Insights Fermat's Last Theorem

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  • #51
TensorCalculus said:
Oh, okay. Thanks that does clarify it - so this is probably not the book I am looking for.
Maybe I'll just find some resources online that explain ring theory and Eisenstein integers, without proving the n=3 case (since I want to do that myself). This is proving much more difficult than I could have even imagined... it's so interesting that such a simple statement requires such (in the eyes of a student like me) complex methods to prove...
Here is a pdf that looks very basic to start with. I would skip the section of continued fractions unless you want to read original Euler papers, but it looks quite interesting.
https://resources.saylor.org/wwwres...-Introductory-in-Elementary-Number-Theory.pdf

Search key: "Introduction to elementary number theory + pdf".

Here is one about Eisenstein numbers. How difficult depends on where you want to stop reading it.
https://www.diva-portal.org/smash/get/diva2:1728553/FULLTEXT01.pdf

I searched but couldn't find FLT for n=3 in it, so it should be safe to read, and it has many interesting facts about Eisenstein numbers. But be warned, things can get complicated. I'm not sure whether I would confront a young man with characters on his first encounter with number theory; however, they are an essential tool.

Search key: "Eisenstein numbers + pdf"

Maybe I should have used "Introduction to Eisenstein numbers + pdf" instead.

Only the "+ pdf" part is really important. You get lecture notes and stuff like that rather than colourful websites.
 
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  • #52
Ah yes I remember you mentioning searching for pdfs on another thread (and it works!) Now time to read them through :)

thank you!
 
  • #53
fresh_42 said:
TL;DR Summary: Why it took 350 years and a genius to prove Fermat's Last Theorem.

Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation
$$
a^n+b^n=c^n
$$
has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy

"Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duas ejusdem nominis fas est dividere: cujus rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet."

"It is not possible, however, to decompose a cube into two cubes, or a biquadrate into two biquadrates, and in general, to decompose a power higher than the second into two powers with the same exponent: I have discovered a truly wonderful proof for this, but this margin is too narrow to grasp here."

which has been found after his death. Everybody understands the problem statement, however, it turned out to be extraordinarily difficult to solve, and nobody today assumes that Fermat had actually found a proof for the general case. For example,
$$
6^3+8^3=9^3-1,
$$
is almost a solution for the cubic case. Imagine how many natural numbers we have to rule out as possible solutions. Numbers we cannot even write down in a lifetime. Such considerations aren't even evidence in number theory since exceptions can always occur, but it shows the principal difficulty of any proof of negative results. Non-existence is hard to prove.

Both aspects of the story - the simplicity of the statement and Fermat's provocative remark - were a blessing as well as a curse. A blessing because, during the 350 years it took before Andrew Wiles and Richard Taylor succeeded in providing a complete proof in 1994, extraordinary developments in mathematics were initiated in the effort to prove the statement, especially in number theory, abstract algebra, class field theory, and the theory of elliptic curves. Its curse, however, lies in the fact that to this day, even after the more than 100-page solution to the problem, full of elaborate mathematics, laypeople still attempt to prove the theorem using simple means. It's safe to say that such attempts are doomed from the start. This article aims to shed light on why this is the case.

Continue reading ...
One thing I have never understood about this problem... this is the case in point below... I always thought these two lines representing all Fermat cases was a simple solution and I genuinely thought this is what Fermat meant in the margin comment...

We have two equations. One as simple as it gets. And a second one just a tiny bit more interesting than that.

a^2 + b^2 = c^2

For all integers.

This is the case for all Pythagorean triples. Any c^3 or c^4 or c^n that is a solution for higher powers is a multiple of the right hand side by c ( by c, c^2, c^3, c^4, etc), it is simple to prove that this will never work for any power, when it starts a^2 + b^2. That's child's play.

ALL other integers must be in the format below if not above...

a^2 + b^2 = c^2 + k where k is an integer

The second equation represents all integers which are not the sum of two squares being multiplied by c to raise c from a power of two to any power higher. We can choose any c that would be a third power, a fourth power, or a fifth power, etc and this one equation covers the ones that are not the sum of two squares before we do that.

These two equations represent ALL integers. BOTH are easy to prove. NEITHER needs worry about powers above 3, since the case for 3 can be extended by induction.

