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Divisibility and p's and q's

  1. Mar 19, 2007 #1
    If 3p^2 = q^2 and p and q are integers, how do I prove that 3 is a common divisor for p and q?
    My attempt: q^2 is divisible by 3, so q is divisible by 3. I can't prove that p is divisible by 3.
     
  2. jcsd
  3. Mar 19, 2007 #2
    Any integer that is not a perfect square has an irrational square root. What can you do with that?
     
  4. Mar 19, 2007 #3

    JasonRox

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    Personally, I have no idea what you can do with that.
     
  5. Mar 19, 2007 #4

    JasonRox

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    Now, just re-write it as p^2 = q^2/3. What does that tell you about p?

    Remember, p is an integer so p^2 is also an integer. And q is divisible by 3, but there is only one 3. :wink:

    Finish it off from there.
     
  6. Mar 19, 2007 #5
    Lol, yeah maybe it's not very relevant. Let me repent:

    A power has the same prime factors than its root.
     
  7. Mar 19, 2007 #6
    3|q, rewrite q=3q'
     
  8. Mar 19, 2007 #7
    oops...I completely disregarded the fact that p and q are integers.

    sorry
     
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