Results 1 to 5 of 5

Math Help - If a is odd, prove ... by division algorithm

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    Virginia
    Posts
    2

    If a is odd, prove ... by division algorithm

    The problem states that if a is odd, then prove 12 | a^2 + (a+2)^2 + (a+4)^2 + 1

    Here is what I did but it seems more by induction than the division algorithm. If someone could let me know if I am on the right track.

    Base case a=1 .... 12 |36 done.

    So, if k is in G then prove (2k-1) is in G

    12 | (2k-1)^2 + ((2k-1)+2)^2 + ((2k-1)+4)^2 +1 - I simply multiplied this whole thing out and got 12 | 12k^2 + 12k + 12

    Then if a|b we know a|bc, so 12|12 or 12|-12 if we want to use if a|b and a|c then a|b-c

    Thus k+1 is in G.

    I guess I just thought I would be looking for k(k+1)(k+2) as we did in the example in class.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member MaxJasper's Avatar
    Joined
    Aug 2012
    From
    Canada
    Posts
    482
    Thanks
    54

    Lightbulb Re: If a is odd, prove ... by division algorithm

    a=odd = a\text{=}2k+1

    a^2+(a+2)^2+(a+4)^2+1 = 12 \left(3+3 k+k^2\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,714
    Thanks
    633

    Re: If a is odd, prove ... by division algorithm

    Hello, janchilds!

    You are trying an inductive proof . . . Okay!


    \text{If }a\text{ is odd, then }a^2 + (a+2)^2 + (a+4)^2 + 1\text{ is divisible by 12.}

    Verify S(1)\!:\;1^2 + 3^2 + 5^2 + 1 \:=\:36\;\hdots\;\text{True!}


    Assume S(k)\!:\;k^2 + (k+2)^2 +(k+4)^2 + 1 \;=\;12p\;\text{ for some integer }p.


    Add (k+6)^2-k^2 to both sides:

    . . (k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + (k+6)^2-k^2

    . . (k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + k^2 +12k + 36 - k^2

    . . (k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;12p + k + 36

    . . (k+2)^2 + (k+4)^2 + (k+6)^2 + 1 \;=\;\underbrace{12(p + k + 3)}_{\text{divisible by 12}}

    We have proved S(k\!+\!1).
    The inductive proof is complete.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2012
    From
    Virginia
    Posts
    2

    Re: If a is odd, prove ... by division algorithm

    I get this... a bit different format from what I was doing but the same bottom line.

    But this is proving it by Induction. If the instructions specifically says prove using the Division Algorithm.... is that the same?

    Thanks,
    Janet
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,845
    Thanks
    715

    Re: If a is odd, prove ... by division algorithm

    You can add the division algorithm into the proof rather easily. Write the statement of the proof, then any time you are discussing divisibility, you just need r=0 where r is the remainder in the division algorithm. So, your proof by induction changes form. You claim that for any a\in 2\mathbb{Z}+1, there exists p\in \mathbb{Z} so that (a^2+(a+2)^2+(a+4)^2+1) = 12p + 0. Now, you are using the division algorithm where your q=12 and r=0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The Division Algorithm 2
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 5th 2011, 12:18 PM
  2. Division algorithm/mod
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: February 13th 2010, 12:51 PM
  3. The Division Algorithm
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 16th 2009, 12:45 PM
  4. Division Algorithm...
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: July 8th 2008, 08:33 PM
  5. division algorithm
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: September 7th 2007, 01:39 PM

Search Tags


/mathhelpforum @mathhelpforum