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Abstract algebra question chapter 1.2

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    prove that if 2 does not divide a then 24 divides a^2-1


    2. Relevant equations

    I know that if 2 does not divide a then a is odd.

    I proved that the square for all odd integers are of the form 8K+1

    I also proved the square of any integer is either in the form 3k, 3k+1 or 3k+2. I know 3k and 3k+2 are even so the odd are in the form 3k+1






    3. The attempt at a solution

    Can I simply say that because both 8 and 3 are in this form then 8*3 is in this form?
     
  2. jcsd
  3. Sep 3, 2009 #2
    That is not what I mean to say. based on before I know that 8 divides a^2-1 and 3 divides a^2-1. Can I then say that since both 3 and 8 divide a^2-1 then 24 divides a^2-1
     
  4. Sep 3, 2009 #3
    Ok, I just realized that 3k and 3k+2 are not necessarily even and 3k+1 is not necessarily odd. So now I am even more confused
     
  5. Sep 3, 2009 #4
    OK, I am going to say this is not true. When a=3 then a^2 is in the form 3k. so I have disproved it I think.
    However the next part says if 2 or 3 do not divide a. That eliminates every third number. I also read my proof wrong and it is either 3k or 3k+1 not 3k+2 so eliminating all the times when 3 divides a will leave me the other two options and for each of those a^2 is in the form 3k+1. OK, I need to write it up but I think I got it.
     
  6. Sep 3, 2009 #5

    Hurkyl

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    If this hypothesis is true, then this conclusion is true, by the Chinese Remainder Theorem.
     
  7. Sep 3, 2009 #6

    HallsofIvy

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    Or, since 3 and 8 have no prime factors in common, yes, any number that is divisible by both 3 and 8 is divisible by 3(8)= 24.
     
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