Results 1 to 5 of 5

Math Help - AMATYC question

  1. #1
    Junior Member
    Joined
    Aug 2010
    Posts
    25

    AMATYC question

    if a^2-b^2=33 and a^3-b^3=817 have the integer solution with a>b, find the value of a-b
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Can you solve (33+b^2)\sqrt{33+b^2}-b^3=817 ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, rai2003!

    I expected a clever solution, but couldn't find it.
    So I was reduced to TinkerToy steps.


    \text{If }\;\begin{Bmatrix} a^2-b^2& =& 33 & (1)\\ a^3-b^3 &=& 817 & (2)\end{Bmatrix}\;\text{ has an integer solution}

    \text{ with }a>b,\:\text{find the value of }a-b.

    In equation [1], the difference of two squares is 33.

    There are several solutions: . (a,b) \;=\;(\pm17,\:\pm16),\;(\pm7,\:\pm4)


    But only one satisfies equation [2]: . (a,b) \,=\,(17,\:16)

    Therefore: . a-b\:=\:1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Here's a slightly quicker solution. Since (a-b)(a+b) = a^2-b^2 = 33, it follows that a-b is a factor of 33, so it must be \pm1,\ \pm3,\ \pm11 or \pm33. But a-b is also a factor of a^3-b^3 = (a-b)(a^2+ab+b^2) = 817. However, 817 is not divisible by 3 or 11, and it follows that a-b = \pm1. You then only have a couple of cases to check in order to conclude that a=17 and b=16.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2010
    Posts
    25
    thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. AMATYC question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: August 9th 2010, 06:02 PM

Search Tags


/mathhelpforum @mathhelpforum