Sunday, August 13, 2017

RECENT DISCUSSION OF THE 1965 PROOF OF FERMAT'S LAST THEOREM

I think that this recent exchange with a friend who has been looking at FLT65 for some time may be of interest to  those who are following the subject of Fermat's Last theorem, which as they may know, is importent to the explanation of quantum phenomena in TDVP. This conversation may be helpful for anyone interested in understanding the logic of the 1965 proof. For this reason, I am posting my latest response here without identification of my friend or his math professor friend.

MY RESPONSE OF AUGUST 12, 2017

I spent some time thinking about Prop. P between meetings and events in LA last week. There really should be no confusion about when P is true and when it is false. I will attempt to clarify this while answering your latest comments, because they do reflect what I see as confusion about what FLT65 says and does. I shall also attempt to show you how distinguishing when Prop. P is true and when it is false, and when the identity sign is appropriate, leads to a better understanding of FLT65. 

From your email earlier today:

      Q: Do you believe that Prop P is ever false?

Let’s consider this. Here is Prop P as you stated it:

     P: Equating a polynomial to a constant, for the purpose of finding a  
        specific solution, automatically turns the polynomial itself into a  
        constant.

This should not be a matter of “belief”, because it is easy to show that P can be either true or false, depending upon the nature of the polynomial, and the circumstances. And, it is an added benefit that the circumstances also allow us to clarify the proper use of the identity symbol ≡.

P is true, if and only if, all variables are specified as constants. For example, we know that, if Y in the FLT equation is an integer, then for any Y there is some integer factor A such that z – a = A. If we should also know that Z and Y are the specific integers Z1 and Y1, then a is determined, and we have: Z1 – a1 A, an identity.

But in FLT65, z is an unknown. Clearly, integers can be assumed for X and Y, and if so, then whether or not z can be an integer is as yet, in the proof, unknown. To assume that it is an integer at the beginning, is premature and leads to circular reasoning. For the polynomial z – a in FLT65, z is a real number, but its specific value is unknown, so when we set the polynomial z –a = A, a known integer factor of Y, P is false, because z and a can take on an infinite number of values. The only requirement imposed by setting z – a = A is that their difference is always A. The value of z can vary, and thus this equation is not an identity.

 I once sent you the judgement of a mathematician who had spent his entire working life as a mathematics professor in a university. You dismissed his judgement that FLT65 was invalid as “something you had seen before.” 

To be clear, I had seen this line of reasoning several times before, and have given it all the serious thought it deserves, so I saw it as the same knee-jerk reaction I’ve seen numerous times, and refuted every time. I felt that you should have recognized the circularity of the argument. But I apologize for my abrupt manner. I should not have been so abrupt, and I certainly should not have been condescending. I’ll copy the math professor’s comment here and try to respond more appropriately.  He said:

“Here is a possible way of putting it that might convince Mr. Close.  In his argument he sets  a = Z - A   and considers the divisor polynomial  g(Z) = Z - a, which he says is a polynomial of degree  1  in Z.  But  g(Z) = Z - a = Z - (Z - A) = A,  which is not of degree  1  but of degree 0.  When the divisor A  is a (nonzero) constant, the polynomial division algorithm over the reals just says there exists a (unique) polynomial  q(Z)  such that  f(Z) = Aq(Z) + 0, where, of course, q(Z) = f(Z)/A.  so there is no  f(a).”

This argument completely misses the point of FLT65 by assuming the definition a = Z – A which is not the case in FLT65. His statement that I set a = Z – A is false. You will not find this anywhere in FLT65. Z is the unknown, the dependent variable. Z – a is defined as a polynomial of the 1st degree, meaning that the value of Z depends on the value of A, an integer factor of Y and the value of a, which is an integer if there is an integer solution of the FLT equation, which is as yet, in the FLT65 chain of logic, unknown. By assuming that Z is an integer by definition, the conclusion that Z –a is a constant, and therefore of 0 degree, is of course, circular reasoning, as mentioned above, and stated in my more abrupt response. By the way, I meant no disrespect to your professor friend by dismissing his comments. A university professor known as a number theory expert made the same mistake, but he quickly acknowledged that it was circular reasoning when I pointed it out to him.

Thank you for your statement of what you see as a disproof of FLT65. It enables me to better understand why you kept coming up with propositions that had nothing to do with FLT65. I think other critics may have had this same misconception about how FLT65 goes about proving FLT.

Your statement has the logic of FLT65 completely backward:

 “The equation of constants, f(Z1) = Ap = (Z1 - a)p, does not imply that the variable (Z - a) is a variable factor of the variable f(Z). 

I agree! However, what FLT65 says is the converse: the fact that f(z) cannot contain z – a as polynomial factor for real number values of the variables, implies that, if there were an integer solution for the Fermat equation, then there would be an integer version of f(z)/(z – a) = q(z) + f(a) where f(a) would equal zero, and that would violate the ‘if and only if’ condition of the division algorithm for polynomials.

It appears to me that the confusion comes from considering Z to be a specific integer before it is known whether z can be an integer or not. Maybe this will be clearer if we go step-by-step:

Dividing f(z), a polynomial of degree p-1 by z - a, a 1st degree polynomial, we have unique polynomials q(z) and f(a) of degree less than p such that:

(zp-1 + zp-2x + zp-3x2 +•••+ xp-1)/(z-a) = q(z) + f(a)/(z-a). Multiplying through by z-a, we have:
f(z) = (zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = q(z)(z-a) + f(a).

From this we see that the polynomial f(z) is factorable into two polynomial factors, q(z) and z-a, if and only if f(a) = 0. But f(a) = ap-1 + ap-2x + ap-3x2 +•••+ xp-1, which cannot equal zero because a and x are positive for any integer solution of the FLT equation.

Therefore, the FLT equation factor polynomial f(z) cannot be factored into two polynomials of degree less than p, one of which is z - a.

If there is an integer solution, then with the term-by-term substitution of the integer variables into the variable polynomial f(z), we obtain the variable polynomial f(Z) for any integer solution of the Fermat equation. But, for any integer solution, f(Z) = (Z-a)p, where Z-a = A, a single integer factor of Yp. So now we have:
The hypothetical integer polynomial f(Z) = (Zp-1 + Zp-2X + Zp-3X2 +•••+ Zp-1) = q(Z)(Z-a) + f(a) = (Z-a)p.

By inspection of this integer polynomial equation we see that f(a) contains Z - a as a factor, and although we don’t know what the specific values of a and X are for an integer solution, we know that they are positive constants. So f(a) = M(Z-a) where M is an integer constant, and we have f(Z) ) = (Z-a)p, = (Zp-1 + Zp-2X + Zp-3X2 +•••+ Zp-1) = q(Z)(Z-a) + M(Z-a), from which we have: f(Z) = [q(Z) + M](Z-a).

And q(Z) + M is a variable polynomial in Z of degree less than p, call it q1(Z). Thus for hypothetical integer solutions of the FLT equation, we have the variable polynomial f(Z) = q1(Z)(Z-a). But this is a violation of corollary III of the division algorithm, which tells us that the variable polynomial f(Z) cannot be divided into the factors Z - a and another polynomial of degree less than p. The only way we can avoid this contradiction is for z to be an irrational real, not an integer. 

I have demonstrated in at least three different ways, including Fermat’s favorite method of proof, infinite descent, that if we ignore this contradiction, and assume that two of the three variables x, y and z are integers, and solve the FLT equation for the third variable, then that third variable cannot be an integer.

I don’t consider adding such demonstrations to FLT65 to be necessary because the contradiction f(a) ≠ 0 versus f(a) = R = 0 is sufficient to prove FLT by itself. This contradiction is valid because it is obtained by applying the division algorithm and corollaries to variable polynomials, not constants, and a single contradiction is sufficient to prove there can be no integer solutions for zp – xp = yp.


I believe the whole confusion for most critics arises from assuming that the division algorithm and corollaries are inappropriately applied to constants. They think this because they jump to the conclusion that all three variables must be treated as integer constants from the beginning of the proof. Thanks to you, this confusion has been made clear with the analysis of Prop P! 

Once you see that P can be true or false, depending upon the nature of the polynomial, and that in FLT65, f(Z) and Z – a are still polynomials of the variable Z, even though for a hypothetical integer solution, they are equal to the constant integer factors of Yp, the logic of FLT65 becomes clear, and the search for counter examples and counter propositions becomes unnecessary and irrelevant.

Edward R. Close  August 12, 2017

3 comments:

  1. hmm i see a methodological problem here, let's assume i (if i would look more into detail into the EC 'proof) that i would find a mistake, and thus refute the 'proof', then there are these scenario's
    1) you EC would not accept my refutation; thus you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect.
    2) you (EC) see the problem, try to fix it, come up with some modification; refuting that again would take time, and in the meantime, you you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect.
    3) but if after a while i would again point out the mistake in your modification, then there are a few scenario's
    a) you EC would not accept my refutation; thus you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect. BACK TO 1) above
    b) 2) you (EC) see the problem, try to fix it, come up with some modification; refuting that again would take time, and in the meantime, you you -and later maybe some who support you- can endlessly claim you have found the 'best' (shorter) proof, even if it would be incorrect.
    c) but if after a while i would again point out the mistake in your modification, then there are a few scenario's: BACK TO 3) above (yes circular and endless spirals of thought)
    CONCLUSION: this could lead to endless discussion. Thus it's safer for a spectator like me to look at what the math community at large is saying, namely, accepting the proof by Andrew Wiles, and dismissing all other 'proofs' as crank stuff. Meanwhile ofcourse it might be interesting for some odd math researchers, or math/sociology researchers to do a broad study on all (hundreds, or thousands) of crackpot 'proofs' but unless this leads to a clear result, another proof, accepted by the math community at large, i'm not going to waste my time on some other unconventional 'proofs', whether they may be true (very small chance) or not.
    PS Now you EC can ofcourse delete this comment again, but i've saved it this time and if you delete it might popup some time elsewhere; lets say eg . on a certain ning forum, or something like that (youknowwhattamean)

    ReplyDelete
    Replies
    1. I don't think I have ever deleted one of your comments, unless, of course it contained crude language, which you seem sometimes to like to use. But your analysis is interesting. I agree that if you start by looking for an error that you believe might lie behind methodological problems, rather than trying to understand the logic of the proof, it would be a waste of time.

      Delete
  2. Dr Ed and Jaqui Close, I have to thank you for sharing your remedies for various toxic molds. My house is clean and my indoor garden harvest had only one spot of mold within the fattest flower top that was leaning into a wall! That is success on any day where I am from . Off subject perhaps, maybe not, but I feel good about the non-caustic, non-threatening solution you re-discovered and shared! It works for me and my indoor garden of flowers. sorry if post is inappropriate, you know what I mean. gwz

    ReplyDelete