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NAME Silvie Coffelt
E-MAIL scoffelt11@hughes.net

COMMENTS

 It seems so logical that two out of three dimensions cannot possibly fit into something three dimensional.
DATE Wednesday, January 18th 2012 - 10:22:18 AM

NAME dan James
E-MAIL danielb.james100@gmail.com
URL http://www.bebopdajig.com

COMMENTS

 interesting. i struggle with maths but it's such an interesting story!
DATE Friday, January 13th 2012 - 06:38:24 AM

NAME Bill
E-MAIL bfahle@bigfoot.com

COMMENTS

 Your triangle replicates Fibonacci's method for generating pythagorean triples. See http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples
He uses sequences of odd integers, 1, 3, 5, 7, ... You can see that your triangles have those odd integers in order starting from the left, because adding another 2 on top of the prior column will form another odd integer in sequence, if the first one is 1. This insight might help you put your proof into more rigorous mathematical terms.
DATE Wednesday, December 21st 2011 - 02:22:00 PM

NAME Tanner
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DATE Sunday, December 4th 2011 - 09:10:49 AM

NAME Richard Jainchell
E-MAIL rajainchell@yahoo.com

COMMENTS

 I will study your proof. By the way, I have developed new methods for addition, subtraction, multiplication, and division. Are you interested in viewing some examples?
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DATE Saturday, November 26th 2011 - 01:46:48 AM

NAME James Barclay
E-MAIL eccl2922@gmail.com

COMMENTS

 Andy did a good job in spite of himself; remember that he is a pure mathematician which means he thinks in terms of complexities instead of simplicity. We all knew his proof could not have been Fermat's and I even went so far as to simplify matters to the shape of the lifts used in hoisting stones in Fermat's day when building was huge; after all, they were basically right triangles. Bravo! You saw it in Diophantine analysis. Showed us all the value of reading good biographies as a part of essential research. Well, I'm humbled.
DATE Wednesday, November 23rd 2011 - 12:01:23 PM

NAME B.S.Prabhakara
E-MAIL prabhasubban@yahoo.co.in
DATE Wednesday, November 9th 2011 - 06:00:00 AM

NAME RamanaJohn
E-MAIL john.minkowski@gmail.com

COMMENTS

 Very nice exposition of P. triplets.
I would suggest adding the following before the introduction of the triangle:
(n+1)^2 = n^2 + (2n+1) for a tad more clarity.
The Fermat part is beyond my abilities as yet.
Have you thought about a^n + b^n = c^(n+j), all integers e.g: 2,11,5?
DATE Sunday, November 6th 2011 - 10:37:28 AM

NAME Mirkhabibov Asiljan
E-MAIL amirhabibov@unicon.uz
URL http://www.unicon.uz

COMMENTS

 I modified (extended) Fermat's Last theorem what was proved by Tom Ballard and sent an article (short article) to Journal of AMS, but the article was rejected. It is interestingly, how can I publish this extension?
DATE Monday, October 17th 2011 - 10:30:43 PM

NAME Stefano
E-MAIL robotec@netsurf.it
URL http://www.maruelli.com/prime_study.htm

COMMENTS

 Sorry what bellow has several refuse/too short conclusion that we provvide to correct onto my blog page.
Thanks for the patience.

Stefano
DATE Monday, October 17th 2011 - 09:24:57 PM

NAME stefano
E-MAIL robotec@netsurf.it
URL http://www.maruelli.com/prime_study.htm

COMMENTS

 FERMAT THE LAST PROOF using the
(pls before trash read twice...)

PART 1) THE "COMPLICATE MODULUS" (DEFINITION)

- What is "complicate modulus" algebra:

- Is a clock with 2 hands: one is the numbers of turn ( for example called with "M" ) the other is the rest "R".

Also on this special "counters" the numbers of hours onto the clock, that is the modulus "M(n)", change (by a known function) each turn.

[of course we can also immagine a more complice, or complex, clock based on the same model...]

