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myininaya
 one year ago
Best ResponseYou've already chosen the best response.2pn => 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

escolas
 one year ago
Best ResponseYou've already chosen the best response.0And 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?

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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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

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

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1IF 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 \[pqn\]
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