Fermat's last theorem

[slick]

I am a latecomer to this problem having watched the program about FLT and Andrew Wiles and I am no mathematician so I am looking for a bit of help understanding this. I have googled and read but most of the stuff is completely impenetrable for a layman. The thing that confuses me most is that the theorem seems fairly easy to 'prove' in descriptive visualisation terms so why is it so difficult to prove mathematically.

 

How I visualise it is as follows:

 

The basic premise is: aⁿ + bⁿ = cⁿ cannot be true where n is greater than 2. My starting point is to visualise pythagorean triples where a² + b² = c² is true. In this case I can describe three squares of which square a and square b will exactly fit into square c. If I now cube a, b and c I can see that although one face will be an exact fit the total volume of a and b must be less than the volume of c.

Notch it up to a⁴ + b⁴ = c⁴ and the differential can only get larger since I will now have c times as many large cubes but only a times as many a cubes and b times as many b cubes. Since n is incremental the differential can only get incrementally larger.

 

For instances where a² + b² < c² then again I can describe three squares, two of which (a and b will fit into c with some remainder. If I now cube a, b and c I can see that the cubes a and b will not only fit into one face of cube c but also then cannot reach or exceed the other faces of cube c. As before, since n is incremental, the volume of c can only get incrementally larger than the total volumes of a and b. Another picture shows what I mean

 

For instances where a² + b² > c² then I don't even need to describe squares since if the area of a and b exceed the area of c then the cube of a and b combined must be larger than the volume of c cubed. Again, since n is incremental, the volume of a and b will grow incrementally larger than the volume of c.

 

So, my questions:

 

What have I misunderstood about Fermat's last theorem?

What have I missed in my description?

If my descriptions are correct why is it difficult to represent mathematically?

[D H]

The numbers a, b, c, and n have to be positive integers. Without this restriction there are uncountably many solutions to aⁿ+bⁿ=cⁿ for any given position value of n.

 

If you do not understand the basics of the problem you will not understand Wiles' proof. I'll venture further: you will never understand it. Very few PhD mathematicians understand it.

[slick]

I was thinking about it last night and realised that there is a more fundamental problem since my third description - "For instances where a² + b² > c² then I don't even need to describe squares since if the area of a and b exceed the area of c then the cube of a and b combined must be larger than the volume of c cubed." is complete nonsense.

As regards "The numbers a, b, c, and n have to be positive integers." - I don't see that as a problem unless you have to allow for negative values of n. In other words, my understanding was that n had to be greater than 2, is that incorrect?

[TonyMcC]

D H is almost certainly correct in what he says. However if you think the journey itself may be interesting even if the destination seems always just out of reach you will join a happy band of pilgrims!

If Fermat really did prove what has become known as his last theorem it is accepted that his proof must have been much more simple than Andrew Wiles'.

If you use the forums search facility and tap in "Fermat" you will find some of us pilgrims. My particular thought at the moment is that he may have found a relatively simple proof that the solution to the nth route of A^n+B^n (n being larger than 2) must be irrational. It is quite easy, for instance, to prove that the square root of many numbers must be irrational by quite unsophisticated means.

If you can accept that this is an extraordinarily simple question with an extraordinarily difficult (probably impossible for the likes of you and me) solution then have a shot at it. Warning - you may become hooked!

[D H]

I was thinking about it last night and realised that there is a more fundamental problem since my third description - "For instances where a² + b² > c² then I don't even need to describe squares since if the area of a and b exceed the area of c then the cube of a and b combined must be larger than the volume of c cubed." is complete nonsense.

Yep, it is. Good that you recognized that.

As regards "The numbers a, b, c, and n have to be positive integers." - I don't see that as a problem unless you have to allow for negative values of n. In other words, my understanding was that n had to be greater than 2, is that incorrect?

That's correct. The important thing is that a, b, and c have to be integers. Relax this constraint and there are an uncountable number of solutions. For example, [math]3^3+4^3 = (\sqrt[3]{91})^3[/math]. However, since the cube root of 91 is not an integer, it is not a solution.

Edited December 20, 2010 by D H

[slick]

Many thanks for the input and advice. I have no intention of becoming hooked and I certainly didn't spend sleepless hours in bed last night thinking about it!

 

Okay, so I feel that I can describe this problem for instances where a² + b² < c² ora² + b² = c². The problem, as I see it, comes with a² + b² > c². Within that, if either a or b is greater than c then I can still describe the solution since c² will fit into either a² or b²,therefore the problem comes specifically with a² + b² > c² and neither a nor b is greater than c. In triangular terms, that would be any triangle where the angle opposite c is less than 90^o but greater than 60^o.

