Let $n\in \mathbb{Z}^+ $ s.t $n$ has no square factors other than 1. Prove that for any $x \in \mathbb{Z}$, $n|x^2 \implies n|x$

Question

I tried to prove above theorem. I used the Theorem:
Let p be a prime number. Then for any $a,b \in \mathbb{Z}; p|ab \implies p|a$ or $p|b$.
Proof:
Since $n|x^2$, $x^2 = an$.
By prime factorization,
$n = p_1^{t_1}p_2^{t_2}...p_k^{t_k}$ ; $t_1,t_2,...,t_k$ are all odd integers.
$\therefore p_i | n$ for some i $\in \{1,2,...,i\}$
$\therefore p|x^2 \implies p|x$ (Theorem)

Then I can write $x^2=bp_i$ and $x = cp_i$.
$\therefore c^2p_i^2 = bp_i$
$c^2p_i = b$
$\therefore c|b$.

Futher by using $x^2 = an$ and $x^2=bp_i$,
$p_i=\frac{an}{b}$
By substituting this to $x=cp_i$,
$x=\frac{c}{b}.an$
Since $c|b$,
$x=kan$
$\therefore n|x$.(Proved)

Is this proof correct? Is there any better approach or other method?

Answer 1

Define $p_{i}$ as prime factor of $x$ and $x=\prod p_{i}^{a_{i}}, a_{i}\geq 1$. Therefore $x^{2}=\prod p_{i}^{2a_{i}}$.

If $n|x^{2}$ then $n=\prod p_{i}^{b_{i}}, b_{i}\in\{0,1\}$ (remember that $n$ has no square factor?). Since $b_{i}\leq a_{i}$ , then $n|x$

Answer 2

This theorem is true when n can be expressed as a product of distinct primes (every prime factor should be raised to the power of $1$) $$ n|x^2 \implies n|x.x$$ By writing x in terms of the product of its prime factors (say $p_1, p_2,...,p_k$), we get $$n|(p_1^{2a_1}p_2^{2a_2}...p_k^{2a_k})$$ $$\implies n=\frac{p_1^{2a_1}p_2^{2a_2}...p_k^{2a_k}}{a}, a \in \mathbb{Z}$$ Since n has no repeating prime factors, it should be expressible as follows. $$n=\prod p_i^{b_i},$$ where $b_i$ can be either $0$ or $1$. For every value of $n$, it can be seen that $n|x$.