Fermats Last Theorem - simple ideas

[TonyMcC]

I am getting on in years and since retirement I have been giving Fermat's Last Theorem some thought. I believe most people are attempting to solve this puzzle in the wrong way. There may be NO good reason why (say) A^3 + B^3 = c^3 should not have a whole number solution. However there is a very simple reason why the relevant "wrong sum" A+B=C cannot logically be deduced. Please note, I am not saying it cannot exist - only that there is a simple reason that it cannot be deduced. The reason applies to all powers>2. Perhaps this was Fermat's starting point!? I should be interested if this thought gets any response. More later - perhaps a conjecture if not quite a proof.

[ajb]

Have a look into the work of Taylor and Wiles from the early to mind 1990's.

[TonyMcC]

I saw the television programme at the time. That was what grabbed my interest. I don't remember anyone suggesting that a simple solution might be found from analysing incorrect sums. I have a transcript of the programme which I will read again. By the way can you see why it should not be possible to obtain the incorrect sum A+B=C which relates to A^n+B^n=C^n where n>2?

I would like to quote Prof. Wiles "sometimes you realise that nothing that's ever been done before is any use at all and you just have to find something that is completely new, and it's a mytery where it comes from"

Edited February 14, 2010 by TonyMcC

[ajb]

Faltings I believe did give a simpler version of Wiles' proof.

 

The original papers you want are

 

Wiles, Andrew Modular "Elliptic curves and Fermat's last theorem". Ann. of Math. (2) 141 (1995), no. 3, 443--551

 

Taylor, Richard and Wiles, Andrew "Ring-theoretic properties of certain Hecke algebras". Ann. of Math. (2) 141 (1995), no. 3, 553--572.

 

As you can tell from the titles the proofs are not trivial and use quite advance mathematics from algebraic geometry and related things.

[TonyMcC]

I have visited your homepage Andrew (If I may call you that) and am very impressed. Perhaps more importantly we are very different in our levels of formal qualifications and life experiences. Much of my thoughts concerning "Fermat" come from pondering the various jobs I have undertaken and the training required to perform them. My experiences have all been of a practical nature and I'm afraid my highest qualification is Higher National Certificate in Electrical and Electronic Engineering. Much of my ponderings have more to do with physics, science and computer programming than pure mathematics. If you have the patience I would value your opinion if I float my ideas bit by bit until you "shoot me down". Out of interest my career involved troubleshooting and teaching others to troubleshoot military radar systems and troubleshooting mainframe computer systems. My final position was as a College Lecturer teaching basic principles of Mechanical, Electrical and Electronic Engineering. As a hobby in the 70's I taught myself computer programming (whilst working as a computer Engineer) and ended up teaching it as a secondary subject at my College. I mention these facts so that you will realise my limitations. For what it is worth Fermat may quite possibly have been more like me than you - or perhaps not !

[ajb]

I doubt I can help you much with Fermat's last theorem and the work of Wiles and others. It is outside my area of expertise and used lots of quite advanced stuff.

[TonyMcC]

For Andrew and any other interested party. As I have said I believe there is something peculiar about all whole number incorrect sums (A+B=C) and that peculiarity precludes the possibility of that sum being corrected by raising the three terms by any power other than 2.

This simple exersize in logic convinced me that this is an idea that was worth developing:-

You stand outside a room with an open window. Inside is a person with a normal schoolboys geometry set. This person will follow a sequence of instructions that you have provided beforehand (i.e. An algorithm).

You want to pass 2 rods of any length through the window (A and B), have the person inside follow your algorithm and pass out to you a single rod equal to the sum of A and B, called C).

This of course is very easy:-

1) Accept A and B

2) Draw a straight line

3) Place A and B on the straight line so that they touch

4) Cut rod C such that it extends from the open ends of A and B

5) Pass rod C out of the window.