What is the flaw in the above?
 
  • #54
Office_Shredder said:
It seems exceedingly unlikely because most of the techniques Wiles used were from branches of math invented well after Fermat died.
If you understand why the Liar's Paradox is nonsense, it goes a long way to show why maths in it's entirety needs to be rewritten in a context free language.
 
  • #55
noelbcornerstone said:
NEITHER needs worry about powers above 3, since the case for 3 can be extended by induction.
There are inifinitely many Pythagorean triplets (##n=2##) and no solution for ##n=4##, why? Euler needed two articles to cover the case ##n=3,## which is a sure indicator that it is not trivial.

But even if we accept the case ##n=3## as given, there is no such thing as an induction. Any induction step fails.
 
  • #56
fresh_42 said:
Things become more difficult for the ##p=3## case. The basic idea is to translate the additive problem
$$
a^p+b^p=c^p
$$
into a multiplicative problem by writing it as
$$
c^p=a^p+b^p=(a+b)\cdot(a+\zeta b)\cdot (a+ \zeta^2 b)\cdot \ldots \cdot (a+\zeta^{p-1}y)
$$
where ##\zeta## is a root of ##x^p-1=0.## This is a complex number. The crucial point, however, is that it is a polynomial equation with coefficients in the ring ##\mathbb{Z}[\zeta]## as @martinbn has mentioned in post #7. Now we have a multiplicative equation in a ring. This ring has primes. A prime is a number that, if it divides a product, then it has to divide one of its factors. E.g. ##6## divides ##3\cdot 4## but it doesn't divide ##3## or ##4,## so it is not prime. On the other hand, ##3## divides ##2\cdot 6## so it has to divide either ##2## which it does not, or ##6,## which it does divide. And this is true for any factorization of ##1\cdot 12=2\cdot 6= 3\cdot 4,## i.e. ##3## is prime. Hence, if a prime element of ##\mathbb{Z}[\zeta]## divides ##c^p## then it has to divide one of the factors ##a+\zeta^kb## for some ##k<p.##

This is the basic idea.

In the case ##p=3## we may set ##\zeta=\dfrac{1}{2}+ i\dfrac{\sqrt{3}}{2}.## Then ##
\zeta^2=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}## and ##\zeta^3=-1\, , \,\zeta^2-\zeta=-1.## Now,
$$
(a+b)(a-b\zeta)(a+b\zeta^2)=(a+b)\left[a^2-ab+b^2\right]=c^3.
$$
##\mathbb{Z}[\zeta]## has unique prime factor decompositions and is a Euclidean integral domain. Its elements all can be written as ##x+y\zeta +z\zeta^2## with integers ##x,y,z.## However, all numbers ##m^2+3n^2## are reducible, i.e. ##3## is not prime. Its units are all sixth roots of unity, i.e., solutions to ##x^6=1.## All positive integer primes ##p\equiv 2\pmod{3}## are prime in ##\mathbb{Z}[\zeta],## too. All positive integer primes ##p\equiv 1\pmod{3}## are a product of two complex conjugate primes in ##\mathbb{Z}[\zeta].## All those, ##1-\zeta^2=2-\zeta,## and their associates are the prime elements of ##\mathbb{Z}[\zeta].## Associate means up to units. Units are those elements that can be inverted. Of course, we can always multiply units to a number without changing its primality. ##-3## is also a prime integer, however, not in ##\mathbb{Z}[\zeta]## anymore. Anyway, this ring has all the nice properties I listed and allows us to consider divisibilities. These properties are not obvious, so either you trust me or you have to prove them.

The point is, that with primes, a unique prime factor decomposition, and a Euclidean algorithm, a norm function ##N(a+y \sqrt{-3})=x^2+3y^3## to distinguish between larger and smaller numbers, we have all ingredients for number theory on ##\mathbb{Z}[\zeta]## and the investigation of the prime factors of ##c^3## in the now multiplicative equation.

##\mathbb{Z}[\zeta]## are called Eisenstein numbers in case you want to look it up. Also, I think the first source I listed (D. Zachow) has more about Eisenstein numbers and the rings ##\mathbb{Z}[\zeta]## and a proof for the case ##p=3,## but it is in German.

