ElijahJones Posted August 25, 2005 Posted August 25, 2005 The famous Last Theorem of Fermat (proven by Andrew Wiles in the early 1990's) http://en.wikipedia.org/wiki/Andrew_Wiles says There are no positive integer solutions for [MATH]n>2[/MATH] to the equation [MATH]z^n=x^n+y^n[/math] We will not prove this theorem here (Wiles method is very advanced). But we will play around with the implications a bit. Something fun for us mortals to do si assume the negative of FLT and then start messing around a bit and see what happens. It remains entertaining only because some people believe Fermat actually solved it with only his knowledge of number theory and that someday the "real" proof will re-emerge. It seems unlikely but stranger things have happened. So assuming the negation, suppose we have found an integer solution to FLT. Notice that we can easily write [MATH]x^n=z^n-y^n=(z-y)\sum_{i=0}^{n-1}z^{n-1-i}y^i[/MATH] It can be shown that if [MATH]n[/MATH] is even then [MATH]\sum_{i=0}^{n-1}z^{n-1-i}y^i=(z^s+y^s)\sum_{i=0}_{n/s}z^{n/s-i}y^i[/MATH] If [MATH]s[/MATH] is odd we can factor further since [MATH]z^s+y^s=\sum_{i=0}^{s-1}z^{s-1-i}y^i[/MATH] In all there are about four different factorization formulas that (no matter what [MATH]n[/MATH] is) let you peel this thing down to irreducible polynomials in [MATH]z[/MATH] and [MATH]y[/MATH]. Is there anything special about irreducible polynomials in this situtation? One thing it would mean here is that [MATH]x[/MATH] is not prime, other wise every irreducible polynomial factor would be a power of [MATH]x[/MATH] contradicting the fact that they are irreducible. We can follow a similar construction for [MATH]y[/MATH] and [MATH]z[/MATH] showing that they cannot be prime. Now somewhere in the dusty literature is a theorem that says that any allowable solutions to FLT must have at least two of the bases being prime. Have we found a contradiction? Why or why not? I'll get back to this eventually. Suffice to say that the theroem is proven (by Wiles) and now the result can itself be used to prove theorems.
the tree Posted August 25, 2005 Posted August 25, 2005 I could have sworn that it couldn't be proven, and that that was proven.
matt grime Posted August 26, 2005 Posted August 26, 2005 Now somewhere in the dusty literature is a theorem that says that any allowable solutions to FLT must have at least two of the bases being prime. Have we found a contradiction? Why or why not? perhaps if you explinaed what you meant by bases then poeple may hve more idea of what you're talking aobut. i can't make head nor tail of it, presumably you eman x,y, or z, but that doesn't make sense since when n=2 the solutions are p^2-q^2, 2pq, q^2+p^2. even if we allow primitive roots (ie when p and q are coprime) you have just said that there is an easy way to generate infinitely many primes. or are you referring only to n>2?
ElijahJones Posted August 30, 2005 Author Posted August 30, 2005 Now somewhere in the dusty literature is a theorem that says that any allowable solutions to FLT must have at least two of x,y and z being coprime. Have we found a contradiction? Why or why not? I'll get back to this eventually. Suffice to say that the theroem is proven (by Wiles) and now the result can itself be used to prove theorems. There changes have been made. Well Fermat's Last Theorem stipulates that n>2. Yes you can parametrize the n=2 solution thats pretty nifty, its pythagorean triples. The point is that x cannot be prime under the construction below (that is really the question that needs to be proven or not I think and by me too, I am passing it off on anyone). If it is true how does that imply I can generate infinitely many primes from an elementary function? Thats already been shown impossible as far as I know. This thread is just for throwing around some ideas, its not intended to be the return of Frenchy's proof. So yeah that is right for course being able to show that neither x nor y is prime means they are composite and coprime (we say relatively prime over here in the States). What I think is interesting is that this maps {x^n}->Z[y,z]. {x^n|n is a non negative integer} is closed under multiplication and contains 1, it does not have inverses. I have not had time yet but I am hoping that this map will give some further insight into the problem. But you should be able to say something about that Matt if your research is in algebraic topology.
matt grime Posted August 30, 2005 Posted August 30, 2005 Your knowledge seems a little off. there are many ways of generating primes from functions, see wolfram for prime generators. there is one in 26 or 27 variables that whose positive values are primes, of course we don't know when the function will take positive values based upon the inputs. And there is a function f(n)=p_n that gives the n'th prime, though there is no simple way of working out what f(n) is for any given n that doesn't stop it being a function from N to N. I'm glad you made the coprime amendment, but no, allowable solutions do not need to have coprime x,y,z (they are all three coprime, if d fivides any two of x,y,z then it obviously divides the third) and the solutions you mention are the primitive solutions, eg in the n=2 case 3,4,5 is primitive, but 6,8,10 is also a solution, albeit not primiitive. Z[y,z] as in the polynomial ring in two variables over the integers? in what way do we have a map that sends x^n to Z[y,z]? This has nothing to do with topology. Sorry, but the what is closed under multiplication and contains 1? the polynimial ring? Yes, and? it is also commutative and has a Spec, but...? And who the bugger is Frenchy?
