# Fermat's Last Theorem Proof in WSEAS

#### jcfdillon

In fact:

(z_xya)^a = x^a + y^a

#### jcfdillon

Also:

(z_xya)^a = (z_xyn)^2 = x^a + y^a

the value "a" is a primordial component of n. so as n ranges from near zero to 3, for example, with x, y held constant, the process "produces" the value "a." This is no different from saying that in the range from 0 to 100, there is a number 27 resulting from going from 0 to 100 without skipping any integers, except "a" would be termed a "real" number, not an integer.

#### jcfdillon

a < 2 < n

z_xyn < z_xy2 < z_xya

(with x, y held constant according to stipulations in the proof method)

then,

for example let x = 3, y = 4, and if we have n very large, we have z very close to y = 4, we have a huge number for z_xya.

These are just suggestions on how to test the method.

#### jcfdillon

It's hard to think of the diagram size increasing or decreasing with x = 3, y = 4 as constants, or x, y being any particular constants one may choose, however, this is the method used in the proof. The diagram size varies by changing the exponent value, but all x^n, y^n, z^n values, for all true FLT form equations, can be placed into an averaged Pythagorean-styled diagram of this sort. Then as the endpoint of z varies along the 45 degree axis of the system, we have a series of nested geometrically equivalent Pythagorean-style diagrams featuring the constants x, y, and these diagrams then create geometric conflict when n > 2. This geometric dichotomy or contradiction does not exist in the FLT form equations featuring exponents 1 or 2, because in those diagrams, the z value endpoint is not internal to any diagrams based on higher exponents.

#### jcfdillon

Ok suppose x, y are not integers, but are real numbers or rational numbers but not integers. But make them constants. If they are held constant, and the proof method is used as described above, the same contradiction will arise. The Fermat problem requires that we prove this for integers, but more generally, we can prove it for any number, simply noting that when x, y are held constant positive real or rational values not necessarily integers, the same dichotomy will arise; i.e., when n > 2 in the proof method, then we find a contradiction causing the dependent value z to become chaotic, and not being able to provide a solution to the given equation when the exponent is greater than 2 (even fractionally). The existence of calculated "real" solutions to FLT when n > 3 and with constant x, y positive reals, producing some real z value, does not mean that these numbers can escape the dichotomy; when these calculated values are reiterated in the given equation, they diverge and exhibit chaotic behavior. ln my opinion this is due to the fact that the geometric model is correctly analogous to the chaotic system of the given equation of Fermat's Last Theorem. By the geometric model then we can correctly show the contradiction inherent to the given equation, and this constitutes proof by contradiction. (As found by the simultaneous required solutions to the given and derived equations given above.)

#### zefram_c

I'm not a mathematician so feel free to ignore me, but did you just claim that for any integer n>2 and (real) numbers x,y there is no solution to x^n + y^n = z^n ?

lol

That's what Fermat's Last Theroem states. It's already accepted to be true. This person just has an aternate proof.

Although I believe x, y and z must also be integers.

#### Hurkyl

Staff Emeritus
Gold Member
Yes, the requirement that the solutions be integers is rather crucial to the statement of FLT -- when the four parameters are allowed to range over real numbers, or even just one of them, it's a simple exercise to find nontrivial solutions to the equation $x^n + y^n = z^n$.

e.g.

$$1^3 + (7^{1/3})^3 = 2^3$$

and you can use the intermediate value theorem to prove

$$4^n + 5^n = 6^n$$

has a solution with n real. (Because $6^2 - 5^2 - 4^2 < 0$ and $6^3 - 5^3 - 4^3 > 0$)

So if jcfdillon's argument does indeed conclude that the equation has no solutions over the reals (and it seems it must, because he nowhere makes use of the requirement that x, y, and z be integers, and he explicitly permits n to vary over the reals), then there is either a fatal flaw with his argument, or with the mathematical field of arithmetic and geometry.

#### CrankFan

Although I believe x, y and z must also be integers.
Yea that's just a minor detail...

