Fermat's last theorem

[TonyMcC]

There are definitely no rational solutions to aⁿ+1=cⁿ, it's a fairly trivial special case.

 

How exactly did you infer, from me pointing out that I'd said it first, that I was disagreeing?

 

 

Hi tree,

Sorry if I misinterpreted some of your responses in post 11.

When I suggested that there no NON integer values I was trying to suggest one of the terms a, b, c was infinitely long when expressed in decimal form and could not be simplified. i.e. I was suggesting irrationality.

Your response that it quite easy to find examples I took to mean you could find examples which were not irrational. You seem to suggest that the infinitely long term could be of the form x=a/b.

That is why I challenged you to provide an example of a^n +b^n = (c/d)^n. (usual provisos)

You have not commented on my reasoning using similar triangles so I assume that is sound?

As I say, sorry for any misunderstanding.

 

Incidentally I first floated the idea of irrationality in "Fermat's Last Theorem - simple ideas" on 18th March.

 

D H says it is the height of hubris to think an amateur could solve Fermat's Last Theorem. I have to agree with him, but if people find it an interesting problem to chew on and exchage views with like minded people then what is the harm?

 

It should not be forgotten that if Fermat did indeed have a proof it would have had nothing to do with Wiles' proof.

 

I think I'll demonstrate a little hubris myself:-

TonyMcC's conjecture:-

a) In all equations of the form a^n + b^n = c^n (where n is a positive whole number >2) one or more of the terms a, b, or c is irrational.

b)This results in the impossibility of forming an equation of the form a^n +b^n = c^n where a, b, and c are positive whole numbers and n is a positive whole number >2.

 

I stand before the firing squad!!!

Edited January 3, 2011 by TonyMcC

[D H]

TonyMcC's conjecture:-

a) In all equations of the form a^n + b^n = c^n (where n is a positive whole number >2) one or more of the terms a, b, or c is irrational.

All I need to disprove this is one counterexample, such as n=3, a=0, b=2, c=2.

 

Exclude trivial solutions (one or more of a, b, or c is zero) and that is essentially just a restatement of Fermat's Last Theorem.

 

 

 

[TonyMcC]

All I need to disprove this is one counterexample, such as n=3, a=0, b=2, c=2.

 

Exclude trivial solutions (one or more of a, b, or c is zero) and that is essentially just a restatement of Fermat's Last Theorem.

 

Thank you D H . I need to revise my conjecture. I had not meant (I think obviously) to exclude part of the equation by the use of zero.

So here goes :-

 

Tony McC's conjecture (Revised)

(a) In all equations of the form a^n + b^n = c^n (where a, b and c are positive numbers >0 and n is a whole number >2) one or more of the terms a, b or c is/are irrational.

(b) A consequence of (a) above is that it is impossible to form an equation of the form a^n + b^n = c^n, where a, b and c are whole numbers >0 and n is a whole number >2.

 

You seem to accept that part (a) is true (essentially a restatement of Fermat's Last Theorem).

You also seem to believe irrationality exists in all relevant examples ( post 13). Have you a proof of this?

Do you think part (b) above follows?

 

I'm still in front of the firing squad, but not shaking quite as much!!

[D H]

From this thread:

You also seem to believe irrationality exists in all relevant examples ( post 13). Have you a proof of this?

 

From your simple ideas thread:

III) The only way to prove that there is always a solution to A^2+B^2=C^2 is via geometry and the only way to find that there are integer solutions is by some means of "trial and error".

 

 

These two statements are, to me, a signs that you are pursuing things in the wrong order. You really should try to understand the basics of algebra and number theory before you dive off the deep end into Fermat's Last Theorem.

 

The first one: Given positive rationals a and b, define sets A and B to be the sets of all rationals x such that [math]a^n+b^n-x^n[/math] is negative (set A) or positive (set B). This is a Dedekind cut that defines the real number c that is the positive real solution to [math]a^n+b^n=c^n[/math].