Some examples: 3ⁿ + 4ⁿ = 5ⁿ is okay since 3² + 4² = 5². 3ⁿ + 4ⁿ = 6ⁿ is also okay since 3² + 4² < 6². 10ⁿ + 5ⁿ = 8ⁿ is okay since 10² is greater than 8². 8ⁿ + 9ⁿ = 10ⁿ is where the problem starts since 8² + 9² > 10².

Does this make sense? Is there any value in reducing the problem or is this a well worn path?

[TonyMcC]

I don't think we are on quite the same wavelength but you might find the following action relevant:-

Search "Fermats Last Theorem Simple Ideas" and then look at entry 23. This shows curves for a, b and c for different powers of a^n+b^n=c^n. These were built up from triangles formed from a, b and c.

[slick]

I don't think we are on quite the same wavelength but you might find the following action relevant:-

Search "Fermats Last Theorem Simple Ideas" and then look at entry 23. This shows curves for a, b and c for different powers of a^n+b^n=c^n. These were built up from triangles formed from a, b and c.

 

It will take me a while to fully understand your posts but thanks for pointing me to them, there seems to be some resonance to what I have described. Specifically I can conclude from my description that if a, b and c cannot form a triangle then either a² + b² must be less than c² or a or b must be greater than c. In either case I can demonstrate FLT. If a triangle can be formed then for all triangles where the angle opposite c is less than 60^o or greater/equal to 90^o then I can demonstrate FLT. I will try to see how this relates to your curves.

[the tree]

That's correct. The important thing is that a, b, and c have to be integers. Relax this constraint and there are an uncountable number of solutions. For example, [imath]3^3+4^3 = (\sqrt[3]{91})^3[/imath]. However, since the cube root of 91 is not an integer, it is not a solution.

Well, if you relax the restraint a tiny bit and allow for rational a,b,c then there are still no solutions (if (a/b)³ + (c/d)³ = (e/f)³ then adf³ + cbf³ = ebd³) - I think there might be some room for relaxing the condition on n - perhaps even a way to get countably many solutions. Edited December 30, 2010 by the tree

[TonyMcC]

I wonder if Fermat was implying that there is no NON integer values for a, b and c for the equation a^n+b^n=c^n.

The reason I say this is that if you could obtain complete values in decimal form you could get a whole number equation by multiplication.

For eample let us say that 6^3 + 8^3 = 8.995883^3. If this was true then we could say 6000000^3 + 8000000^3 =8995883^3.

I think Fermat found (or thought he had found) that at least one of the values for a,b or c had to be infinitely long and this is the reason whole number values could not be obtained.

[the tree]

I wonder if Fermat was implying that there is no NON integer values for a, b and c for the equation a^n+b^n=c^n.

Well clearly not, because there clearly are - it doesn't take an particular knowledge to find them.

 

I think Fermat found (or thought he had found) that at least one of the values for a,b or c had to be infinitely long and this is the reason whole number values could not be obtained.

That is (sort of) what I meant, when I said that a,b,c could be rational and there would still be no solutions.

 

To be clear on that, a rational number x is one that can be described as x=a/b where a and b are integers and b is not 0.

 

Every rational number has either a finite or an eventually periodic decimal expansion, and every number with a finite or periodic decimal expansion is rational.

 

For eample let us say that 6^3 + 8^3 = 8.995883^3. If this was true then we could say 6000000^3 + 8000000^3 =8995883^3

Not quite. (6x1000000)³+(8x1000000)³=1000000³x(6³ + 8³), so your hypothetical example (as we've covered, is impossible anyway) is out by a factor of 1000000².

[TonyMcC]

Firstly - Happy New Year!

the tree - I would like to concentrate on your last comment as mathematics is not my strong point.

However I can't see a flaw in my mathematical reasoning and would appreciate any required correction.

 

6^3 + 8^3 = 728

= 8.995883^3 ( Approx of necessity!)

 

An acute angled triangle of sides 6, 8, 8.995883 could be drawn to represent the equation (same idea as right angled triangle)

 

If all sides are multiplied by the number of digits following the longest number of digits after any decimal point we can make a similar triangle of sides 6000000, 8000000, 8995883 (i.e. whole numbers)

Because of this similarity we can say :-

6000000^3 + 8000000^3 = 8995883^3 (Approx of necessity)

7.28E20 = 7.2800002E20 (difference due to approximation)

 

This is an example of what I posted in "Fermats Last Theorem - simple ideas entry 23"

 

It seems to me that if you can present me with an equation using finite numbers following all three decimal points I could easily give you a whole number solution.

 

I would like to consider your other points further before replying - thank you for your interest.

Edited January 1, 2011 by TonyMcC

[D H]

The flaw in your reasoning, Tony, is that [math]\sqrt[3]{728}[/math] is not a rational. There are no rational solutions (a, b, c) to [math]a^n+b^n=c^n[/math] where n is a positive integer > 2.