#### jcsd

Gold Member
lol

That's what Fermat's Last Theroem states. It's already accepted to be true. This person just has an aternate proof.

Although I believe x, y and z must also be integers.
An alternative proof, especially a comparitvely simple one would however be something of note, most mathemaricians believe there is no 'simple' proof of FLT.

Of course as jcfdillion has claimed that his proof proves it for the more general case of x,y,z are real numbers, staright away his proof must be incorrect as we already know that it does not hold for such a genral case.

#### Hurkyl

Staff Emeritus
Gold Member
To be fair, that doesn't necessarily imply his proof is incorrect -- the other possibility is that arithmetic/geometry are logically inconsistent.

Either way, it cannot be counted as a proof of FLT.

#### cogito²

matt grime said:
Probably because he cannot for legal reasons redistribute now the journal has it.
Not that anyone here of sufficiently high standing in the math community is going to bother looking at it. A perhaps sad state of affairs but as it's a proof by picture by the sound of it I doubt it is at all rigorous, and probably badly written by maths standards - we have a notoriously short attention span for such things - after all if it's so easy why is the proof so disguised sort of attitude. See sci.math responses to james harris
I figured this much. I only said it in an attempt to get him to detail the proof a little more (and apparently it worked).

#### jcfdillon

I'm not sure I interpret everything correctly in my own method. But yes if x, y are held constant, the method should work because there is divergence found immediately in the "z" value when exponent n is allowed to vary over the positive reals. The use of positive real range for the exponent, in my approach, was just a way of exploring how z varies as a dependent variable, with x, y both constants as any given positive integers, or perhaps as any given positive real values, and with exponent n allowed to vary as an independent variable. I found that the z value varies inversely to n, but also found a simple method of illustrating these relationships geometrically in 2D, using a Pythagorean-styled diagram. Therefore, if the diagram is drawn and interpreted as stipulated, then it may represent a simple proof method in 2D geometry. This is not a small claim of course... (cont'd)

#### jcfdillon

So I understand that most mathematicians will obey statistical probability and ignore my work, and also will have extreme hostility toward it. In the above critique one person wrote an expression using the number 1 as the value for x, which I had thought was usually thought to be a "trivial" solution, whereas most people working on FLT concern themselves with "nontrivial" solutions and would not even consider the use of x = 1 in the given equation of FLT.

I had thought that most likely, the testing of real values as constants for (nontrivial) x, y will lead to the same contradiction found in my paper. However, I may be overreaching.. My paper did set x, y as constant integer values, as the first step in the stipulations of my proof method. I am not sure I can properly make all the claims I have mentioned on this website and my interpretations may be flawed. So if that is the case I apologize.

I was hoping some people might want to take my approach and run with it, test it out, and see if they can validate it. However, if people pop in, say here's the fallacy and use "trivial" solution values in combination with attacking my own overstatements, there's no way to make any progress.

The published paper is available but I am sorry it is not readily accessible to everyone. I think it might draw some critiques from actual readers of the paper but so far no one has presented me with any actual written critiques in any organized fashion.

If people here want to test it out on a blackboard and see where the method leads please feel free. I think if anyone draws these diagrams as I have described and then tests the behavior of the given and derived equations, the same divergence and dichotomy (contradiction) can be readily seen.

The general approach can be seen if you look at http://www.geocities.com/jcfdillon/crx.doc Then imagine that as diagrams increase in size with constants x,y and increasing real exponent n, for all resulting true FLT-form equations, ...(cont'd)

#### matt grime

Homework Helper
There are seemingly non-trivial solutions in real x,y,z that your paper says can't exist, and in fact x=1 is not a trivial solution in any sense.

Your method has been shown to be wrong since it apparently implies, for instance that there is no cube root of 16 in the real numbers, which is obviously nonsense.

So not only is there a stastical knee-jerk reaction to your work, but said reaction seemingly turns out to be justified.