- A simple example of how powerful is the "complicate modulus" algebra:

EXAMPLE: CONSIDERING POWERS OF INTEGERS:

I understood that we can develop any power of integer as:

C^n = Sum (from P=1 to P=C) of [ P^n - (P-1)^n ]

(same for A^n and B^n changing just the limit of P)

We call M(n)=[P^n -(P-1)^n] "complicate modulus" since:

| C^n | modulus [ Pn - ( P-1) n ] = C

So this can be an easy method to extract the "n" root by hand (or just to understand if an integer "N" is a perfect power "n")

How to do: for the square root of 9 (for example)

We cal (2x-1) = Complicate Modulus M2

simply table:

x 2x-1 9
1 1 9-1=8
2 3 8-3=5
3 5 5-5=0

So yes 9 has perfect root 3, rest 0 or

|9|mod(2x-1) = 3M2 + 0

For cubic root of 28 ? Simply use M3= 3x^2-3x+1

x 3x^2-3x+1 28
1 1 28-27=1
2 7 27-7=20
3 19 20-19= 1

Ok 28 = 3M3+1 non perfect root.

How to have this "magic" functions for any "n": simply use

M(n) =P^n-(p-1)^n that comes from Tartaglia:

PART 2) INVOLVING TARTAGLIA'S TRIANGLE

It helps us to understand what is, and the values of our "complicate modulus" operator since:

is enough to keep the Tartaglia's interested (n) line, minus the first value that is ever "1"

Remembering the triangle (calculated for x-1) considering the value with sign :

n - TARTAGLIA - check -

My complicate modulus M(n) operator is:

0 1
1 1-1 = 0
2 1-2+1 = 0 so M2=[X^2-(X-1)^2] = 2x-1
3 1 -3+3-1 = 0 so M3=[X^3-(X-1)^3] = 3x^2-3x+1
4 1-4 6 -4 1 = 0 so M4=[X4-(X-1)4] = 4x^3-6x^2+4x-1

ETC.....

PART 3) HOW TO CALCULATE BY HAND "n" ROOT:

If we wanna know if an integer "m" has an integer (n) root

Exampe: n=2 SQUARE ROOT:

m = 9

We would like to calculate by hands the square root:

So we first wrote Tartaglia's line for (x-1) and n=2:

1 -2 1

we remove the first value

-2 1

we change the signs a re-put the x value (with their correct powers) then we obtain our M2 modulus :

M2= 2x-1

so we can tabulate (and calculate the square root):

x 2x-1 9 rest
1 1 9-1= 8
2 3 8-3= 5
3 5 5-5= 0 so rest = zero 9 is a perfect square.


Or: | 9 | M2 = 3 M2 + 0


Example for n=3 , cubic root by hand:

We have, for example the number "28" and we would like it as an integer bubic root and witch is.

From Tartaglia, line n=3 for (x-1)3

3 1 -3+3-1 we remove the first

-3+3-1 and we change the sign so:

M3 = [X^3-(X-1)^3] = 3x^2-3x+1

And we can calculate:

x (3X^2-3X+1) 28
1 1 28-1 = 27
2 7 27-7 = 20
3 19 20-19 = 1 (rest)

so 28 has no perfect cubic root or:

| 28 | Mod(3x^2-3x+1)= 3M3 + 1

ETC......


PART 4) New M(n) simbols, some examples:

27 MODULUS M3 = 3 M3+ 0 OR:

|27| Mod [3x2- 3x +1] = 3 M3 + 0

or as in the final shorter form:

|30| M3 = 3 M3 + 3

|625| M4 = 5 M5 + 0

|627| M4 = 5 M5 + 2 etc....

We leave this symbol M(n) into the result to clarify what we did.

PART 5) FEATURING:

The "complicate modulus algebra" simplify (or easy solve) the life in Power problems (including Fermat the last...)

SO FERMAT THE LAST PROOF (? HOPE CORRECT)

We have a direct proof of 1 line only, but since I won't to be considered a stupid I explain you what we can do step by step than, once experienced with this kind of complicate modulus, just accept the fact as simple as they are...

PROPERTY OF X^n POWERS:

If X (and n) are integers so X^n mus follow this property:

derivate of [X^n/(derivate of X^n)] = 1/ n

We prove that this is not respected by C^n -B^n , except for some (C,B) for n=2

Now it's easy to be proved using the complicate modular algebra:

C^n-B^n =? A^n

If yes, since the above property of x^n that is (shortly):

[x^n/(x^n)']' = 1/n

Where (') is the derivate, and because:

Putting for simplest example B=C-1 (and finally B=C-K)

we have:

[C^n-(C-1)^n] =? A^n if yes than:

(just long calculation):

(1) {[C^n-(C-1)^n] /[C^n-(C-1)^n]'}' =? 1/n

Result: (you can check it and jump onto the chair too..)
We have that the only possible integers valid solution are for n=2.
In all the other case n>2 and B=C-K, the result is C=0/1/K that are not admitted by the

Fermat's conditions A < B < C and A,B,C, integers.

