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The Lost Proof of Fermat

  1. Apr 12, 2004 #1
    5^1 = 1*0 + 5

    5^2 = 2*10 + 5

    5^3 = 3*40 + 5

    5^p = p*a + 5

    x^p = p*a + x



    x^p = p*a + x

    y^p = p*b + y

    z^p = p*c + z



    ...x^p + y^p = z^p



    p*a + x + p*b + y = p*c + z

    p*[a + b - c] = z - [x + y]


    p = [z - (x + y)]/[a + b - c]



    http://www.maa.org/mathland/mathtrek_12_8.html

     
  2. jcsd
  3. Apr 12, 2004 #2
    how does this prove that x,y and z can't be integers when p>2? Also a,b and as far as I know can be anything you want.
     
  4. Apr 13, 2004 #3

    p, x, y, a,b,c, are always integers > 0, when p is a prime number. When p is 1, a = 0, b = 0, and c = 0.


    z and c cannot be a +integer for p > 2.


    [x^p - x]/p = a

    [y^p - y]/p = b

    [z^p - z]/p = c


    for example:


    [3^2 - 3]/2 = 3

    [3^3 - 3]/3 = 8

    [3^5 - 3]/5 = 48

    [+integer^p - +integer]/p = another +integer.

    It works for all prime numbers.


    3^2 + 4^2 = 5^2

    2*3 + 3 = 3^2

    2*6 + 4 = 4^2



    2*3 + 4*3 + 3 + 4 = 6*3 + 7 = 5^2
     
    Last edited: Apr 13, 2004
  5. Apr 13, 2004 #4
    Russel Rierson,

    I don't know why this is here and no in the number theory section of the math board.

    But as long as it's here, you lost me on your ifrst line: 5^1 = 1*0 + 5
     
  6. Apr 13, 2004 #5
    z and c can be integers when p>2. you also have to prove that statement you can't just say it.

    Also fermat's theorem as far as I know isn't just for primes, it's for all numbers. Unless this is some other theorem by fermat. In which case you need to state it so we know what your talking about.
     
    Last edited: Apr 13, 2004
  7. Apr 14, 2004 #6

    If p = 2 :

    x^2 = 2*[1 + 2 + 3+...+ x-1] + x

    2^2 = 2*[1] + 2

    3^2 = 2*[1 + 2] + 3

    4^2 = 2*[1 + 2 + 3] + 4

    5^2 = 2*[1 + 2 + 3 + 4] + 5

    etc...

    This congruence does not hold for p > 2
     
  8. Apr 14, 2004 #7
    5^2 = 2*[1 + 2 + 3 + 4] + 5


    5^2 = 2*[1 + 2] + 2*[3 + 4] + 5



    5^2 = 2*[1 + 2] + 2*[3 + (4 - 1)] + 5 + 2*1



    5^2 = 2*[1 + 2] + 2*[3 + 3] + 7



    5^2 = 2*[1 + 2] + 3 + 2*[ 1 + 2 + 3] + 4



    5^2 = 3^2 + 4^2
     
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