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Fermat's Last theorem

  1. Aug 28, 2007 #1
    We all know of this theorem which was finally proved in the 1960's. It says that we cannot find any real integral solution for n>2 when an integer is expressed to a power of 'n' and is equal to the sum of two numbers which individually are raised to the power 'n'.

    x^n=a^n+b^n

    Well for n=2, we are familiar with the pythagorean(3,4,5 etc.) combinations, but there is indeed is no solution when n>2...check it out.

    I recently ran a program to find this solution, but couldn't find for n=3,4,5... This certainly validates the theorem but how do we prove it mathematically?
     
  2. jcsd
  3. Aug 28, 2007 #2

    cristo

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    I thought it was proved in the '90s?
     
  4. Aug 28, 2007 #3

    mgb_phys

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    Either try all infinite 'n' - might take a while (actually only need to test all the infinite number of prime n's)
    or learn an awful lot of seriously complicated number theory about elliptic curves and some conjecture that I can't even spell as Andrew Wiles did. There isn't a proof understandable by mere mortals.
     
    Last edited: Aug 28, 2007
  5. Aug 28, 2007 #4
    Yes, thats a mistake, it was done so in the 90's but that's apart from the point...
     
  6. Aug 28, 2007 #5

    mgb_phys

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    There is a good book by Simon Singh, but since it is pretty impossible to even outline the numerical techniques to someone without grad school maths it mainly concentrates on the stories of the people involved.

    It would be interesting to know what Fermat's original proof was and where he went wrong!
     
  7. Aug 29, 2007 #6

    HallsofIvy

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    There was no original proof. After he wrote his comment in the margin of a book, he gave completely different proofs for n= 3 and n= 4. He wouldn't have done that if he had a proof for all n.

    What happened to Fermat is what happens to mathematicians all the time- he saw a way of extending result he already had and wrote that "marginal" comment. Later he realized it didn't extend as he thought.
     
  8. Aug 29, 2007 #7
    Also known as the Taniyama–Shimura conjecture. Don't even bother starting unless you already know what elliptic curves and modular functions have to do with each other. (I don't. Not a clue.)
     
  9. Aug 29, 2007 #8

    HallsofIvy

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    I wouldn't say it is "also known as". Yes, Wiles proved the Taniyama-Shimura conjecture. It has already been proven that Fermat's last theorem was true if and only if theTaniyama-Shimura conjecture was true.
     
  10. Aug 29, 2007 #9
    The German mathematician Frey announced that he believes that Fermat's equation, if false, will imply that the Taniyami-Shimura Conjeture to be false. This became known as the Eplison Conjecture. Soon the mathematician Kenneth Ribet actually prove this idea in detail. Wiles immediately realized that all he needs to do know if prove the Taniyami-Shimura Conjecture.

    In the old days, Fermat's problem was attacked by Kummer Ideal Complex numbers. But this does not work succesfully. There is a book (I forgot the name) which teaches algebraic number theory while simultaneously working with Fermat's equation in a classical approach.
     
  11. Aug 30, 2007 #10
    How did you "run a program" to check for all n= 3, 4, 5?
     
  12. Aug 30, 2007 #11

    CRGreathouse

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    I thought Taniyama-Shimura was stronger than Fermat's last theorem -- that Wiles proved special case of T-S first, enough to prove Fermat's last theorem, and later (with several others) proved the whole Taniyama-Shimura conjecture. Am I confusing this with something else?
     
  13. Aug 30, 2007 #12
    It was bit of an accident that I stumbled upon Fermat's Last Theorem. I was originally coding a program to find the smallest number, which when squared, can be expressed as the sum of two different sets of individually squared numbers, which is:

    65 = 1^2+8^2 = 4^2+7^2

    in terms of three three numbers, you have:

    325 = 1^2+18^2 = 10^2+15^2 = 6^2+17^2

    in terms of four numbers, you have:

    1105 = 23^2+24^2 = 4^2+33^2 = 9^2+32^2 = 12^2+31^2

    This list goes on...but what I tried to do next is replace the power of n=2 with n=3 or greater integral values. What I encountered is that the program kept looping till infinity i.e. the program never terminated for the first set of two numbers cubed individually(a^3+b^n=c^n);this led me to doubt the program. I checked the program but found no errors. So I concluded that there was no solution for a power greater than 2. I incidentally also came across Fermat's last theorem.So my conclusion was indeed justified.