Reason:

[C^n-(C-1)^n] or

[X^n-(X-1)^n] is my "complicate modulus" that cannot be the same (n>2) of the main function (X^n).

Tartaglia's develope helps us to immediately have value of the modulus for the "n" we are considerning

So we keep Tartaglia's develope for (x-1)^n, without the first element, than we change all the sign.

For example n=2

1 -2 1 so keep -2 1, change signs and put x:

Modulus M2= (2x-1)

f.e. n=3

1 -3 3 -1 so keep -3 3 -1, change signs and put x^ :

Modulus M3= (3x2- 3x +1 )

etc...

And Fermat Proof come from:

We change: C^3-B^3, in the simplest case B=C-1
with our modulus (3x^2-3x+1).

USING (') SIGN AS DERIVATE:

|(3x^2- 3x +1 ) |'
|------------- | =? 1/n
|(3x^2- 3x +1 )'|

That , after boring calculation give as result X = 0 / 1 so X= C = 0 / 1 are trivial non admitted solutions

SO NO solutions for B= C-1

Will be the same with (C-K) since this is one of the characteristic of the "complicate modulus", just gives:

C=X =0/ K solutions

But if C=K then C-K=0 is trival, so NO SOLUTIONS TOO...

Not necessary to go ahead for other powers for reason that we can prove too.

n=4) Same concerning for n=3:

[X^n-(X-1)^n] become for B=C-1 and C=X

4x^3-6x^2+4x-1 =? y^4

if we make the derivate of that:

12x^2 - 12X + 4 = 4 y^3 that dividing by 4 is again:

3x^2-3x+1=y3 that is what we just proove is false...

Again for n=5 for any n>2.

So we (think to have) proved that:

X^n is never equal to its modulus M(n) = [X^n-(X-1)^n], or

|X^n| Mod [X^n-(X-k)^n] <> Integers (for n>2) or

|X^n| M(n) <> Integers (for n>2)

PART 7) FERMAT THE LAST PROOF "the most elegant":

If you understood how "complicate modulus work is now simple to solve Fermat with this short trick (!):

Why cannot be:

n.root of (complicate modulus) = Integer ?

[this is exactly what Fermat said]

Example: n=3 (in simple Case: B=C-1)

Cubic Root ( 3x2- 3x +1) <> integer since (as shown up )
modulus M3 value of C^n-B^n with B=C-1)
x 3x^2- 3x +1 of 3x^2- 3x +1
1 1 1-1=0 (end of calculation !!!)

Example: n=3, In generic Case: B=C-K ; our complicate modulus can be the same of the function: 3x^2- 3xK +K^2

Cubic Root ( 3x^2- 3xK +K^2 ) = Integer ? W

ith my method for the "n" roots:

we know that any possible value of C^3- B^3 = 3C^2- 3CK +K^2 (putting B=C-K) and putting X=C (just more clear)

x 3x^2- 3xK +K^2 of 3x^2- 3xK +K^2
1 1 1-1=0

That is true for any integer K...

So the calculation stops at it's first turn, so there are no other integer solution except the trivial one.

(the same for all powers >2)

REMEMBER THAT THIS PROOF IN NOT JET ACCEPTED, SO, PROBABLY, IS WRONG !

Finally) But why n=2 works ?

It works since the derivate of the complicate modulus is a constant.

Hope not boar you more than necessary!
DATE Wednesday, October 12th 2011 - 10:08:29 AM

NAME Matheus Fernandes
E-MAIL matheus_f88@hotmail.com

COMMENTS

 I was thoroughly amazed by that proof.
It simply departs from simple statements to more complex ones, that ultimately proves what it says at the beginning it's going to prove.
Yes, Fermat's Last Theorem might have not been thought entirely by him, and he might had been teasing the mathematical community, but, I must say, he created a puzzle that was truly a challenge - thus, this proof is sheer brilliancy.
DATE Thursday, September 15th 2011 - 04:48:28 PM

NAME stefano
E-MAIL robotec@netsurf.it
URL http://www.maruelli.com:prime.htm

COMMENTS

 Just little correction on my previous post:

p = prime if z= (p-1)!/p is an NON integer

Note: the above Rufus liks are to pillows... not math !(to the postmaster: delete if possible !)

Ciao
Stefano
DATE Tuesday, September 13th 2011 - 10:48:41 PM

NAME Stefano
E-MAIL robotec@netsurf.it
URL http://www.maruelli.com/prime_study.htm

COMMENTS

 Shortest crazy proof:

Fermat is right since:
Z= n/(n-1) is an integer just for n=2 and because
n! = n just for n = 2 and because
First derivate of X^n is a line just for n=2
(I remember p = prime if z= (p-1)!/p is an integer )

I stop flooding this kindly place.

