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Wierd Mathematical Proof

  1. Feb 8, 2005 #1
    I'm given the statement: if m^2 is of the form 4k+3, then m is of the form 4k+3. I don't even know how to begin proving this. I'm guessing by contraposition.
     
  2. jcsd
  3. Feb 8, 2005 #2
    Just look at the four cases of integers: 4k, 4k+1, 4k+2, 4k+3.
     
  4. Feb 8, 2005 #3
    Ok, so it is even, odd, even, odd. What are you getting at?
     
  5. Feb 8, 2005 #4

    dextercioby

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    Yes,Robert presented you with 2 many numbers:4 instead of 2.So you're left only with the odd numbers.
    Can u do the analysis in this case...?

    Daniel.

    P.S.Relabel k and k' not to create confusion when putting them in the same equation.
     
  6. Feb 9, 2005 #5
    In response to the above:

    m^2 = 4k + 3 = 4(4k'^2 + 6k' + 3/2) + 3 = [16k'^2 + 24k' + 6] + 3 = (4k'+3)^2

    but is k an integer is k' is? I'm not really trying to answer the OP's post just sort of curious, so pardon my hijacking!
     
    Last edited by a moderator: Feb 9, 2005
  7. Feb 9, 2005 #6

    dextercioby

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    It's definitely in the integers.Positive integers,even.

    Daniel.
     
  8. Feb 10, 2005 #7
    Well, O.K., we need only consider the odd numbers: 4k+1 and 4k+3. Suppose we multiply them together: (4k+1)(4k+3) = 16k^2+16k+3. What form is that?
     
  9. Feb 10, 2005 #8

    Zurtex

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    3*3 = 1 mod 4

    Woo! I love modulo arithmetic.
     
  10. Feb 10, 2005 #9

    mathwonk

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    i wish to point out that "weird" is spelled "weird".
     
  11. Feb 10, 2005 #10
    if m and k are positive integer.... then your "THEN" statement is always true, no matter what is following...., here are some examples
    if m^2 is of the form 4k+3, then the world is peace
    if m^2 is of the form 4k+3, then I have 3 hands and 4 legs
    if m^2 is of the form 4k+3, then US government will give us $10billions dollars to build a particle accelatator

    can you see the reason?
    the answer is in white:

    the "if" statement can never be true
     
  12. Feb 10, 2005 #11

    mathwonk

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    in that case i guess if m^2 = 4k+3, then weird is spelled wierd, and m = 4n.
     
  13. Feb 10, 2005 #12
    I don't think too mathematically, but is this entire proof moot since the condition is never true? (alluding to my previous post where k and k' can't both be integers)
     
  14. Feb 10, 2005 #13

    Zurtex

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    *sigh*

    This thread is not coherent at all and hardly a help to anyone.

    This is quite simple, take any 2 numbers:

    a=4p + 3
    b=4n + 3

    [tex]a,b,p,n \in \mathbb{N}[/tex]

    Define c such that:

    c = ab

    We know that:

    a ≡ b ≡ 3 (mod 4)

    Therefore:

    c ≡3*3 ≡ 9 ≡ 1 (mod 4)

    Therefore there exists some t in N such that:

    c = 4t + 1

    And looking at the original post we can safely say that m2 is not of the form 4k + 3.

    Is this not valid in some way?
     
    Last edited: Feb 10, 2005
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