Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
escolas
Group Title
Prove that for two distinct primes p and q, if pn and qn, then (pq)n.
 one year ago
 one year ago
escolas Group Title
Prove that for two distinct primes p and q, if pn and qn, then (pq)n.
 one year ago
 one year ago

This Question is Closed

myininaya Group TitleBest ResponseYou've already chosen the best response.2
pn => pa=n for some integer a qn => qb=n for some integer b we want to show pqk=n for some integer k we also have that p and q are two distinct primes if we were to give the prime factorization for n it would look something like p*q*some other prime integers possibly
 one year ago

escolas Group TitleBest ResponseYou've already chosen the best response.0
And we know that a and b have prime factorizations as well, so we could say that a and b are some product of primes and then that n^2 is pq times some other primes, but how do we know that pq isn't larger than n?
 one year ago

escolas Group TitleBest ResponseYou've already chosen the best response.0
Oh, is it because if p and q both divide n then they are both part of n? Kinda an Euler's Totient type thing?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Yeah since both p and q divide n then we know the prime factorization for n=p*q(some other primes possibly) For example: Let n=972 n=3(324)=2(3)(162)=2^2(3)(81)=2^2*3^5=2*3*(2*3^4) 2972 => 2a=972 3972 => 3b=972 But as we can see in the prime factorization we also have 2*3*k=n
 one year ago

escolas Group TitleBest ResponseYou've already chosen the best response.0
Thanks, that makes a lot of sense. Much more clear now.
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
IF you want a more rigorous way you can do the following: since p and q are distinct primes then theie gcd is 1 and so they can be written as a linear combination \[\alpha p+\beta q=1\] and using the notation from above \(n=ap\) and \(n=bq\) we have \[n=n\cdot 1=n(\alpha p+\beta q)\] \[=n\alpha p+n\beta q=bq\alpha p+ap\beta q\] \[=(b\alpha+a\beta)pq\] so \(n\) is an integer multiple of \(pq\). Thus \[pqn\]
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.