 

The second one: "Trial and error" is not needed. Pick any positive integer n and some other positive integer m that is greater than n. Then [math]a=m^2-n^2,\,b=2mn,\,c=m^2+n^2[/math] is a Pythagorean triple. This is Euclid's construction (i.e., it is ancient).

Edited January 3, 2011 by D H

[slick]

Exactly. Or you could eliminate c, yielding [math]a^n+b^n=1[/math]. This is the portion of the L_n-norm unit circle (aka squircle) that lies within the first quadrant. Fermat's Last Theorem is equivalent to saying that there are no rational solutions to [math]a^n+b^n=1, a>0, b>0[/math] for integer n>2.

 

How does that relate to The Tree saying - "There are definitely no rational solutions to aⁿ+1=cⁿ, it's a fairly trivial special case." Can The Tree elaborate on that.

 

 

 

Not that you are going to get far with this. You will not find a simple proof of Fermat's Last Theorem.

 

Nor am I trying to. My interest is in why such a simple looking problem is so difficult and to do that I am trying to understand where the actual problem lies. In other words, what is the point at which anybody attempting to resolve this realises they have hit an obstacle that cannot be overcome? I should declare that it is a professional interest for a business solutions methodology I a working on for which I am using FLT as an unambiguous reference case.

[the tree]

How does that relate to The Tree saying - "There are definitely no rational solutions to aⁿ+1=cⁿ, it's a fairly trivial special case." Can The Tree elaborate on that.

Elaborate on that one statement? No, not really.

 

1). FLT states that there are no integers [a,b,c] (all not equal to zero) that satisfy aⁿ+bⁿ=cⁿ for a positive integer n>2.

2). A corollary of (1) is that there are no rationals [a,b,c] (all not equal to zero) that satisfy the same equation.

3). 1 is both an integer and a rational number

4). (2) and (3) mean that there are no rational pairs that satisfy any of these equations 1+aⁿ=cⁿ, aⁿ+1=cⁿ or aⁿ+bⁿ=1.

 

 

Nor am I trying to. My interest is in why such a simple looking problem is so difficult and to do that I am trying to understand where the actual problem lies. In other words, what is the point at which anybody attempting to resolve this realises they have hit an obstacle that cannot be overcome?

Well it's fairly simple nowadays - with automatic proof programs, you can brute force your way through various axioms and known results and it will indeed turn out that no conventional method is going to get you there. And at the same time you can set another computer to look for a counterexample and fail just as miserably. If you're doing it by hand then I guess the typical mathematician knows that they are in trouble once the coffee runs out.

 

I should declare that it is a professional interest for a business solutions methodology I a working on for which I am using FLT as an unambiguous reference case.

o___________O

[D H]

My interest is in why such a simple looking problem is so difficult and to do that I am trying to understand where the actual problem lies.

One simple poke at this simple looking problem reveals that it is an extremely hard problem. [math]a^3+b^3=c^3[/math] is a high-order Diophantine problem. High-order Diophantine problems are notoriously hard to solve. Unlike linear Diophantine problems, there is no generic technique for solving high order ones. [math]a^4+b^4=c^4[/math] is a different high-order Diophantine problem. [math]a^5+b^5=c^5[/math] is yet another. Fermat's Last Theorem in one swell foop says that an infinite number of high-order Diophantine problems have no solution. That makes this an extremely hard problem.

 

 

In other words, what is the point at which anybody attempting to resolve this realises they have hit an obstacle that cannot be overcome?

When they read the enormous amount of work done on the problem. Or when they read Kummer's proof that the theorem is true for all regular primes. Of course, it is now a solved problem, but the solution was not to look at equations of the form [math]a^n+b^n=c^n[/math]. The solution came from tying the problem to an apparently unrelated problem of modular equations.

 

 

 

[uncool]

There are definitely no rational solutions to aⁿ+1=cⁿ, it's a fairly trivial special case.

