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I'm learning Real Analysis by myself and I wanted to prove that if a prime $p$ divides $n^2$ where $n$ is an integer, then $p$ divides $n$ itself. I saw that proving this is the same as saying that the prime factors of $n^2$ are "the same" than those of $n$.

In this case, assume there exists a prime factor $a_q$ in $n^2$ which is not in $n$ (proof by contradiction). By Fundamental Theorem of Arithmetic, we can represent $n^2$ as $n^2= a_1 * a_2 * ... * a_x * a_q$, where every $a$ is prime and $n=k_1 * k_2 * ... * k_y$ (every $k$ is also prime). We know that $frac{n^2}{n}=n$. So (by susbstituting):

$frac{n^2}{n}= frac{a_1*...*a_x*a_q}{k_1*...*k_y}=n rightarrow frac{a_1*...*a_x}{k_1*...*k_y}=frac{n}{a_q}$

Because $frac{n}{a_q}$ it's not an integer (initial statement), there's at least one $a_s$ in the left side of the equation that is not divisible by any $k$. Now we have two factors of $n^2$ that don't divide $n$. We can repeat this process,

$frac{a_1*...*a_x*a_s}{k_1*...*k_y}=frac{n}{a_q} rightarrow frac{a_1*...*a_x}{k_1*...*k_y}=frac{n}{a_q*a_s}$,

for each prime factor of $n^2$ which means that there's no factor $a_i$ in $n^2$ divisible by any factor $k_j$ of $n$. But we knew that $frac{n^2}{n}=n$, hence we have a contradiction ($n^2$ is not divisible by $n$). So there's no prime factor $p$ in $n^2$ that does not divide $n$ itself $rightarrow$ every prime factor $p$ in $n^2$ also divides $n$.

I'm obviously not an expert in demonstrations so I want to know if this is a valid argument. I'm also aware of Euclid's Lemma, but for this case, ignore it.

A shorter way would be to write $n=p_1^{alpha_1}cdots p_k^{alpha_k}$ (where each $alpha_i>0$) by the Fundamental Theorem of Arithmetic. Then $n^2= p_1^{2alpha_1}cdots p_k^{2alpha_k}$. The only primes that divide $n^2$ are $p_1,cdots,p_k$ and these clearly also divide $n$.

Here is a different approach. Unique prime factorization of $n,$ for $1<nin Bbb N,$ follows from the Lemma that if $a,b,cin Bbb Z$ and $frac {ab}{c}in Bbb Z$ while $gcd (a,c)=1$ then $frac {b}{c}in Bbb Z.$ The Lemma follows from something called Bezout's Identity, although it is implicit in Euclid's algorithm for computing a $gcd$ : For any $a,bin Bbb Z$ (not both $0$) there exist $x,y in Bbb Z$ such that $ax+by=gcd (a,b).$

Let $p$ be prime and let $nin Bbb Z$ such that $p|n^2.$ Let $m=gcd (p,n).$ Since $m$ divides the prime $p$ and $mge 1,$ we have $m=1$ or $m=p.$

Now there exist $x,yin Bbb Z$ such that $px+ny=m,$ so $frac {(m-px)^2}{p}=frac {n^2y^2}{p}=frac {n^2}{p}cdot y^2in Bbb Z.$ But if $m=1$ then $frac {(m-px)^2}{p}=frac {1-2px+p^2x^2}{p}=frac {1}{p}-2x+px^2in Bbb Z,$ implying $frac {1}{p}in Bbb Z,$ which is absurd.

So, since we can't have $m=1,$ we have $m=p$. And since $m=gcd (p,n)|n,$ we have therefore $p|n.$