Page 2 of 2 FirstFirst 12
Results 16 to 17 of 17

Math Help - Simple Proof of Beal's Conjecture

  1. #16
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by MrAwojobi View Post
    MrAwojobi
    Hi all, I have added an extra equation to this revised proof to show why the Pythagorean triples do not obey Beal's equation

    SIMPLE PROOF OF BEAL’S CONJECTURE
    Beal’s Conjecture
    Beal’s conjecture states that if A^x + B^y = C^z where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.
    Examples
    ...............................Common Prime Factor
    2^3 + 2^3 = 2^4 => 2
    2^9 + 8^3 = 4^5 => 2
    3^3 + 6^3 = 3^5 => 3
    3^9 + 54^3 = 3^11 => 3
    27^4 + 162^3 = 9^7 => 3
    7^6 + 7^7 = 98^3 => 7
    33^5 + 66^5 = 33^6 => 11
    34^5 + 51^4 = 85^4 => 17
    19^4 + 38^3 = 57^3 => 19

    Primitive Pythagorean Triples
    A primitive Pythagorean triple is one in which the integer lengths of the right angled triangle do not have a common prime factor. Examples are….
    ( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17)
    ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85)
    (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
    (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
    Hence the reason why x, y and z in Beal’s conjecture equation have to be greater than 2. These triples do not obey Beal's equation because of
    the unique factorisation of a^2+b^2=c^2 to a^2 = (c+b)(c-b)

    Simple Proof
    It should be clear that each term in the equation A^x + B^y = C^z can be broken down into the product of its prime factors after numbers have been substituted into it. Thus the equation could be rewritten as abcde + fghij = klmno for instance, where a,b,c,d,e,f,g,h,i,j,k,l,m,n,o are prime . It isn’t difficult to see that
    the 1st product + the 2nd product = the 3rd product
    if and only if the left hand side of the equation can be factorised, i.e. rewritten in the form P(Q + R) where P,Q and R are positive integers. This will therefore guarantee that A, B and C share a common prime factor.
    You have a fundamental flaw in your proof. You need to start over from scratch. Also what if x=y=z=1 or x=y=2, z=1, etc?

    If you're so confident in your proof, why are you telling us it if it can win you $100,000? Aren't you afraid of your "proof" being stolen?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Newbie
    Joined
    Aug 2010
    Posts
    24
    Thanks
    1

    Cool

    chiph588@

    how can it be stolen when the evidence is right here with dates and time? Maybe if the thief is a time traveller then that would be possible. But then if he was a time traveller I'm sure he or she will be up to something more sinister than stealing my proof.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. abc-conjecture proof
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: January 12th 2012, 02:08 AM
  2. Goldbach's conjecture proof?
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 16th 2011, 12:22 AM
  3. Revised Simple Proof of Beal's Conjecture
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: August 13th 2010, 12:30 PM
  4. How can I proof my conjecture?
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: June 4th 2010, 07:17 AM
  5. A Proof & Goldbach Conjecture
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: April 19th 2007, 09:00 PM

Search Tags


/mathhelpforum @mathhelpforum