Results 1 to 5 of 5

Thread: 2n^3+3n^2+n\equiv 0 (mod 6)

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    2n^3+3n^2+n\equiv 0 (mod 6)

    $\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

    $\displaystyle 6|(2n^3+3n^2+n)\rightarrow 6k=2n^3+3n^2+n$

    ???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Hint: $\displaystyle 2n^3+3n^2+n = n(n+1)(2n+1)$

    This is a rather easy exercise - using either induction or direct proof (unless I am missing something in the question)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

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

    I have a back-door approach to this one . . .


    $\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

    As aman-cc pointed out: .$\displaystyle 2n^3 +3n^2 + n \:=\:n(n+1)(2n+1)$


    I recognized this is as part of the formula for the sum of consecutive squares:

    . . $\displaystyle 1^2 + 2^2 + 3^2 + \hdots + n^2 \:=\:\dfrac{n(n+1)(2n+1)}{6}$

    Since the left side is an integer, the right side must be an integer.
    . . That is, $\displaystyle n(n+1)(2n+1)$ is a multiple of 6.


    Therefore: .$\displaystyle 2n^3 + 3n^2 + n \:\equiv\:0\text{ (mod 6)}$ .for all integer values of $\displaystyle n.$

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Consider $\displaystyle 2n^3 + 3n^2 + n = 2(n^3 - n ) + 3n(n+1) = 2(n-1)(n)(n+1) + 3n(n+1) $ , use the property of product of few consecutive integers .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2010
    From
    Israel
    Posts
    148
    $\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

    This a solution I saw once:
    $\displaystyle 2n^3+3n^2+n=n(2n^2+3n+1)=n(n+1)(2n+1) $. One of $\displaystyle n $ and $\displaystyle n+1 $ is divisible by $\displaystyle 2 $.

    $\displaystyle 2\equiv -1(mod\ 3)$;
    $\displaystyle n(n+1)(2n+1)\equiv -(n-1)n(n+1)(mod\ 3)$. Then at least one of $\displaystyle n-1 $, $\displaystyle n $, and $\displaystyle n+1 $ is divisible by $\displaystyle 3 $.

    Since $\displaystyle 2 $ and $\displaystyle 3 $ are relatively prime $\displaystyle 2n^3+3n^2+n $ is divisible by $\displaystyle 6 $.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] if ra\equiv rb (mod m), then a\equiv b (mod \frac{m}{gcd(r,m)})
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Jun 15th 2011, 08:04 AM
  2. [SOLVED] Let p be odd. Then 2(p-3)!\equiv -1 (mod p)
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: Jul 12th 2010, 03:42 PM
  3. Replies: 2
    Last Post: Jul 10th 2010, 05:14 PM
  4. [SOLVED] a \not\equiv b (mod m), then b \not\equiv a (mod m)
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: Jul 5th 2010, 02:44 AM
  5. [SOLVED] if a^2 \equiv 1, then a \equiv \pm 1 (mod p)
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: Jun 27th 2010, 03:43 PM

/mathhelpforum @mathhelpforum