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Prove that when 8 does not divide n^2 - (n-2)^2, then n is even

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that when (n^2-(n-2)^2) is not divisible by 8, n is even

    2. Relevant equations



    3. The attempt at a solution

    Could I just plug in numbers or is there a different way of doing it?

    Let n=2 2^2-(2-2)^2/8=1/2

    When n=2, 8 does not go evenly into the function.

    Thank you very much
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2

    Vid

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    Use difference of two squares and go from there.
     
  4. Mar 2, 2008 #3

    HallsofIvy

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    How long would it take you to try every even integer?

    Even if you did that you would have proved the wrong thing. Your example is an example that "if n is even then 8 does not divide evenly into n^2- (n-2)^2". What you are asked to prove is the converse of that: "if 8 does not divide evenly into n^2- (n-2)^2, then n is even". "If A then B" is not the same as "If B then A".

    What you could do is prove the contrapositive. "If A then B" is the same as the contrapositive, "if NOT B then NOT A". Here that would be "if n is NOT even, then 8 does NOT divide n^2- (n-1)^2". If n is not even, it is odd: n= 2m+1 for some integer m. Then n^2= (2m+1)^2= 4m^2+ 4m+ 1 and n- 1= 2m+1-1= 2m so (n-1)^2= (2m)^2= 4m^3. n^2- (n-1)^2= 4m+ 1. Now there are 2 posibilities: either m itself is even, in which case 4m+ 1= 8k+ 1 for some k and 8k always has a remainder of 1 when divided by 8 or m is odd, in which case 4m+ 1= 4(2k+1)+ 1= 8k+ 5 which has a remainder of 5 when divided by 8.
     
  5. Mar 2, 2008 #4

    rock.freak667

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    You can also do proof by contradiction...not too sure if proof by contrapositive and contradiction are related though.
     
  6. Mar 2, 2008 #5
    Thank you very much

    Regards
     
  7. Mar 2, 2008 #6

    HallsofIvy

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    Proof of the contrapositive is a special case of proof by contradiction.


    In a general "proof by contradiction" or "indirect proof", you assume the conclusion is false and prove two statements that contradict one another- neither necessarily has anything to do with the hypotheses (I have seen example where neither of the two contradictory statements seemed to have anything to do with the whole theorem!). In "proving the contrapositive", you assume the conclusion is false and prove the hypothesis is false- which, of course, contradicts the hypothesis itself.
     
  8. Mar 2, 2008 #7
    Thank you very much
     
  9. Mar 3, 2008 #8

    Vid

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    n^2 - (n-2)^2 = (2n-2)(2) = 4(n-1)

    8 only divides even multiples of four thus when n is even n-1 is odd and 4(n-1) is not divisible by 8.

    The straightforward proof seems a lot easier to me.
     
  10. Mar 3, 2008 #9

    HallsofIvy

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    Yeah, but that's no fun!
     
  11. Mar 3, 2008 #10

    rock.freak667

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    Isn't that proof by contradiction? Assume that if n^2-(n-2)^2 is divisible by 8 then n is even. so that n^2-(n-2)^2 =8A for all +ve integers. and after some algebra you get n=-(2A-1) and 2A-1 is odd for all +ve integers, so that contradicts the assumption, so it is false. Hence if n^2-(n-2)^2 is not divisible by 8, then n is even (since if you are not odd, you are even)
     
  12. Mar 3, 2008 #11

    Vid

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    I don't think so since I'm not assuming the negation of the original statement.
     
  13. Mar 5, 2008 #12
    Thank you very much

    Regards
     
  14. Mar 6, 2008 #13
    Ok let me get this straight, If A then B is the same as proofing if not B then not A,
    so in this case A is 8 does not divide evenly into n^2- (n-2)^2 and B is n is even and you try to proof if not B then not A, isn't the inverse of A be 8 does divide evenly into n^2- (n-2)^2 and B being n is not even?
     
  15. Mar 7, 2008 #14

    HallsofIvy

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    Yes, I misspoke. I believe my (outline of a) proof did do it correctly however.
     
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