 

Next, and only slightly more difficult, you want to pass rods A and B into the room and have rod C returned such that A^2+B^2=C^2. (e.g. If A=3, B=4, then C=5)

1) Accept A and B

2) Draw a straight line

3) Draw another line that intersects this line at 90 degrees.( Use compasses and you don't even need numbers!)

4) Use a right angle from the intersection to align A and B so that they touch.

5) Cut rod C so that it extends from the open ends of A and B.

6) Pass rod C out of the window.

 

Next you want to pass rods A and B through the window and have rod C returned such that A^3+B^3=C^3( eg A=6, B=8, C=9(approx))

SUDDENLY WE ARE PRESENTED WITH AN IMPOSSIBLE SITUATION.

The reason for this is that each pair of input rods A and B of differing ratios requires that the person inside the room sets a different angle with which to align those rods. What you want the person inside the room to do is construct a triangle of a particular shape. Basic geometry requires 3 parameters in order to do this ( 3 sides or 2 sides and included angle). However you are only inputting 2 of the required parameters.

 

This argument also applies to all powers >2.

 

O.K. - This doesn't prove that that the required triangle doesn't exist - only that the method described would demand 3 inputs. Perhaps any method of calculating (A^n+B^n)^(1/n) (where n>2)

would require 3 inputs.

 

For powers n=1 and n=2 one of the parameters needed to form the required triangle is a constant ( 180 degrees and 90 degrees respectively) and a constant can be accomodated in the algorithm. (For n=1 the "triangle" is very very flat, its angles being 0, 0, and 180 degrees!)

 

Perhaps this realisation gave Fermat food for thought.

 

If my simple ideas interest anyone then there is more where this came from.

 

I would welcome any comments

Edited February 16, 2010 by TonyMcC

[uncool]

But it is possible to get a rod C such that A^4 + B^4 = C^4. Here is how:

 

Draw a line l, and choose any point O on it. Find the point M such that the length of OM is A. Find the point N such that the length of ON is B. Draw another line m through ON, and find the point OP such that the length of OP is B. Draw the line between M and P. Find the line parallel to MP that goes through N, and find the point R where the line intersects M. Then we know that OR has length B^2/A, by similar triangles.

 

Now construct a right triangle with side lengths equal to A and to B^2/A (we can do the latter because we have constructed it). Then the length of the hypotenuse is sqrt(A^2 + B^4/A^2) = sqrt(A^4/A^2 + B^4/A^2) = sqrt(C^4/A^2) = C^2/A.

 

We now have a segment of length C^2/A. Construct a line, and choose a point E. Find F such that EF = A, and G on the other side of E such that EG = C^2/A. Draw the circle such that FG is a diameter. Draw the perpendicular to FG that goes through E, and find the point on the circle H where the line intersects. We know that FHG is a right triangle, so we can get by similar triangles that EH has length equal to the geometric mean of the lengths of EG and EF. That is, the geometric mean of C^2/A and A. But that is just C - the desired length.

=Uncool-

[TonyMcC]

Thanks Uncool for your message. I'm a bit rusty on geometry and having trouble with your construction. Any chance of a diagram or reference I can look up. You are probably right and as long as your solution only involves ratios and right angled triangles thats OK. I was a bit hasty by not stopping at powers of 3. I think you just may have introduced a "wobbly" by ending up using a right angled triangle where the required (i.e. unknown) output or a fraction of it finds its way into more than one side. In any case I am still quite happy with further work I have thought out. I would like to follow your argument so if you can make it clearer in any way I should be grateful - Tony

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Is there anybody out there who can explain more clearly the construction used by "uncool" to obtain a rod C in response to nothing except rods A and B that gives the incorrect sum A+B=C which can be corrected by raising each term by the power of 4?

It would be nice if the necessary steps could be written down as an algorithm which I could incorporate in a simple computer program - but I would enjoy doing that for myself.

I feel that "uncool" has probably made a mistake somewhere, or perhaps is "pulling my leg".