Ok, can we please stop this and start with what I said. You are walking down the wrong way, and you have asserted there is no way by induction without a proof. Please follow the below...

All integers that are candidates for ANY solution a^n + b^n = c^n can be rewritten in the form...

1) a^2 + b^2 = c^2

or

2) a^2 + b^2 = c^2 + k

Can we agree these two equations cover every potential solution to Fermat's last theory. If there is an a^3 + b^3 = c^3 solution or a solution with the power 4,5,etc, can all be rewritten to one of the two equations above. i.e. EVERY number that isn't the sum of two squares can be written as the second equation. It really isn't rocket science. This is the first step. We can represent any of the potential solutions without needing to raise any powers above two. Does THAT at least make sense?

If you cannot understand that 1) and 2) cover all integers, can you please explain why?
 
  • #57
noelbcornerstone said:
Ok, can we please stop this and start with what I said. You are walking down the wrong way, and you have asserted there is no way by induction without a proof. Please follow the below...

All integers that are candidates for ANY solution a^n + b^n = c^n can be rewritten in the form...

1) a^2 + b^2 = c^2

or

2) a^2 + b^2 = c^2 + k

Can we agree these two equations cover every potential solution to Fermat's last theory. If there is an a^3 + b^3 = c^3 solution or a solution with the power 4,5,etc, can all be rewritten to one of the two equations above. i.e. EVERY number that isn't the sum of two squares can be written as the second equation. It really isn't rocket science. This is the first step. We can represent any of the potential solutions without needing to raise any powers above two. Does THAT at least make sense?

If you cannot understand that 1) and 2) cover all integers, can you please explain why?
How does this help?
 
  • #58
That is Step 1.

Once we agree that any solution of Fermat's Last Theorem can be written

as

1) a^2 + b^2 = c^2

and

2) a^2 + b^2 - k = c^2

Then by induction for 1 we show any integer solution to FLT must be of the form:

1) c^n(a^2 + b^2) = c^n(c^2)

or

2)c^n(a^2 + b^2 -k) = c^n(c^2)

where n is an integer greater than 0 or we have just the solution for an indice of 2

To raise the indices of a^2 + b^2 by 1, c is required to multiply a and b by different amounts if a and b are not equal, assuming we understand how exponentiation works, so a must equal b for a solution in equation one for a power greater than 2. That covers one set of integers and explains why given all the sum of two squares, why there are NO sums of powers of 4. c^2 would need to multiply a and b by different amounts, any increase in indice for a,b and c does. VERY simple to show for 1)

For 2) To raise indice a and b by one or more, we need an integer for c that multplies -k out and multiplies a and b by different amounts, we cannot do this for 3, and if we could do it for 3, we can show it would mean we couldn't do any higher indices.

This is a simple proof by induction when all written out, without needing the nonsense from before. It looks like this was hard for people because they don't understand that ANY equation can be written in lower powers like this in two forms that covers all possibilities.

I think the nonsense that was spawned from this conjecture just shows how broken maths is and how poorly the language is fit for purpose and how few real mathematicians there are left in the world. This is child's play and simple. I only looked at it for a laugh a couple of days ago, as it was simple to prove and we are looking for something else of ours to go viral, so I thought that showing why maths was broken and rewriting it so we could then solve integer factorisation, would start here. We will be posting all the factors of unfactored RSA numbers too. It's related to why mathematicians struggle with this and also N vs NP, and auto-innovators etc. Start with the two equatons above and figure out why it covers any solution that can exist. This simple change to the way you think will help.

Good luck!
 
  • #59
noelbcornerstone said:
That is Step 1.

Once we agree that any solution of Fermat's Last Theorem can be written

as

1) a^2 + b^2 = c^2

and

2) a^2 + b^2 - k = c^2

Then by induction for 1 we show any integer solution to FLT must be of the form:

1) c^n(a^2 + b^2) = c^n(c^2)

or

2)c^n(a^2 + b^2 -k) = c^n(c^2)

where n is an integer greater than 0 or we have just the solution for an indice of 2

To raise the indices of a^2 + b^2 by 1, c is required to multiply a and b by different amounts if a and b are not equal, assuming we understand how exponentiation works, so a must equal b for a solution in equation one for a power greater than 2. That covers one set of integers and explains why given all the sum of two squares, why there are NO sums of powers of 4. c^2 would need to multiply a and b by different amounts, any increase in indice for a,b and c does. VERY simple to show for 1)

For 2) To raise indice a and b by one or more, we need an integer for c that multplies -k out and multiplies a and b by different amounts, we cannot do this for 3, and if we could do it for 3, we can show it would mean we couldn't do any higher indices.

