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Divisibility problem

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the only ##n \in \mathbb{N}-\{0,1\} ## such that ##2n-1|(3n^2-3n+1)(3n^2-3n+2)## is 3.

    2. Relevant equations
    ## P_n = (3n^2-3n+1)(3n^2-3n+2) ##
    Addition and multiplication in ## \mathbb{Z}/(2n-1)\mathbb{Z} ##

    3. The attempt at a solution

    Hello, I'm not 100% sure my proof is correct for this problem, can you check it please ? Do you see a simpler way of doing it ?

    I want to show that in ## \mathbb{Z}/(2n-1)\mathbb{Z} ##, ## \hat {P_n} = \hat 0 \Rightarrow n = 3 ## .

    First, I rewrite ## P_n = ( (2n-1)^2 + n(1-n) ) ( (2n-1)^2 + n(1-n) + 1 ) ##.

    Using the fact that ## \hat 0 = \widehat {2n-1} ## and ## \hat n = \hat 1 - \hat n ##, I get that ##\hat {P_n} = \hat {n^2} ( \hat {n^2} + \hat 1) ##

    Now, assuming that ## \hat {P_n} = \hat 0 ##, I get that
    ##\hat 0 = \hat{4} \hat {P_n} = \widehat{4n^4} + \widehat{4n^2} = \widehat{ (2n^2)^2 } + \widehat { (2n)^2 } = \widehat{ (2n)^2 } .\widehat{n^2} + \hat 1 . \hat 1 = \hat{n^2} + \hat 1 ##

    Multiplying by ##\hat{2}## left and right :
    ##\hat 0 = \widehat{2n}.\hat{n} + \hat 2 = \hat n + \hat 2 = \hat 1 - \hat n + \hat 2 = \widehat{3 -n} ##

    So ## 2n-1 | 3 - n ##. It is always true that ## 2n-1 | 0 ## so ## n = 3 ## is a solution. It is the only possibility because ##2n-1\ge |3 - n| ## so it does not divide ## 3 - n ## for any other ##n\in \mathbb{N}-\{0,1\}##
     
  2. jcsd
  3. Apr 22, 2015 #2

    mfb

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    Staff: Mentor

    Looks right.
     
  4. Apr 22, 2015 #3
    Thank you !!
     
  5. Apr 22, 2015 #4
    Do you know a simpler way of solving this problem ?
     
  6. Apr 22, 2015 #5

    mfb

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    2016 Award

    Staff: Mentor

    I don't see one.
     
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