Prove that if $p, q$ are relatively prime, then so are $p^n, q^n$.

Question

Trying to prove that if $p$ e $q$ are relatively prime and integers then $p^n$ and $q^n$ are prime numbers among themselves too, I wrote:

If $p$ e $q$ are prime numbers among themselves $p\neq m.q$ for any $m\in \Bbb{N}$, then $p^n \neq (m.q)^n = m^n . q^n$. As $m\in\Bbb{N}$ is arbitrary and $m^n\in \Bbb{N}$, so $p^n$ and $q^n$ are prime numbers among themselves too.

But then I remembered that a natural number can be power of an irrational number, for example $(\sqrt{2})^2 = 2$. Therefore, how can I prove this statement?

As you seem to intuit in your last paragraph, it is at issue what kind of numbers $p,q$ are assumed to be. If we assume $p,q$ are integers (or more narrowly, are natural numbers), then you can apply the Fundamental Thm. of Arithmetic. Are you asking about ordinary integers or (perhaps) about extending the claim to more general "numbers"? — hardmath, Jan 13 '20 at 15:15
I forgot to say that $p$ and $q$ are integer numbers. @hardmath — Rebeca Lie Yatsuzuka Silva, Jan 13 '20 at 15:20
I saw this question now but somehow I didn't really get the explanation.@JoséCarlosSantos — Rebeca Lie Yatsuzuka Silva, Jan 13 '20 at 15:23

score 2 · Accepted Answer · answered Jan 13 '20 at 15:11

2

Euclid's Lemma states that if a prime number divides a product of numbers then it must divide one of the factors of that product. Let's do this exercise by reductio ad absurdum. Suppose there was a prime number $$x\not= 1: x|p^n , x|q^n$$ By Euclid's Lemma: $$x|p,x|q$$ But this is absurd because $p,q$ are coprimes.

answered Jan 13 '20 at 15:11

Kandinskij

3,407

Nice proof. Maybe we might also be able to prove this, not as easily, with Bezout's Lemma. Since $\gcd (q,p) = 1$ , then there exist $qs + pt = 1$, then $(qs + pt)^n = 1^n $, and by the binomial theorem – john Feb 01 '21 at 09:48

Prove that if $p, q$ are relatively prime, then so are $p^n, q^n$.

1 Answers1