This is a simple proof by induction when all written out, without needing the nonsense from before. It looks like this was hard for people because they don't understand that ANY equation can be written in lower powers like this in two forms that covers all possibilities.

I think the nonsense that was spawned from this conjecture just shows how broken maths is and how poorly the language is fit for purpose and how few real mathematicians there are left in the world. This is child's play and simple. I only looked at it for a laugh a couple of days ago, as it was simple to prove and we are looking for something else of ours to go viral, so I thought that showing why maths was broken and rewriting it so we could then solve integer factorisation, would start here. We will be posting all the factors of unfactored RSA numbers too. It's related to why mathematicians struggle with this and also N vs NP, and auto-innovators etc. Start with the two equatons above and figure out why it covers any solution that can exist. This simple change to the way you think will help.

Good luck!
All this is nonsense!!

If what yiu wrote made sense it would work for n=2 as well.

Suppose there is a solution to a^2+b^2=c^2, writw it as

1) a+b=c

Or

2) a+b=c+k

Multiply by c and run the same argument. Thus you have proven that there no pythagorian triples.
 
  • #60
One tip for anyone on this thread. The result of what we are doing with integers and integer factorisation will remove the ability to use any form of modular exponentiation, or elliptic curve prime field for encryption. It is very likely the outcome of our work will result in it not being possible to use AES or anything like it for encryption or one way hashing (i.e. if I know what I need to hash to, I can pick any of the numbers that hash to it easily). We are not seeing a mathematical way to encrypt or hash one way after we publish.

This will result in the collapse of bitcoin and other coins in likelihood since proof-of-work is trivialised.

We are not the only party that has figured this out. It is likely actors are going to rug pull bitcoin for sure in the near future.

Example: RSA 100. if we take (d+n)(d+n) - (x+n)(x+n) = c (diff of two squares, d is the root of c)

See that 2a/x is almost exactly equal to x/n, it is only the remainder that fills in the blanks.

There is a class of integers like this. Essential the ratio of the remainder of dividing x by n is in proportion exactly to the ratio of dividing 2a by x. And 2d is 2a + 2x. So for RSA100 we can divided 2d by n. We don't know what n is so we use the number 1 to 10,000 as the amount of n in 2d. We get to 5424. Easy compared to the general number field sieve. At this number we have a chunk of the start of numbers...

a must start with 379... b must start with 400... x must start with 1054 and n must start with 1438... and theta must start with 72.655 (where theta is 2a/x and x/n rememeber that for all c, that if 72.655 x = 2a, then 74.655.. will equal 2d, since (73.655... x 73.655... x n ) squared equals c.

What is so good about that?

The values of x, n, and theta are not random and only possess certain characteristics that are related to the golden ratio, the squre root of two (you can see that in 1438... for n) and the decimal unit 1.

Because these numbers are fixed, it is very simple and very quick to figure out the expansion of the natural log of c, and which are the squares in the exponential function definition that are the ones that give the exact result for c, half of that exponent is obviously the square root of c, and if we know the five contributors in the natural log of c (five dimensions is always the max needed for any problem), then like finding the place of Pi without those in front, we spigot the solution. Works best for the product of two primes.

There are a fixed number of theses types of integer and it's like whether the square is odd and the remainder is even, the other way round. If n is smaller or larger than x and how does that affect theta. And the 'balance number' for that type of integer. In other words, what is the constant that all our variable must equal. This is what maths has been missing for integer factorisation. That pattern that is fixed. The larger the numbers are the easier it is to find.

What we are seeing evidence of is that P = NP. If a problem statement is complete, it is also the solution.
 
  • #61
noelbcornerstone said:
One tip for anyone on this thread. The result of what we are doing with integers and integer factorisation will remove the ability to use any form of modular exponentiation, or elliptic curve prime field for encryption. It is very likely the outcome of our work will result in it not being possible to use AES or anything like it for encryption or one way hashing (i.e. if I know what I need to hash to, I can pick any of the numbers that hash to it easily). We are not seeing a mathematical way to encrypt or hash one way after we publish.