Ciao
Stefano
DATE Friday, August 26th 2011 - 09:43:30 AM

NAME Stefano
E-MAIL robotec@netsurf.it
URL http://www.maruelli.com/prime_study.htm

COMMENTS

 I was interested by the fact that (just) for n=2

d(x^2)/dx = 2x so putting x= A, B, C into the system

and is possible to confuse the derivate with the tg gonio in some integer number:

d(x^2)|b /dx = Tg(X^2)|b = 2B
d(x^2)|c /dx = Tg(X^2)|c = 2C

And putting into Fermat:

|2C^2-2B^2+2BM-2CM = C^2 - B^2
|
| C^2 - B^2 = A^2

It has solutions. So for n=2, for special B & C, is possible to confuse the derivate with the tangent gonio and obtain Fermat's "A".

2) Obtain Pitagora triplets with square as reference:
Keep any m^2 with m from 0 to infinite.
Keep other 2 numbers: m+1 and m-1
Each ot this 3 numbers will give you a Fermat's A
m-1 is B
m+1 is C

Function that gives A^2 is

A^2= 2C^2-2B^2+2BM-2CM

A is ever an Even if M is a square.

This function will give you ALL even A

To obtain the Odds A you have to keep different value, for example:

B= m-1= 28
m=29
C= m+19 = 48
A= 56 (of course is not soted by value)

Not so likly since there is no reference for m, as for Even .

Hope you'll find interesting too.
DATE Friday, August 26th 2011 - 09:26:12 AM

NAME Leslie
E-MAIL lal4@rice.edu

COMMENTS

 Skepical from a@b.com, good work and thanks! You have a gift for details and I am very grateful that you have spared me future headaches working through them. Also...I can't believe I never noticed that about figure 1(a).
DATE Thursday, August 25th 2011 - 02:15:36 PM

NAME Brian
E-MAIL cayuga@iprimus.com.au
URL http://cayuga-nest-too.blogspot.com/

COMMENTS

 Very interesting
DATE Wednesday, August 17th 2011 - 04:20:37 PM

NAME Corey Stephen Howell
E-MAIL mistercoreyhowell@yahoo.com

COMMENTS

 What an interesting read...
DATE Wednesday, August 17th 2011 - 12:27:41 PM

NAME drew
E-MAIL drewmcu@aol.com

COMMENTS

 I am not a mathematician and need to look at this a few more times but it looks brilliantly simple - many congratulations!
DATE Wednesday, August 17th 2011 - 04:35:15 AM

NAME Stefano
E-MAIL robotec@netsurf.it
URL http://www.maruelli.com/prime_study.htm

COMMENTS

 Nice, can you take a look at more easy solution hypotesy:

http://www.maruelli.com/prime_study.htm

Thanks
Stefano
DATE Thursday, June 30th 2011 - 01:14:07 PM

NAME Skeptical
E-MAIL a@b.com

COMMENTS

 @Leslie: The Pythagorean triple reasoning is actually correct. The earliest fallacy I found was the one I pointed out. Even the comments defending the proof (e.g. "Some reviewers have maintained that the proof has departed from an integral nature, but this is not true.") up to that point are correct. I'll flesh out my explanation of the fallacy more.

In the "higher powers" case, the key equation is r^2 = 2(z-x)(z-y). This first shows up implicitly in the discussion immediately following Figure 10. He defines r, x, y, and z pretty well in the preceding figures, so I won't repeat what they are. In any case, the equation r^2 = 2(z-x)(z-y) explicitly first appears immediately after the section titled "Each r^2 triangle [...]", which itself vaguely justifies this equation by appealing to Figure 3. However, Figure 3's justification relies upon the equation x^2 + y^2 = z^2 for the "spilling down" process to result in equality.

The "spilling down" process of Figure 3 is used implicitly in the fallacious equation and in Figure 10's discussion. Even in Figure 1(a), you can see that this process does not always result in the triangle and the rhombus being equal: the rhombus there has value 8, while the triangle has value 16. Justifying this step rigorously is never actually done, even in the Pythagorean reasoning, but it relies crucially upon the relationship x^2 + y^2 = z^2, and is in fact equivalent to it. A purely algebraic exposition would have caught this mistake quickly.

