Prove or disprove : if $x|y^2 $then $x|y$

Question

How can I prove such statement?

I think that if $x|y^2$ then $x|(y*y)$ so $x|y$ or $x|y$ which means that in any case $x|y$. Am I correct?

I ask this question as such template because I think that it doesn't matter which $x$ or $y$ represents.

However, the original question is: if $6|n^2$ then $6|n$ and if $12|n^2$ then $12|n$.

Answer 1

False. $4|36$ but $4\!\! \not|\, 6$. You should require $x$ to be a prime number (or a product of distinct primes). $x=6$ will work, $x = 12$ won't.

Answer 2

The claim $x \mid y^2 \implies x \mid y$ is clearly false. Take $x = y^2 > 1$ a perfect square and see for yourself. What is true is: $$\begin{cases} x \mid yz \\ \gcd(x,y)=1\end{cases} \implies x \mid z.$$ Having this in mind, if $\gcd(x,y)=1$, and taking $y = z$...

Answer 3

Hint $\ $ It fails for $\,x = y^2 > 1.\ $ It is true iff $x$ is squarefree.