    Well, here is a program in c++ for reference:

    Code (Text):
    #include <iostream.h>
    #include <conio.h>
    #include <math.h>

    int main()
    {
         long n=1,i,j,flag,s,k;  //initializing
         long p=3;               //replace 'p' to check for any other power

         while(1)                //loop till infinity
        {
            flag=0;k=pow(n,p);

            for(i=1;pow(i,p)<k;i++)         //looping first number
            {
                for(j=i+1;pow(i,p)+pow(j,p)<=k;j++) //looping second number
                {
                  s=pow(i,p)+pow(j,p);
                  if(s==k){flag++;}
                  if(flag==2){break;}           //break out of loop if found
                }
                if(flag==2){break;}
            }
          if(flag==2){cout<<n;break;}  //print number if found and the program terminates
          n++;
         }
    return 0;          
    }
    Replace 'p' for power of the expression, and you will find that for p=2, answer is 25 which rightly is the smallest pythagorean triplet; but for n>2 and n having an integral value, the program doesn't terminate i.e there is no solution.
     
  14. Aug 30, 2007 #13

    mgb_phys

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    Have you it for an infinite time yet?


    To prove all odd numbers greater than 2 are prime numbers :
    Mathematician: 3 is prime number, 5 is prime number, 7 is prime number, by induction, all odd numbers greater than 2 are prime numbers
    Physicist: 3 is prime number, 5 is prime number, 7 is prime number, 9 is experiment error, 11 is prime number,......
    Engineer : 3 is prime number, 5 is prime number, 7 is prime number, 9 is prime number, 11 is prime number,......
    Computer Programmer: 3 is prime number, 5 is prime number, 7 is prime number, 7 is prime number, 7 is prime number,......
    Statistician: Let us try some random numbers: 17 is prime number, 23 is prime number, 11 is prime number,......
     
  15. Aug 30, 2007 #14
    ha...Alrighty. Agreed that the program will not run for an infinite time but if there was a solution to it, the program would have given it in a relatively short period of time.Wouldn't it? So you could conjecture that the equation is unlikely to have a solution.
     
  16. Aug 30, 2007 #15

    matt grime

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    No. You cannot say that. The behaviour for a small, finite time, is no indicator of the over all behaviour.
     
  17. Aug 30, 2007 #16
    I don't mean to say that it is universally true that for a small behaviour, the overall behaviour is truly indicated. It was only in my case that the unavailability of the solution by the program for a small time led me to faintly suspect that there might be something peculiar about the solution. It was a thought pertaining only to the situation and was not a generalisation of sorts for all other mathematical concepts.

    So what exactly is the Taniyama-Shimura conjecture?
     
  18. Aug 30, 2007 #17
  19. Aug 30, 2007 #18

    matt grime

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    If you're generalization to 'all integers' is based merely on some searhc not terminating, rather than examining the reason why it does not terminate quickly, then you've clearly done something rash.
     
  20. Aug 30, 2007 #19

    CRGreathouse

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    Sure, you can conjecture that -- and you did, and you were right. There are lots of examples of problems with large smallest counterexamples, though, so matt took issue with your guess. Eh.
     
  21. Aug 30, 2007 #20
    No. It's not rash at all.The search did not terminate for a reason, did it? And it is because according to the theorem there is no solution to an integral power above 2. So, the reason was examined and that is how I came across this ingenious theorem.

    One way to go about solving it would be that a power of 4 for example can be expressed in terms of 2*2 or algebraically, 'p=mn'

    (a^m)^n + (b^m)^n = (c^n)^n

    Perhaps we could somehow use this identity(though I don't know how).
     
    Last edited: Aug 30, 2007
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