My string of inequalities just points out that the equation immediately leads to a contradiction, without all the extra Pythagorean reasoning he goes through or using equations (1)-(3), suggesting that it is fallacious. I had said "r^3 >= 2(z-x)(z-y)z > 2(z-x)(z-y)r = r^3", which is correct. The "detrius" is constructed geometrically in going from Figure 7 to Figure 8 by "shaving" off part of the orange L. Part of what is shaved is a trapezoidal prism of width (z-y), height (z-x), and length z. This prism contains twos, so the detrius is at least 2(z-y)(z-x)z. Since z > r, the detrius is greater than 2(z-y)(z-x)r. If the offending equation were true, this would be r^2 r = r^3. Under the assumption that x^3 + y^3 = z^3, the detrius must be r^3, so that this reasoning gives r^3 > r^3. Specifically, the >= sign is correct.


@Rob Minteer: Rereading my previous post, it did sound elitist. I believe it's also entirely true. Mathematicians would love a short and correct proof of Fermat's Last Theorem. "Elementary proofs", as they're called, are something of a holy grail in math. I would love to be able to write up this proof in a way that would convince the mathematical community--I would be remembered for generations just as the popularizer of an elementary proof of FLT. But, this "proof" has an enormous and essentially unfixable hole, as I've now pointed out at length. It hides in the proof's sketchy geometric descriptions, length, and fuzziness, all of which makes the proof somewhat convincing.

I took the time to turn the Pythagorean case into algebraic reasoning since it was correct and the idea is actually rather insightful. Sorting through the mess of geometry to find the essential algebra took significant time. I will not do the same to the higher dimensional case, since it is incorrect and not at all insightful. As I said before, crank proofs are usually ignored since it takes so long to spot the errors. Even then the crank probably won't believe you, or if they do they'll just make their proof even more complicated with an even more hidden error. To be honest, I'm here because I was curious about exploring reactions to a crank proof in-depth. In the future I'll probably end up letting most cranks pass without comment, because this has been a (worthwhile and interesting, but significant) time sink.
DATE Wednesday, June 22nd 2011 - 10:38:15 PM

NAME Leslie
E-MAIL lal4@rice.edu

COMMENTS

 Yes, I am convinced by the contradiction that Skeptical from a@b.com has provided. Such is certainly undeniable...I am confused as how you derived that r^3 is >=...why not < ?

The root of the problem as he explains it is that an arbitrary triangle need not be equal to a rhomboid of twos. (although the other direction seems to be well defined) Hence the premise of the equations is absurd to begin with and no proof results since if we assume two false things there is no reason to attribute the result to one of the falsities, in particular.

Still, I would be more satisfied if someone could provide a contradiction before we get to the higher model ... Perhaps Skeptical already has one or just knows it to be an absurd task but mathematics is rigor! Such would show the contradiction right away.
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DATE Wednesday, May 25th 2011 - 12:53:29 PM

NAME RCleary

COMMENTS

 I would like to say that, while inteesting to read, thisd proof appears to have a few flaws.

1. It appears only to apply to pythagrean triples, and thus fails to account for the even variances amongthe exponential values and also fails to account for n= prime numbers greater than 3, which was one of the largest hurdles until Wiles published his famous proof.

2. This proof is not laid out algebraicly and supposes too many things.

Its a nice try though!
DATE Monday, May 23rd 2011 - 05:39:36 PM

NAME Rob Minteer
E-MAIL rnrminteer@bresnan.net

COMMENTS

 To the overly-educated person who claims that a reduction to algebra would prove Tom Ballard wrong: why don’t you show us? Tom’s simple “generator” can be reduced algebraically to cases where a(2x+a) is the square of a positive integer, and a and x are positive integers. The algebraic question which must be answered for cubes is: Can a(3xx+3ax+aa) ever be a cube? (Superscript font is apparently unavailable here, so I'm using the old xx for "x squared.") Underlying that is the question of whether any cube of a positive integer can be factored into three integral factors other than the cube itself. Obviously, some can be, but the properties of those factors no doubt correlate with Tom’s claim that such lead to absurdities. As to fourth and higher powers, similar algebraic expressions may no doubt be formulated, although having no geometric basis in reality. You mathematicians work on it, and leave us simple folk alone here in the real world. “Remember Harrison’s Chronometer!”
DATE Monday, May 9th 2011 - 09:53:59 AM

NAME Skeptical
E-MAIL a@b.com

COMMENTS

 The proof of the general case relies on the equation r^2 = 2(z-x)(z-y), which is its fallacy. If this equation were true, then the content of the detrius (in the n=3 case) = (by definition) r^3 = 2(z-x)(z-y)r. In forming the detrius (going from Fig. 7 to Fig. 8), the top of the orange L was shaved. The upper portion of this shaving includes z stacked rhomboids of dimension (z-y) and (z-x). The detrius then has content r^3 >= 2(z-x)(z-y)z > 2(z-x)(z-y)r = r^3, which makes no sense. That is, the author miscalculated the size of the detrius and derived a contradiction from this mistake.

