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If n = 4k+3, does 8 divide n^2-1? Can u check if i did this right?

  1. Sep 17, 2006 #1
    ello ello!

    I'm pretty sure this is right but not 100%. Directions are as follows:
    GIve a reason for your answer in each of 1-13. Assume that all variables represent integers.

    If n = 4k + 3, does 8 divide n^2-1? Here is my answer:
    No. n^2-1 = (4k+3)^2-1 = (4k+3)(4k+3)-1 = (16k^2+24k+9)-1 = 16k^2+24k+8 = 16(k+1/2)(k+1). K+1 is an integer but k+1/2 is not necessarily an integer, because 1/2 is not an integer.

    Is my reasoning correct? thanks!
    :biggrin:
     
  2. jcsd
  3. Sep 17, 2006 #2

    shmoe

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    That doesn't work. Failing to write n^2-1 as 8*(some integer) doesn't mean it can't be done.

    You got down to:

    n^2-1=16k^2+24k+8

    just fine. For what values of k is the thing on the right divisible by 8?
     
  4. Sep 17, 2006 #3
    well if i divide (16k^2+24k+8)/8 = (3k^2+12k+4)/4.
    Is what what you ment? It does divide evenly, and those are all integers still. So can i say, 3k^2+12k+4 is an integer and 4 is an integer because the sums and products of integers are integers?
     
  5. Sep 17, 2006 #4

    shmoe

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    Something funny happened here, check this carefully.
     
  6. Sep 18, 2006 #5

    HallsofIvy

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    Why did you factor out 16? You only need to show that it is 8(an integer).
     
  7. Sep 18, 2006 #6
    Sorry i had a typo it should have been, (3k^2+12k+4)/2. But i'm lost on what they want.

    If n = 4k+3, does 8 divide n^2-1, well yes i just did it above. Nut i'm not sure how that showed 8 is an integer.

    THe book did an example like this, but instead of n = 4k+3 its n = 4k+1. They did the following:

    n^2-1 = (4k+1)^2 - 1 = (16k^2+8k+1)-1 = 16k^2+8k = 8(2k^2+k), and 2k^2+k is an integer becuase k is an integer an sums and products of integers are integers.
     
  8. Sep 18, 2006 #7

    shmoe

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    [tex]\frac{16k^2+24k+8}{8}=\frac{16k^2}{8}+\frac{24k}{8}+\frac{8}{8}\neq\frac{3k^2+12k+4}{2}[/tex]

    To show n^2-1 is divisible by 8, you can show (n^2-1)/8 is an integer. Equivalently, you can show there exists an integer x where n^2-1=8*x, like they did in your example (no real difference).
     
  9. Sep 18, 2006 #8
    My bad! thanks shmoe that makes sense, now I can just show:
    [tex]\frac{16k^2+24k+8}{8}=\frac{16k^2}{8}+\frac{24k}{8 }+\frac{8}{8}[/tex] = 2k^2+3k+1 = (k+1)(2k+1). (k+1) and (2k+1) are both integers becuase k is an integer and the sums and products of integers are integers.

    Thanks alot! :D
     
  10. Sep 19, 2006 #9

    HallsofIvy

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    I was tempted to post that not being able to divide by 8 is a handicap in number theory! (but I won't!)

    You don't need to factor that: Obviously, 2k2+3k+1 is an integer whether it can be factored or not.

    I think it is somewhat simpler just to write 16k2+ 24k+ 8= 8(2k2+3k+1) rather than dividing by 8. The fact that it is a multiple of 8 is sufficient to prove divisibility.
     
  11. Sep 19, 2006 #10
    hah I think i am handicapped, I don't enjoy Descrete Math. This course is what is keeping me in Computer ENgineering and not Computer Science.

    Thanks for the help!
     
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