[slick]

Happy new year! I have been having a bit of fun with this over the holiday break and I am getting nowhere quite slowly. My lack of knowledge seems to be a bit of a stumbling block for some reason but it has been fun re-learning basic trig so I thought I would give an update on where I have got to.

 

My conclusions so far are that there is only one scenario that is problematic: where a² + b² > c² and neither a nor b are greater than or equal to c. For the other scenarios such as a² + b² = c² ora² + b² < c² or where either a or b are greater or equal to c then it is quite easy to see that aⁿ + bⁿ cannot = cⁿ. I concluded that the difficult scenario could be represented as any triangle of three integers where the apex of a and b is in the shaded area in the picture below. This is any triangle with an angle C between 60^o and 90^o.

 

 

 

I then started endless messing about with the area of the triangle (which would have to be 0 for a positive result), the height of the triangle (which would also have to be 0), the angle C (which would have to be 180^o) and various other permutations. The area/height option looked promising since zero is such a good target but I couldn't really find any other way to predict the expansion/contraction other than by standard equations and there would be little point proving that the area can only be zero if aⁿ + bⁿ = cⁿ, it kind of gets things the wrong way round.

 

So now I am looking at the basic equation a²ⁿ + b²ⁿ - c²ⁿ / 2ab which must equal -1 for a positive result. What I have seen is the progressive value of the result is an arc (not surprising) which passes through -1 but I cannot see yet how I can determine why it will never equal -1. The four graphs below show the progression of the result for n = 2 to 54 for four sets of values. I am also finding it difficult to get good results since my laptop seems to run out of computing ability once I get past n = 50 for values such as 99,99,100 for which I have yet to get to -1.

 

 

 

What I think I really need now is a different way to calculate the expansion/contraction of the area of the triangle. Any suggestions?

[TonyMcC]

The flaw in your reasoning, Tony, is that [math]\sqrt[3]{728}[/math] is not a rational. There are no rational solutions (a, b, c) to [math]a^n+b^n=c^n[/math] where n is a positive integer > 2.

D H - That is exactly why I had to use an approximation. If what you say can be proved then surely Fermat's claim is justified when looked at in the way I describe?

Edited January 1, 2011 by TonyMcC

[D H]

Surely not.

[TonyMcC]

Well, as I see it, if you tell me that c is irrational and a and b are rational in a^n + b^n = c^n (n>2) then it follows that c is a number followed by an infinite number of digits after the decimal point. Either that or we are dealing with infinite numbers!

You can imagine drawing a triangle using the values of a, b and c but you would have to multiply all sides of the triangle by an infinite amount to achieve a whole number similar triangle.

You can prove with geometry that if a^2 + b^2 = c^2 then c(a^2) + c(b^2) = C^3. (Which won't come as a surprise!), but doing this provides enough data to produce a different triangle appropriate to a^3 + b^3 = c^3. In fact this is the way I drew my curves in" Fermats Last Theorem - simple ideas, post 23".

[TonyMcC]

Well clearly not, because there clearly are - it doesn't take an particular knowledge to find them.

 

That is (sort of) what I meant, when I said that a,b,c could be rational and there would still be no solutions.

 

To be clear on that, a rational number x is one that can be described as x=a/b where a and b are integers and b is not 0.

 

Every rational number has either a finite or an eventually periodic decimal expansion, and every number with a finite or periodic decimal expansion is rational.

 

Not quite. (6x1000000)³+(8x1000000)³=1000000³x(6³ + 8³), so your hypothetical example (as we've covered, is impossible anyway) is out by a factor of 1000000².

 

Hi tree (no pun intended!),

Thinking about your first point, I think the following thought experiment (or fairy story) is in agreement with Fermat's comments (whether he could prove it or not is another matter):-

 

You have two cubic containers of liquid and wish to fit them exactly into a single cubic container.

You come across an elf who specialises in making cubic containers. In fact he has spent eternity making such containers and in his yard he has in infinite number, each of different size (He also has an infinite sized yard of course!).

You and the elf agree that there must be a container in his yard that is the correct size.

He starts to find the required container by trial and error. If he selects one that is too large then he tries a smaller one and vice-versa.

After spending more time than you will ever have (for you will have ceased to exist eons ago) he finds he has two containers one is too large and one is too small although the size difference is infinitely small! The elf mutters to himself "This doesn't seem to make sense!"

 

I would very much like to see an example of a^n + b^n = (c/d)^n where a, b, c and d are whole numbers.

[the tree]

Well, as I see it, if you tell me that c is irrational and a and b are rational in a^n + b^n = c^n (n>2) then it follows that c is a number followed by an infinite number of digits after the decimal point.

Yes. Not exactly the most rigorous, formal or unambiguous phrasing in the world - but yes.

 

The elf mutters to himself "This doesn't seem to make sense!"