 

Err. No, it is not trivial. I'd like to see your proof of it, because proof of that is entirely equivalent to Fermat's Theorem, as follows:

 

Assume a solution to a^n + b^n = c^n. Then let a' = a/b, c' = c/b. Then a'^n + 1 = c'^n. Therefore, if what you say is true, we have a contradiction, and we are done.

=Uncool-

[the tree]

I did say that it was a special case, so yes - proving it would be equivalent to FLT, that's kind of the point - and the fact that you put that into one line is what makes it trivial.

 

[slick]

Okay, here is an attempt at a simple proof.

 

I said before that my primary concern is with positive integer values of a, b and c that can form a triangle with an angle C < 90^o and >= 60^o. What I realised fairly early on was that for aⁿ + bⁿ = cⁿ the resulting area of the triangle would have to be zero. My thought then is to take a portion of the triangle such that both b' and c' are both 1 which would mean that for that portion of the triangle to have zero area a'ⁿ would have to result in zero. The picture below shows what I mean.

 

 

 

There are, of course, no circumstances in which a'ⁿ can equal zero therefore the whole triangle can never itself have an area of zero. The only exception to this is for a right angled triangle since in those circumstances a'ⁿ would always be zero.

 

I await your brickbats!

[uncool]

I did say that it was a special case, so yes - proving it would be equivalent to FLT, that's kind of the point - and the fact that you put that into one line is what makes it trivial.

 

Ahhh, I misunderstood. I thought you meant that it was trivial without assuming FLT; .

=Uncool-

[TonyMcC]

From this thread:

 

 

From your simple ideas thread:

 

 

 

These two statements are, to me, a signs that you are pursuing things in the wrong order. You really should try to understand the basics of algebra and number theory before you dive off the deep end into Fermat's Last Theorem.

 

The first one: Given positive rationals a and b, define sets A and B to be the sets of all rationals x such that [math]a^n+b^n-x^n[/math] is negative (set A) or positive (set B). This is a Dedekind cut that defines the real number c that is the positive real solution to [math]a^n+b^n=c^n[/math].

 

The second one: "Trial and error" is not needed. Pick any positive integer n and some other positive integer m that is greater than n. Then [math]a=m^2-n^2,\,b=2mn,\,c=m^2+n^2[/math] is a Pythagorean triple. This is Euclid's construction (i.e., it is ancient).

 

DH - Thank you for pointing out Euclid's construction, I wasn't aware of it and so that can be my "learn a little every day" for today. I have amended my simple thoughts thread.

 

Also I an not familiar with the Dedekind cut. I will read up on it soon, but at first glance it just seems to say that there is a gap between a set above "+0" and a set below "-0". I assume that any number in the gap will be irrational. If I have this completely wrong please don't concern yourself - I shall have an interesting time studying it.

 

I can also understand "the wrong order" comment. Because I set out from the beginning to use a novel approach (analyse incorrect sums) geometry seemed to be a way to do that.

 

Having said all that I note you have not commented on whether, in your opinion, my conjecture is valid. I will add a little clarification to it,

 

Tony McC's conjecture:-

 

In every correct sum a^n +b^n = c^n, if n is a whole number >2 and a, b and c are positive numbers >0 then one or more of the terms a, b or c is irrational.

 

A consequence of this is that if a triangle is constructed using a, b and c at least one side will be irrational.

Because of the irrationality no similar triangle can be constructed with all three sides integers.

Therefore there is no integer solution to the equation x(a^n) + x(b^n) = x(c^n) where x is any number.

This proves Fermat's Last Theorem.

 

In summary if the irrationallity mentioned above can be proved then Fermat's Last Theorem would be easily solved.

 

Please comment.

Edited January 5, 2011 by TonyMcC

[slick]

 

Tony McC's conjecture:-

 

In every correct sum a^n +b^n = c^n, if n is a whole number >2 and a, b and c are positive numbers >0 then one or more of the terms a, b or c is irrational.

 

A consequence of this is that if a triangle is constructed using a, b and c at least one side will be irrational.

Because of the irrationality no similar triangle can be constructed with all three sides integers.

