Results 1 to 6 of 6

Thread: Mod 5

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    13

    Mod 5

    I am having trouble with this homework question and was hoping someone could help. Cheers

    Prove that 5|n(n2-1)(n2=1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Look for every possibility of n if 5 divides the product.

    Try $\displaystyle n \equiv 0 mod 5$, then $\displaystyle n \equiv 1 mod 5$ and so on
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by asw-88 View Post
    I am having trouble with this homework question and was hoping someone could help. Cheers

    Prove that $\displaystyle 5|n(n^2-1)(n^2+1)$ (at least, I think that's what is meant)
    If you multiply out the brackets then you get $\displaystyle 5|n^5-n$, which is true by Fermat's little theorem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848

    Wink

    Hello, asw-88!

    Prove that: .$\displaystyle 5\,|\,n(n^2-1)(n^2+1)$
    Let $\displaystyle N \:=\:n(n^2-1)(n^2+1)$


    $\displaystyle n$ must one of five possible forms: .$\displaystyle 5k-2,\:5k-1,\:5k,\:5k+1,\:5k+2$


    $\displaystyle [1]\;n \,= \,5k-2\!:\;\;N \:=\5k-2)([5k-2]^2-1)([5k-2]^2 + 1)$
    . _ . . . . . . . . . . $\displaystyle =\;(5k+2)(25k^2-20x+3)(25k^2-20k + 5)$
    . _ . . . . . . . . . . $\displaystyle = \;(5k+2))(25k^2+20k + 3){\color{red}5}(5k^2 - 4k + 1)$ . . . a multiple of 5

    $\displaystyle [2]\;n \,=\,5k-1\!:\;\;N \:=\5k-1)([5k-1]^2-1)([5k-1]^2+1)$
    . . . . . . . . . . . . $\displaystyle = \;(5k-1)(25k^2 - 10k)(25k^2-10k + 2)$
    . . . . . . . . . . . . $\displaystyle = \;(5k-1){\color{red}5}k(5k- 2)(25k^2 - 10k+2)$ . . . a multiple of 5

    $\displaystyle [3]\;n \,=\,5k\!:\qquad N \:=\5k)([5k]^2 - 1)([5k]^2+1)$
    . . . . . . . . . . . . $\displaystyle = \;{\color{red}5}k(25k^2 - 1)(25k^2+1)$ . . . a multiple of 5

    $\displaystyle [4]\;n\,=\,5k+1\!:\;\;N \;=\;(5k+1)([5k+1]^2-1)([5k+1]^2+1)$
    . . . . . . . . . . . . $\displaystyle = \;(5k+1)(25k^2 + 10k)(25k^2+10k + 2)$
    . . . . . . . . . . . . $\displaystyle = \;(5k+1){\color{red}5}k(5k + 2)(25k^2 + 10k + 2)$ . . . a multiple of 5

    $\displaystyle [5]\;n\,=\,5k+2\!:\;\;N \;=\;(5k+2)([5k+2]^2-1)([5k+2]^2+1)$
    . . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k + 3)(25k^2 + 20k + 5)$
    . . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k+3){\color{red}5}(5k^2 + 4k + 1)$ . . . a multiple of 5

    . . . . . . . . . . . . . . . . . Q.E.D.

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Henderson's Avatar
    Joined
    Dec 2007
    Posts
    127
    Thanks
    2
    Quote Originally Posted by topsquark View Post
    I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.

    -Dan
    I'll do the initial condition:
    $\displaystyle
    5|1^5-1$
    $\displaystyle 5|0
    $

    Because I care, I'll even show the next iteration:
    $\displaystyle
    5|2^5-2 $
    $\displaystyle 5|30
    $

    The rest of it is almost as easy as this.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum