I am working on a question that involves showing that $x^4=6$ has no positive rational solutions. Part of my solution involves using $6\mid p^4 \implies 6 \mid p$. The answer given by the professor instead uses $2\mid p^4\implies 2\mid p$.
I am wondering if $6\mid p^4$ is also a correct solution. I think it is correct but my reasoning for why is wrong. I think that $6\mid p^4$ can imply that $2\mid p$ since $2\mid 6\implies 2\mid p^4\implies 2\mid p$. However since 6 is not a prime number, I am not sure if this is always true. Using Euclid's Lemma (link), I know that $2\mid p^2\implies 2\mid p$ but that is only because 2 is prime. I have tried replicating this using 6 but I am stuck on the following:
$6\mid p^3p\implies 6\mid\gcd(6,p^3)p$
This is saying that $\gcd(6,p^3)=1$ so that 6 doesn't divide into $p^3$ but it does divide into $p$? I feel like this doesn't make sense and I am applying Euclid's Lemma in the wrong way. I would appreciate any hints on what I am doing wrong here.
As a side note, I am also curious on how to prove that $x\mid p^n\implies x\mid p$ where $x$ is prime, because the linked post only proves it for the case $n=2$ and I would like to understand why this also applies to $n=4$.