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OK let us try a different tack. If you have read this far you will know that I think a proof of FLT should emerge from analysis of INCORRECT sums of A+B=C. One way of attempting this is to form triangles from A,B and C. For example a triagle of sides 6,8,and 9 represents the incorrect sum 6+8=9 which when all terms are cubed becomes the (almost) correct sum 6^3 + 8^3 = 9^3.

Next simple thought:-

To many people the triangle described above is a very useful (almost) CORRECT sum! If you, dear reader, are intersted enough perhaps you would like to give me an example and how the triangle might be used in the "real world".

Edited February 20, 2010 by TonyMcC
Consecutive posts merged.

[TonyMcC]

Well I guess this is "dying a death".

People who might find a triangle a useful tool include navigators, electrical engineers and even the people who hang pictures on walls!

(triangle of forces adapted from the parallelogram of forces)

[TonyMcC]

Perhaps I should get nearer to the beginning of my thoughts.

It seems to me that the implications of Fermat's Last Theorem are rather weird.

He seems to imply that I could give you 2 cubes made of something malleable. You could roll them together to make a new cube. However you will not be able to measure the side of the new cube accurately. If you could do so then simple multiplication would provide an integer result.

Example - given two cubes 6 units and 8 units the resultant cube might be measured as 8.99 units. If the result was accurate it would mean that 600^3+800^3=899^3.

However many digits followed the decimal point would not matter, you would just end up with very large numbers.

You could follow this procedure with any two cubes and thus there is ALWAYS a solution to A^3+B^3=C^3, but in all cases (according to Fermat) there will be an infinite number of digits following a decimal point.

Fermat seems to imply that this would happen regardless of the number base used (binary, octal etc.etc.)

This all seems so strange that perhaps we should look for irrationality. It would be nice if any irrationality found broke some simple mathematical rule or law. However perhaps an irrationality could be found in the laws of physics, mechanical principles, electrical principles, electronic principles, logical reasoning or even in the way the universe "works". Is it reasonable to think that irrationality might lie at the heart of any proof?

I would expect any such irrationality, if found, to accommodate all powers above 2.

[Malteaser]

For what it's worth Tony McC, I found your workings interesting.

 

What's really tough with anything like this is that as soon as you start looking at other people's work then you get in to some really complicated Maths that you quite quickly can't understand. I have a degree in Maths and Stats and I can't follow the kind of explanations that are given on some websites and texts either. Perhaps it's a function of the human mind (or my mind!) that once someone starts explaining why an idea is not a goer then there is less of an inclilation to try and understand what they are saying!

 

Having said that. Thanks for your thoughts. I would be interested in some of your other thoughts too, for what it's worth. Even if they don't lead anywhere, I'd still be interested. But we may have to put up with some acedemic/intellectual goading in the process!

 

As for Uncool's work, I couldn't quite follow it and can also see merit in seeing the description in visual form. if this could be provided, that would be great.

 

I think about this when i go to sleep, and don't get very far! My simple steps go along the lines of the following:-

 

X^n + Y^n = Z^n

 

rearrange to get : X^n = Z^n - Y^n

 

which can, I think, always be factorised by (Z-Y)

 

Hence, for example, using n=3: Y^3 = (Z^2 + ZY + Y^2)*(Z-Y).

 

At this stage I then tried to work out what I can say about Z^2+ZY+Y^2.

 

Firstly what doesn't it factorise by?

 

Z? Y?, Z-Y?, Z+Y?,

 

We can also draw up some rules about mutual exclusivity of X, Y and Z and potentially deduce that it won't factorise into 2 and 3.

 

It gets more complicated with factorising where n>2 as higher powers of n mean my equation can have more factors than just (Z-Y), but I think you can always say that a part of the factorising has a larger power than 1 in it and this, therefore means we can draw similar conclusions as above on those parts of the equation

 

What I would then be looking to do would be to come up with a set of things that we can say about X, potnetially then proving that X cannot have a set of factors and therefore limiting would X could be - and contradicting what we originally said that X could be.

 

But evidently I'm far from that with the above.

 

Any thoughts? Has this all been done and discarded before?

 

As a further thought can we do the saem kind of deduction on X^n +Y^n.

 

Can we say X isn't a factor? neither is Y? Neither is X+Y, Neither is X-Y? Neither therefore is 3 and by putting some exclusivity constraints on X and Y, neither is 2.

 

Can we say this or can we say something similar and is it important that we can or can't say any of this?

[Malteaser]

Can anyone help us here?

 

Anyone any views/comments?

[TonyMcC]

I have also worked at factorisation and just got more confused. My ideas are really more simple!

Look up a list of formulae for any branch of physics or engineering and you will find that "squared" and "square root" appear regularly. Off hand I cannot think of a formula that uses "cubed" or "cube root" except in the case of three phase electricity. Three phase electricity I would suggest is not really part of the natural order of things.

I wonder why "squared" and "square root" do feature so prominently?

Do you think that it has anything to do with Einsteins formula E=Mc^2? What would be the effect if Einstein had proved E=Mc^3 for instance? Is Fermat's declaration more to do with how the universe "works" than it is to mathematics?

I guess this just a passing thought.

[ajb]

What would be the effect if Einstein had proved E=Mc^3 for instance?

 

 

Such a result could never be true, the units are not consistent. Just from the units we get (with no other justification) [math]E \propto mc^{2}[/math].

[TonyMcC]

Thank you Physics Expert. I do realise that the formula fits the units and so could not be true for E=Mc^3. What I was trying to float was the idea that possibly the linkage between mass, energy and the speed of light was somehow of crucial importance. This just might be a means to proving that a cubic mass of a particular element when added to another cubic mass of the same element could not form another cubic mass of that element. I am thinking along the lines of the three imagined cubes being converted into energy. The three relevant amounts of energy would each have an amount cubed. It might not actually occur but you could imagine the raw energy as having shape - and each shape would be a cube. These "cubes" of energy would shrink to make the three cubes of matter. If either the raw energy "cubes" or the cubes of matter could not be defined (i.e. the side of any of the six cubes always proves to be irrational) then there is a situation which supports "Fermat". I like to imagine the three cubes of matter being transformed into three "cubes" of energy and back again. What does the mathematics say about the three "cubes"of energy and their relationship to the three cubes of matter?

What would be the cube root of c^2?

I do realise that Einstein was born much later than Fermat!

Edited April 16, 2010 by TonyMcC

[khaled]

geometric representation can let you notice possible relations easily,

 

geometric representation allows transformation,scaling,..etc

 

sometimes translating one formula into different field expand its enhancement scope,

[TonyMcC]

Here's another simple thought.

Fermat seems to imply that the nth root of A^n + B^n is irrational.

Have you ever wondered how far you could get without using numbers at all?

Let's consider the case of two cubes.

I give you two cubes which I have measured at 3 cubed and 4 cubed, but instruct you not to measure them except to determine if there is a whole number ratio. In other words you wouldn't know whether they had sides of 3 and 4 or perhaps 6 and 8.

Since there is a whole number ratio you could slice the cubes into equal thickness slices. You would then have 3 solid pieces 3*3*1 and 4 solid pieces 4*4*1.

Using only a right angle you could form the 3 solid pieces into a single solid piece which would be sqr(27)*sqr(27)*1 and the 4 solid pieces into a single solid piece which would be sqr(64)*sqr(64)*1.

Similarly using only a right angle you could form a single solid shape which would be sqr(91)*sqr(91)*1.

So you could transform any 2 cubes into a solid which has a square face and a depth which is smaller than the side of the square and this can be done without using numbers at all.

It would seem that we are half way there!

BUT if I give you a solid shape which has a square face and a depth less than the side of the square, what manipulations are needed to reshape it so that it becomes a cube?

Just thinking!!

[TonyMcC]

In the book "The Hitch Hikers Guide to the galaxy" by Douglas Adams a super computer, after millions of years of deep thought, answered a question.

The question was "What is the answer to the Ultimate Question of Life, the Universe and Everything".

The anwer given was 42.

I'm afraid the computer got it wrong - the correct answer is "The Sine Wave" .

I am quite serious, after all you can't get far in mathematics, physics or any branch of engineering without it appearing in your calculations.

I have given this some thought (a bit less time than millions of years!) and will float those thoughts later.

[TonyMcC]

The following describes a unique property of a sine wave:-

Take two sine waves of any amplitude but of the same frequency (i.e same periodic time or using the same distance for 360 degrees on the X axis of a graph).

Draw a graph of each, one above the other. Slide one waveform along the X axis (i.e give it phase shift).

Working point by point vertically add one graph to the other to form a composite waveform which is the addition of the two above.

You will always find the resultant composite waveform is also a sine wave.

By definition each point on a sine wave is the square root of something (SQR(A^2+B^2)).

Thus each vertical calculation made to form the composite waveform is a correct sum of the form SQR(something)+SQR(something)=SQR(something. (Very often 1, 2 or 3 of the "somethings" may well be irrational).

Now, if you take 3 points, one from each of your graphs, not vertically in line, you will make an incorrect sum also of the form SQR(something)+SQR(something)=SQR(something).

Since sine waves are curves you will ALWAYS be able to find an incorrect sum that can be corrected by squaring each term. (As before 1,2, or 3 of the terms may well be irrational).

Where do we go from here? Think triangle!

[TonyMcC]

In the early days of computer development the analogue computer had its following. One of ts attributes was the ability to handle irrational numbers both for input and output. The circuit shown is a simple analogue device which will accept two inputs (VA and VB) and produce an output (VC) which is SQR(VA^2+VB^2).

i.e. If VA=3 and VB=4 then VC=5

If VA=1 and VB=1 then VC= SQR(2)

If VA= X^n and VB= Y^n then VC= SQR(((X^n)^2)+((Y^n)^2)). Or VC= SQR(X^2n + Y^2n)

What simple change to this circuit could be made to make the output (VC) equal to the nth root of VA^n+VB^n ?

e.g. set VA=6, VB=8 and ensure VC= Cube root (6^3+8^3) instead of 10 ?

What are the implications for Fermat's Last Theorem?

Will soon get on to triangles!

Edited August 2, 2010 by TonyMcC

[TonyMcC]

Talking to myself here lol.

You could set VA to a 6V sine wave and VB to an 8V sine wave.

Having done so you could alter VC to any value between 14V and 2V (including, for example, cube root of VA^3+VB^3) in at least two seperate ways.

a) Have a variable phase shift and set it to somewhere between 0 degrees and 180 degrees.

Alternatively:-

B) Have a variable gain summing amplifier and set it to a value of less than 1.

This is another example of a "machine" that can provide you with the nth root of A^n+B^n BUT it demands 3 variable inputs!

Edited August 3, 2010 by TonyMcC

[TonyMcC]

.

It seems to me that you can think of sums such as 3+4=5 and 5+12=13 as incorrect sums that can be corrected by squaring each term.

It also occured to me that "Fermat" is saying that any such sum for any power above 2 must include at least one irrational term.

I wondered how incorrect sums could be analysed and decided that all relevant incorrect sums could be formed into triangles. e.g. since 6^3+8^3 approximately equals 9^3 then a triangle of sides 6,8 and approximately 9 can represent the relevant incorrect sum.

If you form enough of the triangles for a particular power you can produce a curve that can be used in the same way that a semi-circle can be used for power of 2.