Therefore there is no integer solution to the equation x(a^n) + x(b^n) = x(c^n) where x is any number.

This proves Fermat's Last Theorem.

 

In summary if the irrationallity mentioned above can be proved then Fermat's Last Theorem would be easily solved.

 

Please comment.

 

 

I think I understand what you are saying however it is always worth clarifying. To my mind you seem to be saying that for any 3 integers a, b and c where a + b = c, for there to be any common root > 2 at least one of a b or c must be irrational. If I have understood correctly then I think that is highly probable however it doesn't seem fundamentally different to the original problem, just a re-statement in reverse. It may be worth looking at the Golden Ratio to see whether there is any relationship there.

 

 

[TonyMcC]

I think I understand what you are saying however it is always worth clarifying. To my mind you seem to be saying that for any 3 integers a, b and c where a + b = c, for there to be any common root > 2 at least one of a b or c must be irrational. If I have understood correctly then I think that is highly probable however it doesn't seem fundamentally different to the original problem, just a re-statement in reverse. It may be worth looking at the Golden Ratio to see whether there is any relationship there.

I personally don't think I am saying anything startling or particularly new. I certainly can't prove the reqired irrationality. However Fermat didn't specifically mention irrationalty. I believe in Fermat's time many mathematicians in the Western World were sceptical of the idea.

I haven't seen anyone else try to analyse incorrect sums using geometry and use the idea of similar triangles the way I have.

I have found a certain reluctance occurs as soon as irrationality is mentioned and a better mathematician than me seems to doubt that proving irrationality would lead to a simple solution.

I think you understand what I have posted. Whether the irrationallity of the Golden ratio can be somehow be used is something I just don't know. Apparently the term "Golden Ratio" was first used in the 1800's.

 

Okay, here is an attempt at a simple proof.

 

I said before that my primary concern is with positive integer values of a, b and c that can form a triangle with an angle C < 90^o and >= 60^o. What I realised fairly early on was that for aⁿ + bⁿ = cⁿ the resulting area of the triangle would have to be zero. My thought then is to take a portion of the triangle such that both b' and c' are both 1 which would mean that for that portion of the triangle to have zero area a'ⁿ would have to result in zero. The picture below shows what I mean.

 

 

 

There are, of course, no circumstances in which a'ⁿ can equal zero therefore the whole triangle can never itself have an area of zero. The only exception to this is for a right angled triangle since in those circumstances a'ⁿ would always be zero.

 

I await your brickbats!

 

Hi Slick,

I can't really see what you are getting at. However I see you are making isosceles triangles . The base of an isosceles triangle is 2*(the square root of something). Is zero twice the square root of "something". A long time ago I decided an extremely important property of any number is that it is "the square root of something".

I find the following quite satisfying - If I superimpose the curves in my diagram onto your diagram I can see that your large triangle roughly fits my cube root curve. Not only that, but the ratios are ( roughly) 0.8, 0.88, 1.0.

Allowing for approximations due to playing with a ruler on my laptop face I deduce 80^3 + 88^3 = 100^3

and 800^3 + 880^3 = 1000^3 etc. etc.

This illustrates my use of similar triangles.

[slick]

Hi Slick,

I can't really see what you are getting at.

 

Hi Tony, let me try to explain a bit better, it was very late last night and I didn't really flesh it out properly.

 

The shaded area of my diagram illustrates triangles that, for any value of c (c being the bottom line), would have an angle C >= 60^o and < 90^o. This is because it is self evident that any other values of a, b and c that do not result in a triangle angle C >= 60^o and < 90^o cannot produce the result aⁿ + bⁿ = cⁿ. For values of a, b and c that do result in the described triangles, increments of n will cause the angle C to widen until it passes through 180^o. Up until it passes through 180^o aⁿ + bⁿ will still be greater than cⁿ however once it has passed through 180^o aⁿ + bⁿ will always be less than cⁿ. The question then is whether, for any increment of n, angle C can ever be exactly 180^o since that would mean that aⁿ + bⁿ = cⁿ. It would also be true at that point that the area of the triangle would be zero. This is, of course, what happens when you start with a right angled triangle and square the numbers.