To be specific, the section titled "Each r^2 triangle is equal to a rhomboid..." is rubbish. If you don't believe me, try showing r^2 = 2(z-x)(z-y) using only the relations x^n + y^n = z^n and r = x-(z-y), for n>2 *algebraically*. It's easy to delude oneself with geometric arguments. Note that geometric arguments can be turned into algebraic ones, so geometric intuition can certainly guide a rigorous algebraic proof (as in my previous post's summary).


Also, to the remarkably large number of people who can't spell it, the guy who actually completed a proof of Fermat's Last Theorem is "Andrew Wiles". He really completed a special case of the Taniyama-Shimura conjecture, instead of directly proving the theorem. It turns out a counterexample to Fermat's Last Theorem would have created a counterexample to this conjecture, so no such counterexample can exist--hence the theorem holds.

Most of the posts in this guest book are as cranky as this page's "proof". I don't think many of you realize how complex the reasoning gets in higher level math--this proof is a joke by comparison, and one that hides is fallacies in fuzzy explanations. If it was written algebraically, its mistakes would be much clearer. This sort of writeup would never be published in a respectable journal (even if it were correct)--not to mention a real "journal style" writeup would have taken less than a page, since there's so little content.

The hubris of complete amateurs thinking they've outdone mathematicians who've spent lifetimes studying problems is annoying. To be honest, cranks are generally ignored by the mathematical community since their errors are time-consuming to spot. http://web.mst.edu/~lmhall/WhatToDoWhenTrisectorComes.pdf is a wonderful article on the subject.
DATE Monday, April 11th 2011 - 10:23:22 PM

NAME Skeptical
E-MAIL a@b.com

COMMENTS

 Below is a summary of the Pythagorean triple reasoning, which is correct (though it's difficult to read).

Briefly, suppose x, y, and z are positive integers where x^2 + y^2 = z^2 for z > x and y. Let r = x-(z-y), which turns out to necessarily be positive. By distributing and applying x^2 + y^2 = z^2, we find r^2 = 2*(z-x)*(z-y). r is called the "root number" of the triplet (x, y, z).

Conversely, let r=2pq. Take x=r+q^2, y=r+2p^2, and z=r+q^2+2p^2. These satisfy r^2 = 2*(z-x)*(z-y), where x, y, and z are positive integers, and again by expanding, they also satisfy x^2+y^2 = z^2. Thus any Pythagorean triplet arises from a factorization of an arbitrary positive even integer, r, which is the generated triplet's root number.

These equations can be interpreted geometrically, but the proof is shorter and correctness is *much* clearer when written algebraically.
DATE Monday, April 11th 2011 - 08:35:53 AM

NAME radomir zhivanovich
E-MAIL kocmoc@comcast.net
URL http://internet explorer

COMMENTS

 curious about Fermat theorem
DATE Thursday, March 24th 2011 - 07:34:44 PM

NAME frank
E-MAIL dasjd@hotmail.com

COMMENTS

 Last time I checked, a proof by example is not a proof at all. You fail maths forever.
DATE Friday, March 11th 2011 - 03:22:37 PM

NAME Mick Rees
E-MAIL reesmb@tiscali.co.uk

COMMENTS

 I posted in November amazed at the proof. I have had the chance to consider it again in detail and sad to say I now realise that this is not a proof. The fallacy lies in the explanation just after figure 10. It assumes that the "r" triangle and the trapezoid are the same size as in the case for the pythagorian case. The whole point is that this must be proved for the the rest of the proof to proceed and it clearly has not been and cannot be with this arguement.
Sorry Tom back to the drawing board for you - but an interesting sleight of hand.
DATE Thursday, March 3rd 2011 - 11:36:43 AM

NAME Marianne
E-MAIL m.marriane@gmail.com

COMMENTS

 Nil - there is NO proof here.
No mathematical society would accept nonsensical "strings of words". Thanks for thinking! At least this makes us humans.
DATE Sunday, February 20th 2011 - 11:40:44 PM

NAME Leslie
E-MAIL lal4@rice.edu

COMMENTS

 hmm i dont know...i retract that
DATE Friday, January 21st 2011 - 12:53:16 PM

NAME Leslie
E-MAIL lal4@rice.edu

COMMENTS

 I think I understand the mistake, and it has been pointed out before.

