma then gives the "true" definition $(1)$ of a prime number. To get this result, you can use the well ordering principle, that any nonempty subset of the positive integers has a least element. Define first

DEF Let $a,b$ be integers. Then the greatest common divisor of $a,b$ is the unique positive integer $d=(a,b)$ such that $d\mid a,b$ and whenever $f\mid a,b$ then $f\mid d$.

Bezout's Lemma then proves both the existence and the uniqueness of $d=(a,b)$, as follows:

PROP Let $a,b$ be integers. Then $(a,b)$ exists, and is unique.

P Consider the set $\Bbb Za+\Bbb Zb=\{xa+yb:x,y\in\Bbb Z\}$. By considering possible cases of the sign of $a,b$, it is seen the set of positive elements of $\Bbb Za+\Bbb Zb$ is nonempty (for example, if $a>0,b<0$ then $a-b>0$ is in the set). Let $d$ be the least positive element of $\Bbb Za+\Bbb Zb$.

First, we show $d$ is a common divisor. Indeed, write $a=qd+r$ and $b=q'd+r'$ with either $r=0,r<d$ and $r'=0,r'<d$. Then $a-dq,b-q'd$ are elements of $ \Bbb Za+\Bbb Zb$ (check they have the form $xa+yb$). Since $r<d$ and $r'<d$ is impossible by the definition of $d$, $r=r'=0$ so that $d$ divides both $a,b$.

Now, we show $d$ divides every element of $\Bbb Za+\Bbb Zb$. Indeed, pick an element $m$ in the set. Then $m=qd+r$ with $r<d$ or $r=0$, by the division algorithm, and again $m-qd=r$ is in $\Bbb Za+\Bbb Zb$ so we must have $r=0$. We have shown thus that $\Bbb Za+\Bbb Zb=\{dz:z\in\Bbb Z\}=\Bbb Z d$. If $f\mid a,b$ then $f\mid ax+by$ for any $x,y$. Since $d$ is of this form by construction, $f\mid d$. Thus $d$ is a greatest common divisor. But if $d'$ is another one, by definition $d\mid d'$ and $d'\mid d$ for they both greatest common divisors. Thus $d=\pm d'$; and since they are both positive, by definition, $d=d'$. $\blacktriangle$.

OBS The above (very, very important) result can be stated as

$$ \Bbb Za+\Bbb Zb=\Bbb Z(a,b)$$

Now Euclid's lemma comes in easily

LEMMA Suppose that $(a,b)=1$ and $a\mid bc$. Then $a\mid c$.

P By Bezout's lemma, we can write $ax+by=1$ for some integers $x,y$. Then $cax+bcy=c$. Since $a\mid ac$ and $a\mid bc$, $a\mid cax+bcy=c$.

Then, you move onto

LEMMA Let $p$ be a prime. Then $(p,a)=1\iff p\not\mid a$.

P Suppose $p\not\mid a$. Since the only divisors of $p$ are $1,p$, then $(p,a)$ can be either $1$ or $p$. Since $p\not\mid a$, $p$ is not a common divisor of $a$ and $p$; thus $(a,p)=1$. If $p\mid a$ then $(p,a)=p\neq 1$.

THM (Defining property of prime numbers) Let $p>1$ be a positive integer. Then $p$ is prime if and only if for any $a,b$ integers, $p\mid ab\implies p\mid a $ or $p\mid b$.

P Suppose that $p$ is a prime, and assume $p\mid ab$. If $p\mid a$, there is nothing to prove. Assume thus that $p\not\mid a$. The $(p,a)=1$, so Euclid's lemma says $p\mid b$. By symmetry the claim follows by replacing $b$ with $a$. Now suppose $p$ is not prime. Then $p$ has a divisor $1<a<p$, and thus $p=ab$ with $b=p/a$. Then $p\mid ab$, but $p\not\mid a$ and $p\not\mid b$.

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.

What you want to prove is more generally Euclid's lemma, namely if a prime $p$ divides a product $ab$ then $p$ divides at least one of $a$ and $b$. In fact what one really wants is its generalisation to any finite number of factors (which I don't think carries a name or is even explicitly formulated very often), namely that if a prime $p$ divides a product $a_1a_2\ldots a_k$ then there is some $i$ such that $p\mid a_i$; this follows from the $ab$ case by an immediate induction.

The traditional proof of Euclid's lemma uses the fact that for any (positive) integers $m,n$ one can write $\gcd(m,n)=sm+tn$, for some $s,t\in\Bbb Z$ (called Bezout coefficients for $m,n$). Actually only the existence of $\gcd(m,n)$ for all $m,n$ is necessary, but in the strong sense of being a common divisor that is divisible by any (other) common divisor; this is indicated in the first proof given in the WP article. While such a proof is useful to explain that a counterpart of Euclid's lemma continues to be true in more general algebraic settings where Bezout coefficients no longer need to exist (but $\gcd$s do), I don't think it really provides a simpler proof for Euclid's lemma (in$\def\Z{\mathbf Z}~\Z$), because the fact that in$~\Z$ greatest common divisors in the strong sense exist ultimately depends on the existence of Bezout coefficients anyway.

For the most direct proof from scratch*, I would proceed as follows. Assume $p$ is prime and $p\mid ab$. Let $M=\{\, sp+ta\in \Z_{>0}\mid s,t\in\Z\,\}$, which is a non-empty (since $p\in M$) set of positive integers, and let $d$ be the minimal element of $M$. Any $m\in M$ is divisible by$~d$, because if it were not, then the remainder$~r$ of the division of $m$ by$~d$ would satisfy $r>0$, therefore $r\in M$ (any positive $m_1-qm_2$ is in$~M$, for $m_1,m_2\in M$ and $q\in\Z$), and so the condition $r<d$ (which holds by definition for a remainder after division by$~d$) would contradict the choice of$~d$.

In particular $d$ divides both $p$ and $a$, since $p$ and $|a|$ are elements of$~M$. The fact that $d$ is a positive divisor of$~p$ implies by the definition of prime number that either $d=1$ or $d=p$. In the latter case $p=d\mid a$ and we are done. In the former case let $s,t\in\Z$ be such that $1=d=sp+ta$, which means in particular that $ta\equiv1\pmod p$. Then $p\mid ab$ implies $p\mid tab$, and one has $b=1b\equiv tab\equiv0\pmod p$, so $p\mid b$; we are done for this case too.

Of course this proof is extracted from things that are usually presented in the form of more general statements about Euclidean division, greatest common divisors, and the Euclidean algorithm.

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*I do assume basic facts about division with remainder, and at the end about modular arithmetic (but the latter could be avoided by using explicit witnesses of the modular equivalences used).