You may claim that x and y are integers (and z) but if you don't actually use any property of them being integer then that supposition is unnecessary.

#### jcfdillon

.. with averaged diagrams as shown, and with these diagrams overlapped with geometric equivalence (as with transparencies), using common origin 0, you can show that there will be two equation/curves requiring simultaneous solution:

z = (x^n + y^n)^(1/n)

z = (x^a + y^a)^(1/2)

such that a < 2 < n.

These curves can then be graphed for any specific given x, y constant values, and with n and z varying inversely to each other.

In the second equation, we find the "constant" 2 appearing in the exponent 1/2; also we have the exponent a, which is some specific positive real value representing the exponent in the system when we consider z^2 for higher exponent n. In other words when the exponent is allowed to increase beyond 2 in FLT, we find by geometric construction and interpretation that the lesser exponent a appears in the system as a result of the exponent ranging smoothly from 0 to 2 to higher exponent n. In the geometric construction with constants x, y we then have a situation in which the dependent variable z is forced to (but cannot) provide simultaneous solution to the given and derived equations.

#### jcfdillon

One interesting fact that I discovered, and which led me to think along the lines of using averages to complete the proof, is that there cannot be any solutions for the Fermat equation when the exponent is an integer 2 or greater, and with x = y.

If x = y in the given equation of FLT, then

2(x^n) = z^n

[2^(1/n)]x = z

2^(1/n) = z/x

But for any solution to FLT with all integers, z/x must be rational, whereas, is it not true that for any n > 1, the value

2^(1/n)

is irrational?

#### jcfdillon

I certainly didnt discover this fact first, but discovered it for myself, i should say..!

#### shmoe

Homework Helper
jcfdillon said:
So I understand that most mathematicians will obey statistical probability and ignore my work, and also will have extreme hostility toward it.
I've read parts of your multiple posts. When I see things like "How can z exist as both square root of z^2 and nth root of z^n? Answer: It cannot." I tend to stop reading, and I suspect you won't find many mathematicians who will take you seriously with statements like this.

#### matt grime

Homework Helper
Your discorver two posts back that you don't claim is original? Well, some of us might have pointed out that it was flippin' obvious otherwise sqrt(2) is rational (integer even). If you didn't sport even that, then as shmoe says, we ain't going to bother taking you seriously.

#### jcfdillon

matt grime said:
There are seemingly non-trivial solutions in real x,y,z that your paper says can't exist, and in fact x=1 is not a trivial solution in any sense.

I think solutions using 1 in certain cases of any problem considering "infinite" or open-ended ranges, will be considered trivial solutions. The following definition of "trivial" is quoted from MathWorld:

Trivial

Related to or being the mathematically most simple case. More generally, the word "trivial" is used to describe any result which requires little or no effort to derive or prove. The word originates from the Latin trivium, which was the lower division of the seven liberal arts in medieval universities (cf. quadrivium).

According to the Nobel Prize-winning physicist Richard Feynman (Feynman 1997), mathematicians designate any theorem as "trivial" once a proof has been obtained--no matter how difficult the theorem was to prove in the first place. There are therefore exactly two types of true mathematical propositions: trivial ones, and those which have not yet been proven.

**********
Your method has been shown to be wrong since it apparently implies, for instance that there is no cube root of 16 in the real numbers, which is obviously nonsense.

I did not say there is no cube root of 16 in the real numbers. But why not test specific real numbers using my method, such as x = SQRT(2) and y = pi, (both held as constants, estimated to some specific degree of accuracy) and allow the exponent n to range through the positive reals in the given equation of FLT. Then z must vary inversely, decreasing as the exponent n increases, and the same dichotomy arises as detailed above.

******************

So not only is there a stastical knee-jerk reaction to your work, but said reaction seemingly turns out to be justified.

***********

I am just saying I dont blame people for not wanting to learn about my simple paper, when they know that so many FLT proof attempts have failed, and the one accepted proof is fiendishly complex. So my paper is akin to blasphemy.

