Results 1 to 7 of 7
Like Tree5Thanks
  • 2 Post By JeffM
  • 3 Post By DenisB

Thread: Natural number

  1. #1
    Newbie
    Joined
    Nov 2015
    From
    Denmark
    Posts
    2

    Natural number

    So the problem is:
    The numbers a, b and c fullfill, that a2+2b2=3c2. Show that (a+2b+3c)/(a+c)*((a+b)/(b+c)+(b-c)/b-a)) is a natural number.

    I hope someone can help me!
    thx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,734
    Thanks
    808

    Re: Natural number

    Is this the problem:

    $Given:\ a,\ b,\ and\ c \in \mathbb R\ such\ that\ a^2 + 2b^2 = 3c^2,\ a \ne c,\ b \ne -\ c,\ and\ a \ne b;$

    $Prove\ that\ \dfrac{a + 2b + 3c}{a - c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) \in \mathbb N.$
    Thanks from topsquark and Honestly
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,671
    Thanks
    313

    Re: Natural number

    Quote Originally Posted by JeffM View Post
    $Prove\ that\ \dfrac{a + 2b + 3c}{a - c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) \in \mathbb N.$
    Typo Jeff: left denominator = a + c

    That expression seems to always equal 4. Prove it does and we're done?
    Thanks from topsquark, Honestly and JeffM
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2015
    From
    Denmark
    Posts
    2

    Re: Natural number

    Will you show me how to prove it?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,671
    Thanks
    313

    Re: Natural number

    Not me!
    But I'll tell you that by rearranging equation:
    b^2 = (3c^2 - a^2) / 2
    it is easy to show that all 3 variables are all even or all odd.
    First 2 are:
    (a,b,c) = (1,11,9) and (2,22,18).

    But I'm sure that mon ami Jeffroi has figured out the proof
    in his head and will ride in soon to save the day !
    Last edited by DenisB; Nov 20th 2015 at 11:41 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,734
    Thanks
    808

    Re: Natural number

    Thanks Denis for catching the typo.

    $Given:\ a,\ b,\ and\ c \in \mathbb R\ such\ that\ a^2 + 2b^2 = 3c^2,\ a \ne -\ c,\ b \ne -\ c,\ and\ a \ne b;$

    $Prove\ that\ \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) \in \mathbb N.$

    $\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{(b + a)(b - a)}{(b + c)(b - a)}+ \dfrac{(b - c)(b + c)}{(b - a)(b + c)} \right ) \implies$

    $\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{2b^2 - a^2 - c^2}{(b - a)(b + c)} \right ).$

    $But\ a^2 + 2b^2 = 3c^2 \implies 2b^2 = 3c^2 - a^2.$

    $\therefore \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{3c^2 - a^2 - a^2 - c^2}{(b - a)(b + c)} \right ) \implies$

    $\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{2(c^2 - a^2)}{(b - a)(b + c)} \right ) \implies$

    $\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{2(c - a)(a + 2b + 3c)}{(b - a)(b + c)} \implies$

    $\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{2(ac + 2bc + 3c^2 - a^2 - 2ab - 3ac)}{b^2 - ab + bc - ac} = \dfrac{2(3c^2 - a^2 - 2ab + 2bc - 2ac)}{b^2 - ab + bc - ac}.$

    $But\ a^2 + 2b^2 = 3c^2 \implies 3c^2 = 2b^2 +a^2.$

    $\therefore \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{2(2b^2 + a^2 - a^2 - 2ab + 2bc - 2ac)}{b^2 - ab + bc - ac} = \dfrac{2 * 2(b^2 - ab + bc - ac)}{b^2 - ab + bc - ac} \implies$

    $\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = 4 \in \mathbb N.\ QED$

    Just lots of fussy algebra.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,671
    Thanks
    313

    Re: Natural number

    Nice wrapup! I gave up at end of my 1st sheet of equations!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. n-th root of natural number
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: Jun 9th 2015, 11:05 PM
  2. Natural Number
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Aug 30th 2012, 01:04 PM
  3. Proof for every Natural number
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: Mar 16th 2012, 07:02 AM
  4. natural number problem
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: May 5th 2010, 12:05 AM
  5. Prove natural number
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Apr 25th 2009, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum