Fermat's Last Theorem with hardly any Mathematics

[Joatmon]

Someone claims to have a whole number solution to A^n+B^n=C^n where n is greater than two.

If they have the three numbers then they can form a triangle. This triangle will always have an apex (junction of A and B) which will be less than 90 degrees. Please consider whether you agree with this statement before proceeding!

This triangle can be considered as a triangle of forces and from which a parallelogram of forces can be drawn.

Using the parallelogram of forces both a vector diagram and a phasor diagram can be drawn and used as reliable tools for navigators and engineers of all disciplines.

A phasor diagram can only be used for adding sine waves and they must be of the same frequency. By definition each point on a sine wave is a multiple of the square root of a proper fraction. (opposite divided by hypotenuse)

If you plot a graph or create a waveform where each plotted point is something different to the square root of the fraction (say the cube root) you will have a graph that is not a pure sine wave.

Only a sine wave has the property of adding two sine waves to make another sine wave. Any other shape of waveform or regularly repeating graph has to be considered as a group of sine waves of different frequencies acting together( fundamental frequency and harmonics). This is Fourier analysis of a waveform.

So each frequency that makes up the waveform that is not a pure sine wave, requires its own phasor diagram, hence its own triangle, and you would need to analyse a full set of these triangles to form the final total. But it was stated earlier that a single triangle would suffice!

The rather strange conclusion that I come to is that s each side of a triangle, where each side is an integer, is the square root of something and the triangle gives a correct addition of these square roots on a plane surface. Only triangles with an apex angle of 90 degrees will form a a phasor or vector straight line if all sides are squared.

The case was made earlier that all triangles that represented A,B,C, in the formula A^n+B^n=C^n where n is greater than 2 will not contain a right angle. That all such triangles can represent a vector or phasor diagram on a plane surface and that the only chance you have of putting the vectors or phasors (which represent sine waves of one frequency) in line is to square each side. It will always fail to do so if the triangle does not contain a right angle.

So the rather strange conclusion is that it is only possible to construct a triangle, where each side is the square root of a number. The the only way of using the triangle to represent a sum is to square the sides. If the triangle is right angled we get a correct sum and if the angle is less then 90 degrees we get an incorrect sum..

So, for example a triangle of sides 6,8,9 which seems to almost represent 6^3+8^3=9^3 represents two phasors separated by approximately 110 degrees, which provide the incorrect sum 6^2+8^2=9^2.

I see this as strong evidence, perhaps even a logical proof, that Fermat's last Theorem is true by absurdum.

[imatfaal]

Nope. Three numbers 1 2 10 - draw the triangle

[Joatmon]

Perhaps I should have realised that someone might think he had a possible solution that was 1^n +2^n=10^n!

I thought the point was covered in the first line. To be clear I should have asked the reader to also satisfy themselves that in all possible candidates for A^n+B^n= C^n, C will be larger than A or B and smaller than A+B.

,

[imatfaal]

Perhaps I should have realised that someone might think he had a possible solution that was 1^n +2^n=10^n!

I thought the point was covered in the first line. To be clear I should have asked the reader to also satisfy themselves that in all possible candidates for A^n+B^n= C^n, C will be larger than A or B and smaller than A+B.

,

 

Yeah -- I go along with that; I was being a bit flippant

[Carrock]

By definition each point on a sine wave is a multiple of the square root of a proper fraction. (opposite divided by hypotenuse)

How about an e, pi, sqrt(e^2 + pi^2) triangle?

I.e. all points between -x and +x are points on a sine wave.

[Joatmon]

I don't see a problem, but perhaps the point you raise was why Fermat stipulated integers, not irrational numbers.

 

However if you could get absolutely accurate values for e and pi then the following would, IMO, be true:-

One of the sine waves to be added would peak at e times 1 (sine 0 degrees), be e times .707 at 45 degrees and e times 0 at 90 degrees.

The other sine wave to be added would similarly peak at pi times 1, be pi times .707 at 45 degrees and 0 at 90 degrees.

If these two are in phase and added the resultant would peak at e+pi and if one is phase shifted with respect to the other the resultant waveform will have a lower peak value. When the resultant phase shift is the correct value it will be SQR(e^2 +pi^2). This will happen at a phase shift of 90 degrees. Therefore squaring each side of the triangle will produce a correct sum e^2+pi^2=(SQR(e^2+pi^2))^2.

 

NB *707 is approx SQR(1/2) .

[Carrock]

By definition each point on a sine wave is a multiple of the square root of a proper fraction. (opposite divided by hypotenuse)

I was just questioning your definition.

It is generally accepted sin x can take any value of y such that -1 <= y <= 1.

Your definition excludes e.g. the infinite number of transcendental numbers between -1 and 1.