***********

You may claim that x and y are integers (and z) but if you don't actually use any property of them being integer then that supposition is unnecessary.

If a dichotomous or divergent phenomenon is proven and shown for the Fermat equation, using constants x, y, then indeed it does not matter whether x, y are integers or positive real numbers of any kind, as long as x, y are held constant.

The proof's requirement for integer values to be considered and used, is simply the way the problem is stated and how it was historically defined by Fermat. However if the same dichotomy causing failure of solutions in integers for higher exponents, can be shown to exist using x, y as positive reals, then there is no need to specify or require that x, y, n, z must be integers simultaneously, in the completed proof. (Prove first for integers, then test the same method for x, y as some constant reals). This makes my proof unusual because via complete 2D geometric generalization we find the same dichotomous or chaotic behavior in the system using x, y as constants whether integers or not (but they must be held constant, by my proof method).

If some specific z value is found, solving FLT for some given positive reals x, y, then we still have the problem associated with the derived exponent a which is associated with an inner-nested diagram. Then one might say that the exponent a is both independent and dependent in the system; because exponent "a" is a primordial component of the exponent range from 0 to 2 to n, and a < 2 < n, and in the geometric construction, this primordial exponent a causes chaotic behavior in the system because z cannot simultaneously satisfy the given and derived equations. (The derived equation is from the geometric construction, which is generalized -- fully generalized -- and thus accurately represents all true FLT-form equations.)

#### jcfdillon

shmoe said:
I've read parts of your multiple posts. When I see things like "How can z exist as both square root of z^2 and nth root of z^n? Answer: It cannot." I tend to stop reading, and I suspect you won't find many mathematicians who will take you seriously with statements like this.
This is hilarious. You act as though I dont know that a number can be square root of its square and also nth root of its nth power. The problem is not with individual terms like z^n, but with its existence in the given equation of FLT, which is more complex.

The given equation of FLT is:

x^n + y^n = z^n

so

z = (x^n + y^n)^(1/n)

Now z is the nth root of z^n.

z is also square root of z^2.

The problem arises in the fact that, in my proof method, z and n are inversely related, due to my stipulation of holding x, y constants.

This is the background of what I said. If you want to ignore stipulations in a proof method, you are not reading fairly at all.

#### matt grime

Homework Helper
But there are infinitely (uncountably) many positive real solutions to the equation

x^n+y^n=z^n

for all n in N.

So your proof which allegedly shows these don't exist, is wrong. That is the reason why no one appears to be taking it seriously your proof can be used to show that something is false when it is actually true.

I don't understand what on earth your getting at with claiming a counter example with x=1 is trivial, and therefore not something to be considered. It is a counter example to a claim you made. It is *trivial* in the sense that it easy to show, however x=1 was not picked for any special reason other than it was a simple small example. (n=1 is trivial in the sense of FLT.)

#### jcfdillon

matt grime said:
Your discorver two posts back that you don't claim is original? Well, some of us might have pointed out that it was flippin' obvious otherwise sqrt(2) is rational (integer even). If you didn't sport even that, then as shmoe says, we ain't going to bother taking you seriously.
I dont have a math degree, it was interesting to me that this relationship was there when x = y--, and it seemed relevant to my proof method when I noticed it in '96-- that was prior to going back to a diagram found in 1980 and simply averaging the thing! A simple step, but it had apparently never been done before, and best of all it led to a fully geometrically generalized approach for illustrating, diagramming and analyzing FLT.

How the heck did my paper get listed #1 in a journal headquartered in Greece. I still find it hard to believe, but the strength of the method is such that it can do all this even with a bunch of typos and inexpert preparation of diagrams and graphs on my part.

#### matt grime

Homework Helper
jcfdillon said:
I dont have a math degree,

No, I don't think that comes as a surprise.

Do you agree that your proof states that there are no real solutions to

z^3=2^3+2^3?

I just want to get this straight, because that's the conclusion some people have drawn.