Results 1 to 5 of 5

Math Help - Prove that a^3 - b^3 = 2011 has no integer solutions.

  1. #1
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539

    Prove that a^3 - b^3 = 2011 has no integer solutions.

    I am a number theory newbie. So please explain all the steps.
    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,
    Quote Originally Posted by fardeen_gen View Post
    I am a number theory newbie. So please explain all the steps.
    You can see that 2011 is a prime number. Thus its only possible factorisation is \pm 1 \cdot \pm 2011. Why do I say that ? It's because we'll factorise a^3-b^3.

    a^3-b^3=(a-b)(a^2+ab+b^2)

    ~~~~~~~~~~~~~~~~~~~~
    Let's see what the signs of a & b are.

    If a<0, then a^3<0 and so -b^3>0 that is to say b<0. And b^3>a^3 --> b>a --> b-a>0, thus a^2+ab+b^2<0, which is not possible because a+b>0 and a & b have the same sign and then ab>0.
    Therefore a has to be positive.
    ------------------------------------
    Let's assume that a & b > 0.

    Then a+ab+b > 1. Thus the only possibility is that a+ab+b=2011 and a-b=1.

    --> a=b+1. Substituting in a+ab+b=2011 :
    b+2b+1+b+b+b=2011.
    3b+3b=2010
    b+b=670
    b+b-670=0

    \Delta=1+4*670=2681. But 2681 is not a perfect square. Thus b is not an integer. Which is not satisfying the conditions. Therefore, there are no a & b, positive integers, such that a^3-b^3=2011.

    ------------------------------------
    What if a>0 and b<0 ?
    Well, try to think about that



    @CB : not sufficient, eh ?


    Edit : my 27th ^^
    Last edited by Moo; July 27th 2008 at 03:42 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,654
    Thanks
    598
    Hello, fardeen_gen!

    A slight variation of Moo's solution . . .


    Prove that .a - b .= .2011 .has no integer solutions.

    We have: .(a - b)(a + ab + b) .= .2011

    Since 2011 is prime, the factors are .1 and 2011


    We have: .a - b .= .1 . . a .= .b 1 . [1]

    . . . .and: .a + ab + b .= .2011 . [2]


    Substitute [1] into [2]: .(b 1) + b(b 1) + b .= .2011


    This simplifies to two equations:

    . . . . .b + b - 670 .= .0
    . . . . . . . . . . . . . . . . . . ... and neither has integer solutions.
    . . 3b - 3b + 2012 .= .0

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Correcting some omissions:
    Attached Thumbnails Attached Thumbnails Prove that a^3 - b^3 = 2011 has no integer solutions.-gash.png  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    A proof by using congruences
    Attached Thumbnails Attached Thumbnails Prove that a^3 - b^3 = 2011 has no integer solutions.-dio.-eq..gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. No integer Solutions
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: July 16th 2011, 03:01 AM
  2. # of integer solutions
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 2nd 2011, 09:43 PM
  3. No integer solutions to x^4 + y^4=100
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 17th 2009, 10:04 AM
  4. Integer Solutions?
    Posted in the Number Theory Forum
    Replies: 44
    Last Post: September 14th 2009, 09:11 PM
  5. Integer Solutions
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: July 31st 2009, 09:18 AM

Search Tags


/mathhelpforum @mathhelpforum