 

I don't see a problem, but perhaps the point you raise was why Fermat stipulated integers, not irrational numbers.

 

However if you could get absolutely accurate values for e and pi then the following would, IMO, be true:-

One of the sine waves to be added would peak at e times 1 (sine 0 degrees), be e times .707 at 45 degrees and e times 0 at 90 degrees.

The other sine wave to be added would similarly peak at pi times 1, be pi times .707 at 45 degrees and 0 at 90 degrees.

If these two are in phase and added the resultant would peak at e+pi and if one is phase shifted with respect to the other the resultant waveform will have a lower peak value. When the resultant phase shift is the correct value it will be SQR(e^2 +pi^2). This will happen at a phase shift of 90 degrees. Therefore squaring each side of the triangle will produce a correct sum e^2+pi^2=(SQR(e^2+pi^2))^2.

 

NB *707 is approx SQR(1/2) .

sin x can't be greater than 1.

 

 

 

In a proof like this, I stop analysing when I encounter (IMO) an unequivocal error as later reasoning may be based on this error.

It's certainly worth looking for a simpler proof of Fermat's last theorem than Andrew Wylie's, but I doubt any such proof will be much simpler.

[Joatmon]

Thank you Carrock for considering my ideas concerning Fermat's last theorem.

Concerning the points you raise you can plot a sine wave by dividing a straight line of numerical value 1 by any number of fractions, comprise a table of the square root of each fraction and make these the points on your graph. In theory you could use an infinite number of plots for an infinite number of fractions.

You can use this data to go up and down and then down and up to produce one cycle of a sine wave. That will cover all possibilities concerning Fermat's mention of integers.

 

I don't understand the comment sin x cannot be greater than 1. I said that the maximum value of sin x was 1 and if all the fractions were multiplied by a greater number you would produce a sin wave with a peak value of that greater number.

 

I'm sorry if my use of IMO put you off. Of course I would like you to look beyond that and let me know where you feel I have gone wrong if you see a misconception. That would be a real help.

Edited March 18, 2017 by Joatmon

[Carrock]

Thank you Carrock for considering my ideas concerning Fermat's last theorem.

Concerning the points you raise you can plot a sine wave by dividing a straight line of numerical value 1 by any number of fractions, comprise a table of the square root of each fraction and make these the points on your graph. In theory you could use an infinite number of plots for an infinite number of fractions.

You can use this data to go up and down and then down and up to produce one cycle of a sine wave.

No you can't.

You can define a countable infinity of points in this way but between each two points there is a (much larger) uncountable infinity of points which you haven't defined.

You can't assume a smooth function when most of the points are undefined.

 

(Near the top top right of Cantor's diagonal argument there is a diagram showing how to generate numbers/points that are not included in your sequence.)

I don't understand the comment sin x cannot be greater than 1. I said that the maximum value of sin x was 1 and if all the fractions were multiplied by a greater number you would produce a sin wave with a peak value of that greater number.

One of the sine waves to be added would peak at e times 1 (sine 0 degrees), be e times .707 at 45 degrees and e times 0 at 90 degrees.

The other sine wave to be added would similarly peak at pi times 1, be pi times .707 at 45 degrees and 0 at 90 degrees.

You seemed to be talking about cosine waves rather than sine waves and my critical faculties switched off at that point!

 

By definition each point on a sine wave is a multiple of the square root of a proper fraction. (opposite divided by hypotenuse)

This is really the claim I can't accept and where I stop.

 

(I've no problem with 'IMO' Even if I was an 'expert' I'd still sometimes get things wrong.)

[Joatmon]

Thank you for giving this thread your thoughts, it is appreciated.

I must think about what you say about the infinity of numbers that exist in the infinitely small space that exists between each of the infinite set of numbers. My feeling is that when it comes to plotting graphs there won't be any surprising jumps - just a continuing of a smooth transition. For example it is surprisingly easy to "show" the square root of two on a plane surface and it fits into the general scheme of things. BTW the link you gave for Cantor's Diagonal Argument took me to the wrong place, but google soon sorted that out so I'll have a slow think about it.

 

Well yes, concerning the cosine wave - I wasn't bothered about that as they have the same shape and can be plotted using the same data. On phasor diagrams etc they are 90 degrees out of phase with each other but adding two cosine waves will produce another cosine wave so this doesn't affect my argument. I'm sorry my laziness put you off.

 

As I round off this post it strikes me to mention that the sine function is a natural product of nature that is follows things like vibrating strings,electromagnetic waves and swinging pendulums. This leads me to conclude that regardless of any number system the progression is continuous and smooth.

 

Anyway, thanks again for your time and your thoughts, I haven't given up yet on what is something of a hobby