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escolas
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Prove that for two distinct primes p and q, if pn and qn, then (pq)n.
 11 months ago
 11 months ago
escolas Group Title
Prove that for two distinct primes p and q, if pn and qn, then (pq)n.
 11 months ago
 11 months ago

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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
 11 months 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?
 11 months 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?
 11 months 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
 11 months ago

escolas Group TitleBest ResponseYou've already chosen the best response.0
Thanks, that makes a lot of sense. Much more clear now.
 11 months 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 \[pqn\]
 11 months ago
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