Well it just does. Limits and irrational numbers and all of that are well established, well defined and possibly more exposed to "excessive rigour" than any other object in mathematics. The fact that they don't stretch neatly into the physical realms with elves and such like is sort of secondary or beside the point.

 

I would very much like to see an example of a^n + b^n = (c/d)^n where a, b, c and d are whole numbers.

Well you wont, as we've established - such a thing doesn't exist.

[slick]

TonyMcC, I was thinking about irrational numbers last night. One of the things that occurred to me while messing about with triangles, although it may be incorrect, is that I could reduce the scale down such that either a or b are equal to 1 which would mean that aⁿ + 1 cannot = cⁿ where a and c are rational numbers and n is an integer > 2. In other words, take any rational number and subtract 1. For any root of the first number, the resulting root of the second number must be irrational however I don't know enough to be able to test whether or not this is true.

[TonyMcC]

Slick - firstly let me say I am no mathematician. However just as you can use a semi-circle to find a, b, and c in the equation a^2 + b^2 = c^2 you can use an appropriate curve to find a, b, and c in the formula a^n + b^n = c^n. I attach curves for n = 2 to 5.

Interestingly, if I had drawn curves for n=2 to n=infinity they would fill the shaded area in your diagram.

It is accepted that many square roots are irrational. It seems to me that we can expect that often a, b, or c will be irrational in the general formula a^n + b^n = c^n. However, it seems to me that Fermat may have proved (or thought he had proved) that ALWAYS at least one of the three terms a, b, or c will be irrational if n is a whole number >2.

As you can read above, "D H" seems to think that is the case and "the tree" seems to think that it may not be the case. In other words the idea is cotentious.

I agree that a^n +1 = c^n cannot have an integer solution.

Sorry about repeated text - I cocked up attaching the diagram.

 

Slick - firstly let me say I am no mathematician. However just as you can use a semi-circle to find a, b, and c in the equation a^2 + b^2 = c^2 you can use an appropriate curve to find a, b, and c in the formula a^n + b^n = c^n. I attach curves for n = 2 to 5.

Interestingly, if I had drawn curves for n=2 to n=infinity they would fill the shaded area in your diagram.

It is accepted that many square roots are irrational. It seems to me that we can expect that often a, b, or c will be irrational in the general formula a^n + b^n = c^n. However, it seems to me that Fermat may have proved (or thought he had proved) that ALWAYS at least one of the three terms a, b, or c will be irrational if n is a whole number >2.

As you can read above, "D H" seems to think that is the case and "the tree" seems to think that it may not be the case. In other words the idea is cotentious.

I agree that a^n +1 = c^n cannot have an integer solution.

Edited January 2, 2011 by TonyMcC

[slick]

 

I agree that a^n +1 = c^n cannot have an integer solution.

 

 

Thanks but do you think that it can also be true for a and c being positive rational numbers?

[TonyMcC]

Thanks but do you think that it can also be true for a and c being positive rational numbers?

I was assuming a, c and n were positive integers. I don't think there is a whole number solution.

[the tree]

I agree that a^n +1 = c^n cannot have an integer solution.

Thanks but do you think that it can also be true for a and c being positive rational numbers?

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

 

However, it seems to me that Fermat may have proved (or thought he had proved) that ALWAYS at least one of the three terms a, b, or c will be irrational if n is a whole number >2.

As you can read above, "D H" seems to think that is the case and "the tree" seems to think that it may not be the case. In other words the idea is [contentious].

How exactly did you infer, from me pointing out that I'd said it first, that I was disagreeing? Edited January 2, 2011 by the tree

[D H]

One of the things that occurred to me while messing about with triangles, although it may be incorrect, is that I could reduce the scale down such that either a or b are equal to 1 which would mean that aⁿ + 1 cannot = cⁿ where a and c are rational numbers and n is an integer > 2.

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.

 

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

• Fermat never did have his mystery proof. He worked on the cases n=3 (successfully), n=4(also successfully), and n=5 (not successfully) after he wrote his famous note in the margin. Now if he did have a universal proof, why did he work on n=3, 4, and 5 after the fact? Why did he fail in his endeavors to prove the conjecture for n=5?
• A huge number of mathematicians, all much more well-versed in number theory than any amateur, worked on the problem and failed. It is the height of hubris for someone lacking any training in number theory to think they can prove something as hairy as this theory turned out to be.

 

What I said above, that you will not find a simple proof of Fermat's Last Theorem, is not really true. Conceptually, the proof of Fermat's Last Theorem is now trivial. Here it is: If the Taniyama-Shimura conjecture is true, then Fermat's Last Theorem is true per Ribet's theorem. The Taniyama-Shimura conjecture is true by the modularity theorem. QED.

 

What does that mean? Not much, at least not to me. I am not a PhD mathematician who specialized in number theory. If you want to understand the proof you will need to pursue higher mathematics. Pure math, not applied math.

Edited January 2, 2011 by D H