"The remaining space is a wedge-shaped void with dimensions "r". This void is thus composed of identical r2 triangles. Note that the foregoing process has retained the "ones" on the bottom of the advanced model, thereby maintaining the integrity of the advanced model..."

Why do we need to fill this void with r^2 triangles? By hypothesis, all we require is that it has the same volume. The triangles could be composed of 1s and 2s in the wrong places, or even have zeros and larger numbers. In fact the triangles in the void need not even be the same.

I am sad, but I learned a lot in finally understanding this original argument.
DATE Thursday, January 20th 2011 - 09:46:22 PM

NAME Leslie
E-MAIL lal4@rice.edu

COMMENTS

 Short proof that if x^2+y^2=z^2 then for n>2,

z^n > x^n + y^n:
z^n = z^(n-2)*z^2=z^(n-2)*(x^2+y^2)

without loss of generality say x > y, (so also z>x>y).

Then the last line
> x^(n-2)*(x^2+y^2) > x^n+y^n.
DATE Tuesday, January 18th 2011 - 10:11:28 PM

NAME Sambasivan G
E-MAIL sambasivan.ganesan@vodafone.com

COMMENTS

 fantastic...
DATE Sunday, January 16th 2011 - 03:50:17 PM

NAME Chirtarun
E-MAIL chirtarun@yahoo.com

COMMENTS

 Having pondered over your simple proof for a few more days, my respect for you has skyrocketed! If I am dead wrong, please forgive me for my effrontery, but this is what I gathered from your proof:

1. If integers x, y, z satisfy x^n + y^n = z^n, where n is a positive integer greater greater than 2, then (x, y, z) MUST be a "Pythagorean" triplet, i. e. x^2 + y^2 = z^2.

2. Granting that (x, y, z) is a "Pythagorean" triplet, it is very easy to see that the triplet simply cannot satisfy
the equation x^n + y^n = z^n: this is your time-honored, sure-fire, supremely slick, Reductio Ad Absurdum proof for Fermat's Last Theorem!

The proof proposed by you has spurred me to visualize a very simple geometrical construction to prove proposition (1) above (my proof is independent of your proof); the construction I have in mind will take up 1/5 th of a page, and is accompanied by five short sentences to nail the proof.

Once proposition (1) is written in stone, anyone who vaguely remembers 9th grade algebra, can prove proposition (2) on a bar napkin, sitting in a crowded bar, sipping Long Island Ice Tea, or any other favorite libation.

Thanks to you, I am no longer spending sleepless nights, trying to prove the theorem using algebraic or number theory based methods. In my opinion, a non-geometrical approach to proving the theorem is akin to trying to catch a butterfly with a sledgehammer.
DATE Tuesday, January 4th 2011 - 08:30:40 AM

NAME Chirtarun
E-MAIL chirtarun@yahoo.com

COMMENTS

 Your proof of Fermat's Last Theorem is clever, elegant, intuitive, and, above all, easy to understand: even a high school senior can comprehend all the steps outlined in your proof. Bravo!
DATE Sunday, January 2nd 2011 - 06:30:45 AM

NAME Dan Djurdjevic
E-MAIL dandjurdjevic@gmail.com
URL http://whatdanhastosay.blogspot.com

COMMENTS

 Tom, correct me if I'm wrong, but your proof seems to predicated upon Pythagorean triplets and hence Pythagoras' theorem. To me this implies that you have restricted your proof to the variables a, b and c (or x, y and z) in relation to right angled triangles only.

If this assumption is correct, you have proved a special case: where a, b and c are sides of a right angled triangle - not in any other case. I have myself proposed a proof in relation to this special case - it is fairly easy to prove algebraically: http://whatdanhastosay.blogspot.com/2010/11/fermats-last-theorem-proof-for-right.html

On the other hand, what about acute and obtuse triangles? To my mind a proof needs to take into account all these.

Accordingly, one needs to factor in the law of cosines.