 

My objective therefore is to find a way to determine if any triangle with an angle C >= 60^o and < 90^o can ever have zero area and/or an angle C of 180^o for increments of n and to determine that I took a portion of my triangle around the angle C that equates to c' = 1, b' = 1 and a' = whatever results from joining c' to a'. Now for any increments of n b' will always remain 1, c' will always remain1 and a' may increase or reduce. For any circumstance a' can never be zero and it is only if a' becomes zero that the triangle can have zero area with angle C = 180^o. Since that portion of the triangle can never have an area of zero it follows that the whole triangle can never have an area of zero consequently aⁿ + bⁿ = cⁿ cannot be true for n > 2.

[the tree]

Tony McC's conjecture:-

 

(1) In every correct sum a^n +b^n = c^n, if n is a whole number >2 and a, b and c are positive numbers >0 then one or more of the terms a, b or c is irrational.

 

A consequence of this is that if a triangle is constructed using a, b and c at least one side will be irrational.

Because of the irrationality no similar triangle can be constructed with all three sides integers.

Therefore there is no integer solution to the equation x(a^n) + x(b^n) = x(c^n) where x is any number.

This proves Fermat's Last Theorem.

 

Please comment.

(1) is a slightly awkward rephrasing of FLT - so you are assuming the thing you're trying to prove. Not a good start at all. DH was right, you are, both you and Slick - approaching this from the wrong direction. You have a much more satisfying experience if you work with some basic algebra before trying to tackle one of the hardest problems in mathematical history.

[slick]

DH was right, you are, both you and Slick - approaching this from the wrong direction. You have a much more satisfying experience if you work with some basic algebra before trying to tackle one of the hardest problems in mathematical history.

 

Not sure what that means, surely any problem that has hung around as long as this one needs a 'wrong' approach since all of the 'right' approaches have been fruitless (with the obvious exception of Andrew Wiles' solution whose own approach was to prove an apparently unrelated conjecture). BTW: Are you saying that my solution is wrong?

[TonyMcC]

(1) is a slightly awkward rephrasing of FLT - so you are assuming the thing you're trying to prove. Not a good start at all. DH was right, you are, both you and Slick - approaching this from the wrong direction. You have a much more satisfying experience if you work with some basic algebra before trying to tackle one of the hardest problems in mathematical history.

 

" A conjecture is a proposition that is unproven but appears correct and has not been disproven".

The first part of what I wrote is the conjecture and the second part attempts to show that if the cojecture is proved then a proof of Fermat's Last Theorem would be easily found.

 

Although written in Latin a translation of Fermat's writing is "It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers". Irrationality is not mentioned!

 

What little I have read and understood of elliptic curves, modular arithmetic and the Tamiyama-Shimura Conjecture seems to concern itself with patterns of numbers. Again, although there is much I haven't read, irrationality seems to get no or little mention so it seems to be a fair assumption that the patterns concern themselves with rational numbers.

 

Andrews Wiles, as far as I know, never said his aim was to prove irrationality and at the end never stated that was what he had proved or that irrationality was an outcome of his proof.

 

You mention D H. Perhaps you should read posts 13, 15 and 16.

 

Anyway, I sggest the addition of claiming that irrationality occurs in all relevant equations is enough of a change to warrant a new conjecture. I think a new conjecture can be claimed even though a proof that may not involve irrationality has been accepted.

 

As the first part of what I wrote is a conjecture it cannot be said that the second part is a proof - just evidence that if the conjecture is correct what could then be proved as a consequence.

[FLT For Fun]

" A conjecture is a proposition that is unproven but appears correct and has not been disproven".

The first part of what I wrote is the conjecture and the second part attempts to show that if the cojecture is proved then a proof of Fermat's Last Theorem would be easily found.