ElijahJones Posted September 1, 2005 Author Posted September 1, 2005 Prenote: There is not way to take a function of the natural numbers and produce only primes. I am not playing semantics here. Hell I can prudce a function that will give lots of primes but not consecutively. In fact Dedekind showed that ax+b cannot be used to generate only primes for any a and b. you know who Dedekind was right? And before you get off and running on your debunking campaign you should know that I have a degree in mathematics. What? Who? By the way at least two of x and y being coprime is a way of getting primitive solutions. Saying that only n=3,4,5 are primitive is pointless, none of the solutions for those cases are general solutions to the problem, they are in effect a waste of time. Are you able to reproduce Wiles argument? I probably could but I can't find a full exposition of it anywhere. So as something to do that involves elementary number theory I started this thread. Furthermore equality of two expressions containing variables is a map. But I am sure that dogmatically I should only see an equation. But if in fact it has any solutions at all it can be thought of as a map between sets. Also wikipedia is a public resource edited by the public and especially on technical topics it is often wrong. Or do you think that they would breach national security on cryptography (as an example) to allow someone to post the latest breakthroughs? I can't help it if you cannot follow the line of argument, but I do have a degree from an accredited university (BS Math/Physics 2000 UWEC) and am about to finish my masters in Environmental Science and Policy with emphasis on resource management and mathematical modeling. My thesis is on mathematical modeling of ecological succession. After that I plan to apply to a Phd program in applied mathematics to study stochastic dynamical systems with applications to environmental modeling. Before I accept from you the old mathematical who? how? what? line of arrogance maybe you should list some credentials. I looked on your website and found none. Please tell me you are not an overread sophomore geek who thinks wikipedia is an encyclopedia of mathematics?
matt grime Posted September 1, 2005 Posted September 1, 2005 Erm, what are you on about? If you have a degree in mathematics then you'll see that the function f: N to N given by f(n)=p_n the n'th prime is a function that gives the n'th prime, of course we have no known algorithm for computing the n'th prime but that is a different matter entirely. You won't be able to reproduce Wiles's Argument, or at least I'm confident you won't understand it; I know I wouldn't if I read it. It shows that all semi-stable ellitpic curves are moduilar, and i seem to recall something about galois representations being required. In anycase that is not important. "Saying that only n=3,4,5 are primitive is pointless, none of the solutions for those cases are general solutions to the problem, they are in effect a waste of time. Are you able to reproduce Wiles argument? I probably could but I can't find a full exposition of it anywhere." RIght, i didn't say n=3,4,5, I said in the n=2 case 3,4,5 is primitive and 6,8,10 isn't as solutions ot x^2+y^2=z^2, nor did i say that anything about theer being no other solutions. The stuff aobut x^n mapping to z[x,y] doesn't make sense since even now since you haven't stated the domain of the map. HOw do I send x^n, (an element of Z[x]?) to something in Z[y,z]? or is it only a map fram n'th powers of elements of N to something? It is not a well defined function! And the expression x^n=z^n-y^n does not define a function to Z[y,z] (function from where?) Better, perhaps is to look at the images of the two maps x--->x^n from N to Z, and (y,z)-->z^n-y^n from NxN to Z and look at their intersection as a subset of Z, wh ich is empty when n>2
ElijahJones Posted September 6, 2005 Author Posted September 6, 2005 Hopefully we can get past the semantics a little bit now that we know each other better. Yes, f:N->N where f(n)=p_n is a function but it doe snot produce prime numbers. In fact is'nt it true to say that it might be useful as a logical construct but in real life because we cannot derive and algortihm for its computation (like we can with say g(x)=x^2) it is of no practical use at all? Here is my idea on that x^n=z^n-y^n. We ar enot talking function we are talking map. On the left see a representative of the set {x^n|n>2, x an integer} on the right see a representative of the set {z^n-y^n|n>2, y, z integers} An assumption of a solution to Fermat says that for at least one pair of numbers in those sets the equation holds. If we can now show one solution implies infinitely many perhaps we can deduce reductio ad absurdum or some other contradiction. The point of the factorization I did on the left was to suggest that we could look at the monoids(?) formed by adding multiplication to the sets I listed above. It's a dead end but I don't think the reasoning was as ridiculuous as you made it out to be. By the way the issue of sequences and factorization is very interesting to me that's why I posted that factorization thread. You of all people should find something worth commenting on there.
matt grime Posted September 6, 2005 Posted September 6, 2005 Hopefully we can get past the semantics a little bit now that we know each other better. Yes, f:N->N where f(n)=p_n is a function but it doe snot produce prime numbers. In fact is'nt it true to say that it might be useful as a logical construct but in real life because we cannot derive and algortihm for its computation (like we can with say g(x)=x^2) it is of no practical use at all? right, then what is pi^2 to 10,000 decimal places? Plus there is an algorithm for working out the n'th prime, it just takes a long time. Existence is different from utility, a distinction you are not making. Here is my idea on that x^n=z^n-y^n. We ar enot talking function we are talking map. On the left see a representative of the set {x^n|n>2, x an integer} on the right see a representative of the set {z^n-y^n|n>2, y, z integers} then why not simply talk about the sets and their intesection? you are completely abusing that equals sign. functions and maps are the same thing. An assumption of a solution to Fermat says that for at least one pair of numbers in those sets the equation holds. If we can now show one solution implies infinitely many perhaps we can deduce reductio ad absurdum or some other contradiction. The point of the factorization I did on the left was to suggest that we could look at the monoids(?) formed by adding multiplication to the sets I listed above. It's a dead end but I don't think the reasoning was as ridiculuous as you made it out to be. it wasn't your reasoning, since it is impossible for me to deduce what you reasoning was. it was the fact that you wrote something that could not be made to make any sense with standard meanings, and I cannot second guess what your thoughts might be. a simple reductio ad absurdum results was fermat's proof that there are no solutions to x^4+y^4=z^4 for positive integers. actually it proves that you can't write it as x^4+y^2=z^4 I think, which is stronger. google for it, there should be an explanation somewhere. I also have to different proofs of this argument (different in only minor ways) that i could write out but that's a hassle.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now