I have just completed a proof in this regard. I'm not sure if it works, since I am strictly an amateur, but I wonder if it gives food for thought.

http://whatdanhastosay.blogspot.com/2010/12/fermats-last-theorem-short-proof.html
DATE Thursday, December 2nd 2010 - 02:32:38 AM

NAME Mick Rees
E-MAIL reesmb@tiscali

COMMENTS

 I have spent months and months on FLT and have an interesting proof based on a conjecture that seems to work for the first couple of hundred primes but that I haven't proved yet.
However, this proof is absolutely brilliant and its author should be up there on Wikopedia in lights with Andrew Wiles.
DATE Monday, November 29th 2010 - 11:00:11 AM

NAME Leslie
E-MAIL lal4@rice.edu

COMMENTS

 I have not been able to find a clear flaw in the argument, including the ones already pointed out. Even on the physics forums thread a flaw did not seem clear to me. But for the last step, could someone explain to me the following claim:

by hypothesis x,y,z are a pythagorean triple, then the nth power of z is greater than the sum of the respective nth powers of x and y.

I can prove it for the case where n is 3 but not for higher powers.

Anyway I like the site, fact or fiction it is very thought provoking.
DATE Tuesday, October 5th 2010 - 01:26:16 PM

NAME F Marchetti
E-MAIL johnhowardliberalsupporter@gmail.com

COMMENTS

 Nuh. Sorry, mate, unfortunately, this is too close-minded. Unfortunately, not substantial enough to be proof.

Keep trying!
DATE Thursday, September 30th 2010 - 09:41:14 AM

NAME Frankl Welsh
E-MAIL fwe1.1@netzero.com
DATE Tuesday, September 28th 2010 - 08:35:00 AM

NAME McSuman
E-MAIL mcsuman@eptsystems.com
URL http://www.eptsystems.com

COMMENTS

 Astounding!
DATE Monday, September 27th 2010 - 04:52:52 PM

NAME Thomas Brosnan
E-MAIL gadro@btinternet.com
DATE Wednesday, September 22nd 2010 - 09:10:26 AM

NAME Bhaskar Jyoti Mahanta
E-MAIL gouratra@gmail.com

COMMENTS

 If you are interested in Mathematics please see VEDIC mathematics where you surly found most of the theorem which was discovered now.As a Indian I proud to say that every branch of mathematics was studied in India many times before Christ. Specially
1.The nine planet,
2.Newtons 3rd law of motion
(you may know about tipu sultans rocket using against British ruler in India and the vehicle using by Meghnaad, the son of King Raavana. which was fly on space using this law.) ,
3.Pythagoras theorem,
4.calculation of days of year(it is exactly 365.4),
5.specify the name of stars.
6.calculation more than 10^10000
7.creation of universe.
8.discovered numbers from 0 to 9.
and lot more.
These were discovered in India in that time when other world using animal skin for wearing,stone for weapon.
DATE Saturday, September 4th 2010 - 05:37:18 AM

NAME Bhaskar Jyoti Mahanta
E-MAIL gouratra@gmail.com

COMMENTS

 My formula is
if x^2+y^2=z^2
Then n>2
x^n+y^n=z^n{(sinA)^n+ (cosA)^n}

an example
If x=1,y=(3)^1/2,z=2
and for n=2

Therefore,
x/z=1/2=sin(45 degree)

=>1^5+{(3)^1/2}^5=2^5{(sin45)^5+(cos45)^5}
DATE Monday, August 30th 2010 - 10:59:07 AM

NAME Bhaskar Jyoti Mahanta
E-MAIL gouratra@gmail.com

COMMENTS

 I hv also a solve to it.
If x^2+y^2=z^2 then
x^n+y^n=z(sin^nA+cos^nA)
Where sin A=x/z
DATE Monday, August 30th 2010 - 09:20:26 AM

NAME ed van der meulen
E-MAIL ameulen@unet.nl
URL http://nnw.berlios.de/html.php

COMMENTS

 I am open source guy

And we promoter new ERA

I have studied pure mathematics and Physics

I have studied the way of Adrew Wiles. And with many paradoxes.

Pure causal deductive mathematics is for me playing and an art. SO Andrew Wiles proof is playing and a n art.

Your proof is already more exact, But also playing and an art. I have used your proof in my proof from fermat. for what we call the INductive mathematics branch, which is avoiding the many paradoxes of causal deductive mathematics.

i can show you the links of it. The many paradoxes and my fermat proof based on yours.
DATE Saturday, August 21st 2010 - 02:07:50 PM

NAME joewalker
E-MAIL dailybread200@yahoo.com
URL http://www.yahoo.com

COMMENTS

 Survey of the proofs of this theorem and to reach a generalization of a more advance concept.
DATE Tuesday, August 17th 2010 - 02:46:37 PM

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