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Interesting lemma

  1. Feb 1, 2007 #1
    Interesting lemma

    If P.Ferma knew the proof of the following lemma
    "Lemma. If numbers a, b, c, have no common factor, numbers c-a and c-b are also mutually-prime and n is odd, then the numbers (c^n-a^n)/(c-a) and (c^n-b^n)/(c-b) are also mutually-prime",
    then with its aid it is possible to briefly and simply prove Fermat's last theorem.

    Actually, in the Fermat’s equality (where numbers a, b, c have no common factor and n is odd) numbers c-a and c-b, obviously, mutually-prime. And then from the Fermat's little theorem it follows that with prime q>2c the numbers c^(q-1)-a^(q-1) and c^(q-1)-b^(q-1) are multiple by q.
    And since, according to lemma, the numbers (c^(q-1)-a^(q-1))/(c-a) and (c^(q-1)-b^(q-1))/(c-b) are mutually-prime (i.e. have no common factor), then one of the numbers c-a and c-b is divided by q (>2c>c-b>c-a), i.e., the solution of the Fermat’s equation is not integer.

    It remains to learn, who and when proved lemma.
     
  2. jcsd
  3. Feb 1, 2007 #2

    Hurkyl

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    I hope nobody proved that lemma; it's not true!

    Here's the smallest counterexample I could find:

    a = 2
    b = 2
    c = 3
    n = 3

    a, b, c don't have a common factor.

    c-a = 1
    c-b = 1

    These don't have a common factor

    (c^n - a^n) / (c-a) = 19
    (c^n - b^n) / (c-b) = 19

    These do have a common factor.


    Here's the smallest I could find with a and b unequal:

    a = 1
    b = 4
    c = 9
    n = 3

    These don't have a common factor.

    c-a = 8
    c-b = 5

    These don't have a common factor.

    (c^3 - a^3) / (c-a) = 91
    (c^3 - b^3) / (c-b) = 133

    These do have a common factor: 7.


    Here's another:

    a = 5
    b = 6
    c = 17
    n = 3

    c-a = 12
    c-b = 11

    (c^3 - a^3) / (c-a) = 399
    (c^3 - b^3) / (c-b) = 427

    These do have a common factor: 7.
     
  4. Feb 2, 2007 #3
    Outstanding counterexamples!

    Therefore I make correction to the Lemma:
    the additional condition: a+b=c.
    By the way, now to prove the lemma it is much easier.
     
  5. Feb 2, 2007 #4

    matt grime

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    Then feel free to post the proof.
     
  6. Feb 2, 2007 #5
    Lemma

    Here are the minimum requirements for the lemma, necessary for the brief proof of the FLT: "Lemma. If integers a, b, c have only one common divisor 1, a+b=c and numbers (c^n-a^n) and (c^n-b^n) have common divisor d>2, then with n>1 one of the numbers c-b, c-a, a+b, c+b, c+a, a-b is divided by d ".
     
  7. Feb 2, 2007 #6

    Hurkyl

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    No it's not.

    a = 3, b = 10, c = 13, n = 3 is another counterexample.
    a = 5, b = 12, c = 17, n = 3 is another.

    If you're guessing, call it a conjecture. You can't call something a lemma until you have actually proven it.



    matt asked you to post the proof. That's not a proof.
     
  8. Feb 3, 2007 #7
    Your counterexamples speak about the accuracy of the Lemma:
    10-3=7,
    12-5=7.
    Consequently, it is possible to begin the search for its proof.
     
  9. Feb 3, 2007 #8
    What do you mean by accuracy of the lemma? If there exists A counter example then it isn't true.
     
  10. Feb 3, 2007 #9
    Here is the Lemma, which easily proves and which is sufficient for the brief proof of the FLT:
    Lemma. If integers a, b, c have only one common divisor 1, a+b=c, number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided into d ".

    to be continued
     
    Last edited: Feb 3, 2007
  11. Feb 3, 2007 #10
    Lemma And Its Proof

    Lemma.
    If integers a, b, c have only one common divisor 1, a+b=c, and number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, and in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided by d.

    Proof.
    Let us assume that number (a^n-b^n)/(a+b) is divided by d.
    Then number S=(c^n-b^n)/(c-b)+(c^n-a^n)/(c-a)+(a^n-b^n)/(a+b), or
    (c^n-b^n)/a+(c^n-a^n)/b+(a^n-b^n)/c = [c^(n+2)+b^(n+2)+(a+b)c^(n+1]/abc=
    =[c^2+b^2+a(c+b)]/abc, is divided by d.
    But since number a^n, b^n, c^n finish by number 1, then the number s = c^2+b^2+a(c+b), where 0<s<d, is single-digit number and, therefore, number S is NOT DIVIDED by d.
     
  12. Feb 3, 2007 #11
    Elementary proof of Fermat's last theorem

    Elementary proof of Fermat's last theorem

    According to known theorem from the theory of integers, for any prime n there exist infinite great number of the prime numbers d of form d=pn+1, where p is integer. (This theorem is the simple consequence from the fact that, if numbers a and b are mutually-prime and n prime, then each divisor of the number (a^n+b^n)/(a+b) takes form pn+1).

    Proof FLT:
    Let us take simple d>c^(4n). Then, according to Fermat's little theorem, the numbers C^p-B^p, C^p-A^p, A^p-B^p, where C=c^n, B=b^n; A=a^n, are divided by d. And since A+B=C, then, according to Lemma, number 3C^2 is divided by d>3C^2. Consequently, the solution of Fermat’s equation is not integer.
     
    Last edited: Feb 3, 2007
  13. Feb 3, 2007 #12

    arildno

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    Dearly Missed

    Can't you just stop posting this nonsense?
     
  14. Feb 3, 2007 #13
    OK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
     
  15. Feb 4, 2007 #14
    Lemma is not true

    Lemma is not true. New version of the lemma:
    If in the simple base q the mutually-simple numbers A^p, B^p, C^p (where C=c^n, B=b^n, A=a^n, prime n>2 and p=/q-1) finish by digit 1, then either A+B and A-B or C+B and C-B, or C+A and C-A have common divisor not equal to 1.
    Today I have not its proof.

    Time out
     
  16. Feb 4, 2007 #15

    Hurkyl

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    I repeat, if you don't have a proof, then the correct term is conjecture. You cannot (correctly) call it a lemma until it has been proven.
     
  17. Feb 22, 2007 #16

    Gib Z

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    This guys tried to post the same thing in scienceforums.net, except there he blatantly said he was proving FLT. He was surprised when no one took him seriously >.<

    Mate its good your trying, but wait till you learn abit more math. Fermat would have known this stuff.
     
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