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(x^2 - y^2)

  1. Mar 26, 2009 #1
    Hi,

    i have just registered to the forum, because this time i study number theory and in some problems i can't figure out how to solve them.

    This time i have to prove: If two integers x,y doesn't divided with 3 then the (x^2 - y^2) always is divided with 3.

    Does any one has a clue how to start?

    Thank you!
     
  2. jcsd
  3. Mar 26, 2009 #2

    Office_Shredder

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    Re: proof

    Break it up into what x and y modulo 3 can be (so four cases)
     
  4. Mar 26, 2009 #3
    Re: proof

    what?? Can you explain it a little bit, please?
     
  5. Mar 26, 2009 #4

    HallsofIvy

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    Re: proof

    If x is not divisible by 3 then it can be written as x= 3k+1 or 3k+2 for some integer k.
    If y is not divisible by 3 then it can be written as y= 3j+1 or 3j+2 for some integer k.

    That gives 4 cases to consider:
    1) x= 3k+1 and y= 3j+1.
    2) x= 3k+1 and y= 3j+2.
    3) x= 3k+2 and y= 3j+1.
    4) x= 3k+2 and y= 3j+2.

    Calculate [itex]x^2- y^2[/itex] for each of those cases.
     
  6. Mar 26, 2009 #5
    Re: proof

    Since +/- 0 = 0 are not cases 2 and 3 equivalent?

    Since 2 = -1 and -1*-1 = 1 , (-1)^2 = (+1)^2 are not cases 1 and 4 equivalent?

    Even better you can rewrite each term of an expression mod n before evaluating the expression mod n. Thus each of the terms can be rewriten mod 3 by substituting 0 for 3n and 1 for -1^2 (or 2^2) to get the equivalent expressions 1-1 = 0 mod 3 which is clearly correct.
     
    Last edited: Mar 26, 2009
  7. Apr 1, 2009 #6
    Re: proof

    Thank you very much for the help!! I'm grateful to you!
     
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