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Direct proof by using if then technique

  1. May 23, 2013 #1
    Hi all,

    I am trying to proof the following question.

    If a is an integer, divisible by 4, then a is the difference of two perfect squares


    now by the definition of divisibility if 4 divides a then there is a natural number k such that

    a = 4k

    Can someone how should I do it with direct proof by using if then technique?
     
  2. jcsd
  3. May 23, 2013 #2

    micromass

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    You want to find ##x## and ##y## such that

    [tex]a = x^2 - y^2[/tex]

    Now, what if you write

    [tex] a = (x-y)(x+y)[/tex]

    Does that give you any ideas?
     
  4. May 23, 2013 #3
    ahhh got it

    proof: if a is divisible by 4 then

    4a = (a+1)^2 - (a-1)^2

    QED

    PS: I have followed the proof method given at http://zimmer.csufresno.edu/~larryc/proofs/proofs.direct.html
    Theorem. Every odd integer is the difference of two perfect squares.
     
    Last edited: May 23, 2013
  5. May 23, 2013 #4

    micromass

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    Why would a = a/4 ??

    Anyway, you can always write ##a = 4k##. So we have

    [tex]4k = (x-y) (x+y)[/tex]

    Does that help? You need to identify some ##x## and ##y## that do that.
     
  6. May 23, 2013 #5

    micromass

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    Something is not right. You say that ##a## is divisible by ##4## and then you say something about ##4a##. Shouldn't you be giving a decomposition of ##a##?
     
  7. May 23, 2013 #6
    Proof:

    now by the assumption and definition of divisibility if 4 divides a then there is a natural number k such that

    a = 4k =(k+1)^2 - (k-1)^2

    QED

    PS. is it correct now?
     
  8. May 23, 2013 #7

    micromass

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    Yes, it's right now! Congratulations!
     
  9. May 23, 2013 #8
    hehehe thanks a lot
     
  10. May 23, 2013 #9

    micromass

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    Anyway, it might beneficial to show my solution.

    I want to find integers ##x## and ##y## such that

    [tex]4k = (x-y)(x+y)[/tex]

    So if I find ##x## and ##y## such that

    [tex]2 = x-y~\text{and}~2k = x+y[/tex]

    then I'm done. So I get

    [tex] x = 2 + y~\text{and}~x = 2k -y[/tex]

    and thus ##2 + y = 2k - y##, which gives us ## y = k-1 ##. Then I set ##x=(k-1) + 2 = k+1##.

    So this gets me that ##4k = (k+1)^2 - (k-1)^2##. This is the same answer you found, but it might be good to see how you can find it.
     
  11. May 23, 2013 #10
    How did you set 2=x-y and 2k=x+y ?
     
  12. May 23, 2013 #11

    micromass

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    It's not that ##4k = (x+y)(x-y)## implies ##2=x-y## and ##2k = x+y##. It might be that ##x## and ##y## are different numbers.

    But I said: if I can find ##x## and ##y## such that ##2 = x-y## and ##2k = x+y##, then ##4k = (x+y)(x-y)## will be satisfied. So I took a guess about what x and y looked like. It might have happened that the guess didn't give anything useful.
     
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