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

  • Thread starter mr_coffee
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  • #1
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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:
 

Answers and Replies

  • #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?
 
  • #3
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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?
 
  • #4
shmoe
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mr_coffee said:
well if i divide (16k^2+24k+8)/8 = (3k^2+12k+4)/4.
Something funny happened here, check this carefully.
 
  • #5
HallsofIvy
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Why did you factor out 16? You only need to show that it is 8(an integer).
 
  • #6
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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.
 
  • #7
shmoe
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mr_coffee said:
Sorry i had a typo it should have been, (3k^2+12k+4)/2.
[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]

mr_coffee said:
But i'm lost on what they want.
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).
 
  • #8
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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
 
  • #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.
 
  • #10
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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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