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Divisibility of n by 6

  1. Dec 12, 2006 #1
    I need to prove that if n is a natural number and n^2 is divisible by 6, then n is divisible by 6.

    I know that I knew how to do this at one point fairly recently, if you could refresh my memory I would greatly appreciate it.
     
  2. jcsd
  3. Dec 12, 2006 #2
    Write n out in terms of it's prime factors, and it should be obvious.
     
  4. Dec 13, 2006 #3

    CRGreathouse

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    Suppose n was not divisible by 2. Then n^2 could not be divisible by 6.
    Suppose n was not divisible by 3. Then n^2 could not be divisible by 6.
    So n is divisible by 2 and 3, hence by 6.
     
  5. Dec 13, 2006 #4
    we know that if p divides ab, then p divides a or p divides b.
    now if 6 divides n.n, then 6 divides n.
     
  6. Dec 13, 2006 #5

    morphism

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    Here's another way to look at it.

    n(n+1)(2n+1)/6 is the sum of the first n squares, and thus must be an integer. Its numerator is 2n^3 + 3n^2 + n. So if n^2 is divisible by 6, n must be as well.
     
  7. Dec 13, 2006 #6

    Hurkyl

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    6 isn't a prime number, so we cannot apply that theorem directly...
     
  8. Dec 13, 2006 #7

    Gokul43201

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    Why are folks writing up complete solutions to what appears to be a textbook problem?
     
  9. Dec 13, 2006 #8
    now if 6 divides n.n, then 3 divides n.n and 3 divides n.
    again, if 6 divides n.n, then 2 divides n.n and 2 divides n.
    so then 3 divides n and 2 divides n. therefore, 6 divides n.
     
    Last edited: Dec 13, 2006
  10. Dec 13, 2006 #9

    Gokul43201

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    Great! That makes 3 complete solutions for the OP to copy down. Anyone else?

    People, please read and follow the Posting Guidelines.

    https://www.physicsforums.com/showthread.php?t=5374

     
  11. Dec 30, 2007 #10
    Hello, my math skills are crap, but can't we simply do that?
    If n is divisible by 6, then it can be writed as 6x where x is an integer.

    [tex]\left(6x\right)^{2}=6^{2}x^{2}=36x^{2}=6\left(6x^{2}\right)[/tex]
     
  12. Dec 30, 2007 #11
    You could do that... if the intention was to show "if n is divisible by 6, then n^2 is too". But the original question was the other way around: "if true for n^2, then it's true for n".
     
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