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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 $n=4$.

parasitical
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  • What's $p$? Note that $\sqrt 6$ is already irrational, which is stronger than the claim that $\sqrt[4]6 $ is irrational. – lulu Feb 06 '23 at 01:23
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    If $6\mid p^4$, then $2, 3\mid p^4$, and by Euclid's lemma, $2, 3\mid p$, and then $6 = 2\cdot 3\mid p$ since $(2, 3) = 1$. – Riemann Feb 06 '23 at 01:27
  • For your side note: if a prime divides a product then it divides one of the factors. That follows easily if you know that integers factor uniquely to primes. It's often an ingredient to proving that, using the extended Euclidean algorithm. – Ethan Bolker Feb 06 '23 at 01:34
  • @lulu

    I'm using $p$ like this: Assume that there is a rational number $(\frac{p}{q})=6$, then $p^4=6q^4$ so $p^4\mid 6$.

    If I claimed that $x^2=6$ has no rational solutions, I would first have to prove that, and then claim that squaring an irrational number always leads to another irrational number, right?

     – parasitical Feb 06 '23 at 02:01
  • @Riemann

    Thank you, that makes sense, it seems kind of obvious now that you point it out.

     – parasitical Feb 06 '23 at 02:07
  • @EthanBolker

    Thank you, I understand!

     – parasitical Feb 06 '23 at 02:11
  • $q\mid n^k\Rightarrow q\mid n,$ is true iff $q$ is squarefree (i.e. a product of distinct primes) - see the linked dupes for this and many other chracterizations of squarefree integers. – Bill Dubuque Feb 06 '23 at 09:03

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