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escolas Group Title

Prove that for two distinct primes p and q, if p|n and q|n, then (pq)|n.

  • 7 months ago
  • 7 months ago

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  1. myininaya Group Title
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    p|n => pa=n for some integer a q|n => 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

    • 7 months ago
  2. escolas Group Title
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    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?

    • 7 months ago
  3. escolas Group Title
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    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?

    • 7 months ago
  4. myininaya Group Title
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    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) 2|972 => 2a=972 3|972 => 3b=972 But as we can see in the prime factorization we also have 2*3*k=n

    • 7 months ago
  5. escolas Group Title
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    Thanks, that makes a lot of sense. Much more clear now.

    • 7 months ago
  6. Zarkon Group Title
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    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 \[pq|n\]

    • 7 months ago
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