 

Although written in Latin a translation of Fermat's writing is "It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers". Irrationality is not mentioned!

 

I am new to this forum and I am not mathematician. I believe that Diophantus dealt with integers only. Furthermore you can't prove conjecture with another conjecture.

 

I tried to prove FLT as a brain excercise a la Sudoku. Here what I can share. A proof using Euclidian geometry won't work. As for proving algebraically, it can be round robin. It will make you lost and give you a false sense of victory, and only few days later you will find yourselve back to square one.

 

FLT is as false as 2=1. Finally if your proof is more than a page then it's no longer that elementary.

 

flt4fun

[TonyMcC]

I am no mathmetician! Just interested in FLT which I have pondered about.

 

It is obvious that there is always a solution from the following idea:- Take two cubes made of wax (call them A and B) and measure their sides. Roll them together and form them into a new cube and measure its side (call it C). From the measurements we can say A^3+B^3=C^3.

 

Everything that I have read so far involves integers. This leads to the idea that perhaps there is a pattern of integers that can be found where the solution C will always be an integer but it always "lands" on an integer which is not a number cubed.

 

I wonder, since that approach hasn't led to a solution over more than 400 years, whether there may be another way of looking at the problem.

 

I wonder why Fermat's original statement appeared against work on triangles.

 

After some thought along these lines I felt that there must be a reason why a triangle formed from A B and C taken from A^n+B^n=C^n cannot have 3 integers (n>2). This led to my conjecture:-

 

"In every correct sum a^n +b^n = c^n, if n is a whole number >2 and a, b and c are positive numbers >0 then one or more of the terms a, b or c is irrational."

 

I have never seen anyone else claim this.

 

Being a conjecture it proves nothing and so cannot be used as part of a proof.

However it can be used in what I would call a "conditional" proof - the condition being that the conjecture can be proved. Of course if the conjecture can be proved false everything developed from it fails as well.

 

So, can anyone prove or disprove my conjecture?

Edited April 5, 2011 by TonyMcC

[FLT For Fun]

I am no mathmetician! Just interested in FLT which I have pondered about.

 

It is obvious that there is always a solution from the following idea:- Take two cubes made of wax (call them A and B) and measure their sides. Roll them together and form them into a new cube and measure its side (call it C). From the measurements we can say A^3+B^3=C^3.

 

Everything that I have read so far involves integers. This leads to the idea that perhaps there is a pattern of integers that can be found where the solution C will always be an integer but it always "lands" on an integer which is not a number cubed.

 

I wonder, since that approach hasn't led to a solution over more than 400 years, whether there may be another way of looking at the problem.

 

I wonder why Fermat's original statement appeared against work on triangles.

 

After some thought along these lines I felt that there must be a reason why a triangle formed from A B and C taken from A^n+B^n=C^n cannot have 3 integers (n>2). This led to my conjecture:-

 

"In every correct sum a^n +b^n = c^n, if n is a whole number >2 and a, b and c are positive numbers >0 then one or more of the terms a, b or c is irrational."

 

I have never seen anyone else claim this.

 

Being a conjecture it proves nothing and so cannot be used as part of a proof.

However it can be used in what I would call a "conditional" proof - the condition being that the conjecture can be proved. Of course if the conjecture can be proved false everything developed from it fails as well.

 

So, can anyone prove or disprove my conjecture?

 

 

Hi Tony,

 

I think your claim should be considered as a corallary. That is an obvious result after a proof or lemma has been proven.

 

It is clear that if a, b, and c are real numbers, then there will a solution. Thus no proof is required. Irrational number is a subset of real number.

 

It's the integer part that made FLT difficult to prove. Many amateur and seasoned mathematicians approached FLT as a^n+b^n-c^n=0. That leads to solve equation of 3 unknowns. Direct approch proof would be impossible. Thus the indirect proofs have been devised. For example, elliptic curve (I called it fancy geometry). I know very little about it. But it is a big thing for math gurus.

 

Cheers,

 

flt4fun