To make the curve universal ( e.g. for power of 3) let the base be 1 and the other two sides be the cube roots of the two fractions that add to produce 1.

As an example cosider the correct sum 0.25+0.75=1 then the required triangle would have sides of approximately 0.63,0.91 and 1. Thus 0.63^3+0.91^3=(approx)1, hence 63^3+91^3=(approx)100^3.

Well,my illustration shows curves for powers 0f 2,3,4 and 5 and examining the curves doesn't help much!

It took me some time to realise that there is a big clue in the maths I employed to draw the curves (I wrote a short computer program). This realisation slowly formed in my mind as I considered practical uses for triangles. I feel I can provide strong evidence that the assertion made in Fermat's Last Theorem is true. Proof? - well that may too much to hope for, or perhaps not!

I have read a few books on the subject, but I am not aware of anyone else taking this approach.

Anyone out there worked along similar lines ?

Does my approach seem reasonable?

Edited August 7, 2010 by TonyMcC

[TonyMcC]

Does anyone dispute these statements:-

i) You can take two squares of the same thickness made from some malleable material and measure the sides (X and Y). You can mix the two together and form a single square of the same thickness and measure its side (Z). The sides will ALWAYS give you an equation X^2+Y^2=Z^2. So there is ALWAYS a solution.

ii) The only way to prove this mathematically is by means of geometry.

iii) This is because most solutions contain at least one irrational term.

iv) The only way to discover the relatively few whole number solutions is by some form of "trial and error".

Edited August 8, 2010 by TonyMcC

[TonyMcC]

.

 

If anyone is interested enough I would value criticism of the following line of reasoning:-

 

I ) An alternative way of stating Fermat’s Last Theorem is “There is no incorrect integer sum A+B=C that can be corrected by raising each term by a power other than 2. )”. For example 6+8=9 is an incorrect sum that is almost corrected by raising each term by the power of 3 (216+512=729).

 

II) Since both A and B must be smaller than C and their sum greater than C every relevant incorrect sum can be formed into a triangle.

 

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”. I discount scale drawing because of the impossibility of physically measuring irrational line lengths.

IV) If given 2 line lengths (say 8 and 9 ) it is not possible to draw a triangle with the third side equal to (8^3+9^3)^1/3 unless you can obtain the required length of the third side by means other than geometry or can determine the angle that joins the two given sides by means other than geometry..

 

V) Any given triangle can be converted into a parallelogram of forces.

For example a triangle of sides 6, 8 and 9 gives rise to a parallelogram of forces showing that a force of 6 units and a force of 8 units acting on a body when separated by an angle of approximately 101.5 degrees are indistinguishable from a single force of 9 units acting at an angle of approximately 61 degrees from the “6” vector and 41 degrees from the “8” vector.

 

VI) If an alternating current or voltage is considered to be providing a force then the value that must be used in any calculation is the rms value (root mean square).

 

VII) Any shape of regularly occurring waveform has an rms value. Different waveforms have different rms values. It is not possible to determine the rms value of a waveform unless you can determine its shape.

 

VIII) If you have two similar shaped waveforms of the same frequency (same periodic time) and introduce phase shift between them you can add them together to form a single composite waveform. The only composite waveform that has the same shape as its two original waveforms is a sine wave.

 

IX) A consequence of VIII above is that for sinewave addition the ratio of peak value and rms value is the same for all sinewave waveforms; the two original waveforms and the composite waveform. However in all other cases the ratio of peak value and rms value is different for the composite waveform than the ratio for the two original waveforms.

 

X) Thus for any condition other than pure sine waves the triangle representing rms value and the triangle representing peak value will be completely different.

 

XI) In order to maintain compatibility between triangles used to form parallelograms of forces ,phasor diagrams using rms values and phasor diagrams using peak values (all of which are valid and commonly used) all straight lines forming triangles must be directly related to phasor diagrams of sine wave shape.

Edited January 5, 2011 by TonyMcC