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Proving p is multiple of 3 iff p^2 is

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that p^2 is a mutiple of 3 ⇔ p is a multiple of 3. We are told p is an integer.

    3. The attempt at a solution

    If p is a multiple of 3 then p=3.a (aEZ) and p^2 = 3.(3.a^2) which is clearly a multiple of 3

    Suppose p^2 is a multiple of 3.
    Then P^2 = 3.a and p=3.(√a/√3) since we know p is an interger (√a/√3) must be an integer and therefore p is a multiple of 3.

    2. Relevant equations

    This isn't the solution that was given to me and I'm wondering whether it is correct or if I've missed something obvious. Or if there is an even easier way to do it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 19, 2012 #2
    If [itex]p^2[/itex] is a multiple of 3, we can write [itex]p^2=3x[/itex]. Taking the square root of both sides yields [itex]p=\sqrt{x}\sqrt{3}[/itex]. We know that [itex]p^2[/itex] is a perfect square and hence p is an integer. However, if p is an integer, the integer x must be of the form 3a for some perfect square a. Then, we obtain [itex]p=\sqrt{3a}\sqrt{3}=3\sqrt{a}[/itex], which is an integer. If x is not of this form, we obtain a contradiction because then p can't be an integer.

    You might be requested to explain why exactly it is that x must be of that form. How can you explain this?
     
  4. Aug 19, 2012 #3
    If I understand you correctly you are claiming that x must be in that form because otherwise we would have [itex]\sqrt{3}[/itex] multiplied by [itex]\sqrt{x}[/itex] which couldn't be a integer (since an irrational multiplied by anything is irrational)but if one claims that aren't we assuming that [itex]\sqrt{3}[/itex] is irrational which we can't do. ie. since we don't yet know that [itex]\sqrt{3}[/itex] is irrational and therefore could be rational and therefore could be an integer when multiplied by [itex]\sqrt{x}[/itex].

    I'm relatively new to proving things and often get my logic in a knot so I do apologise if I've miss understood.
     
  5. Aug 19, 2012 #4
    [tex] p \cdot p = 3 \cdot q [/tex][tex] p = \frac {3 \cdot q} {p} [/tex]Since the left-hand side is an integer, then the fraction must be reducible. Assume p is not a multiple of 3, then q must be a multiple p, so we get [tex] p = 3 \cdot s [/tex]which means p is a multiple of 3.
     
  6. Aug 19, 2012 #5

    HallsofIvy

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    I would recommend an indirect proof. If [itex]p^2[/itex] is NOT a multiple of 3 then it is of the form 3n+1 or 3n+ 2.

    Now use the fact that [itex](3k+1)^2= 9k^2+ 6k+ 1= 3(3k^2+ 2k)+ 1[/itex] and that [itex](3k+ 2)^2= 9k^2+ 12k+ 4= 3(3k^2+ 4k+ 1)+ 1[/itex].
     
  7. Aug 19, 2012 #6
    Thanks
     
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