Results 1 to 2 of 2

Math Help - solve 2^x+3^y=z^2

  1. #1
    Newbie
    Joined
    May 2012
    From
    United Kingdom
    Posts
    4

    solve 2^x+3^y=z^2

    basically i ahve got up to here my taking squares and modulus etc, if you need me to go into more detail i can. but i have basically got to


    2^(k+1)=3^b-1
    where x=2k

    this is meant to to me something, but im not sure what it is, any help would be much appreciated.
    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2012
    From
    United Kingdom
    Posts
    4

    Re: solve 2^x+3^y=z^2

    this is my solution so far


    We are looking for solutions of:
    2^x + 3^y = z^2 [1]
    where x, y, and z are non-negative integers.
    So z^2 ≡1 (mod 3) unless z^2=3m for any integer m, and if x is odd, say x=(2k+1) for some integer k. However when z=3, we have 2^3+3^0=3^2, so one solution is when x=3,y=0 and z=3 When x is odd 2^(2k+1)≡2(mod 3) and when x is even 2^x≡1 (mod3) so z^2 is congrgent to 2^x when x is even.
    We now know x is even, let x=2k. So we have:
    3^y=(z-2^k )(z+2^k )
    Now (z-2^k ) and (z+2^k ) should both be powers of three, so therefore (z+2^k )-(z-2^k ) should equal a power of three, so
    3^w=(z+2^k )-(z-2^k )
    3^w≠2^2k=2^x
    So 2^x should be divisible by three, but it isnít. So let (z-2^k )=3^a and (z+2^k )=3^b then we have
    █((z+2^k )=3^b@(z-2^k )=3^a )/(2^(k+1)=3^b-3^a )
    If a>0 then 3^b-3^a will be a power of three and 2^(k+1) is not divisible by three, so let a=0. We are left with
    2^(k+1)=3^b-1
    Let a=k+1, then we have 2^a=3^b-1.So this means that b≥0, otherwise 2^a wont be an integer, which means a≥1. Also a≤3 because if k>2, then 3^b+1≡0mod(8) which is a contradiction, so therefore a has to be either 1,2 or 3, making k=0,1 or 2.
    When k=0 we have
    2^1=3^b-1 → 3=3^1 so therefore b=1 and
    When k=1 we have
    2^2≠3^b-1 so therefore k=1 doesnít exist.
    When k=2 we have
    2^3=3^b-1 → 9=3^b so therefore b=2.
    So the solutions of 2^x+3^y=z^2 are
    X 0 4 3
    Y 1 2 0
    Z 2 5 3
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 11:29 PM
  2. Replies: 1
    Last Post: June 9th 2009, 11:37 PM
  3. how can i solve this
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 6th 2008, 10:11 AM
  4. how do i solve this?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 2nd 2008, 03:58 PM
  5. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 2nd 2008, 09:17 AM

Search Tags


/mathhelpforum @mathhelpforum