This will result in the collapse of bitcoin and other coins in likelihood since proof-of-work is trivialised.

We are not the only party that has figured this out. It is likely actors are going to rug pull bitcoin for sure in the near future.

Example: RSA 100. if we take (d+n)(d+n) - (x+n)(x+n) = c (diff of two squares, d is the root of c)

See that 2a/x is almost exactly equal to x/n, it is only the remainder that fills in the blanks.

There is a class of integers like this. Essential the ratio of the remainder of dividing x by n is in proportion exactly to the ratio of dividing 2a by x. And 2d is 2a + 2x. So for RSA100 we can divided 2d by n. We don't know what n is so we use the number 1 to 10,000 as the amount of n in 2d. We get to 5424. Easy compared to the general number field sieve. At this number we have a chunk of the start of numbers...

a must start with 379... b must start with 400... x must start with 1054 and n must start with 1438... and theta must start with 72.655 (where theta is 2a/x and x/n rememeber that for all c, that if 72.655 x = 2a, then 74.655.. will equal 2d, since (73.655... x 73.655... x n ) squared equals c.

What is so good about that?

The values of x, n, and theta are not random and only possess certain characteristics that are related to the golden ratio, the squre root of two (you can see that in 1438... for n) and the decimal unit 1.

Because these numbers are fixed, it is very simple and very quick to figure out the expansion of the natural log of c, and which are the squares in the exponential function definition that are the ones that give the exact result for c, half of that exponent is obviously the square root of c, and if we know the five contributors in the natural log of c (five dimensions is always the max needed for any problem), then like finding the place of Pi without those in front, we spigot the solution. Works best for the product of two primes.

There are a fixed number of theses types of integer and it's like whether the square is odd and the remainder is even, the other way round. If n is smaller or larger than x and how does that affect theta. And the 'balance number' for that type of integer. In other words, what is the constant that all our variable must equal. This is what maths has been missing for integer factorisation. That pattern that is fixed. The larger the numbers are the easier it is to find.

What we are seeing evidence of is that P = NP. If a problem statement is complete, it is also the solution.
One tip for you too. If you talk about something you have no clue, the result is pure nonsens.
 
  • #62
martinbn said:
All this is nonsense!!

If what yiu wrote made sense it would work for n=2 as well.

Suppose there is a solution to a^2+b^2=c^2, writw it as

1) a+b=c

Or

2) a+b=c+k

Multiply by c and run the same argument. Thus you have proven that there no pythagorian triples.
a^2 + b^2 = c^2 does have solutions, infinite solutions, that's my point. You would be able to divide both sides of any solution by c, so in any solution dividing by c must remove an indice from the left hand side. In the two cases there are only solutions to 1, and 2.

a + b = c

let's rewrite this as a^2 + b^2 = c^2 + k, where k = 0

I want to raise the indice of c by one and a and b by 1.

We know a + b cannot equal c, since the square root of a^2 + b^2 cannot be a + b

We know a + b = c + k must have a solution then if any solution exists and we state that there is a d^2 + e^2 - k = c^2

If we divide by c, we need to remove -k, and d^2 + e^2 must be divisible by c to give a + b. Infinite solutions to that, especially if k = 0 :)

I think because I didn't write out the examples, now you can see how d and e are employed, and you don't think about the initial a and b. In other words, you are not reading or understanding that the equations:

1. a^2 + b^2 = c^2

and

2) a^2 + b^2 = c^2 + k

represent all integers and any solution that can exist for any powers for FLT

What was confusing is I made it look like I was keep the original integers for a and b and this was a bit lazy, yet depends how you do it, if you understand those two equations represent all integers and all possible solutions, it's kinda obvious really why looking at ALL the indices is a pretty dumb approach when they are all represented by two simple equations. ALL possible solutions are represented above.
 
  • #63
noelbcornerstone said:
If you cannot understand that 1) and 2) cover all integers, can you please explain why?
Because ##k=c^2-a^2-b^2## is so arbitrary that it is useless, furthermore, division by ##c## leaves the realm of integers. And what does it have to do with ##a^3+b^3=c^3##?

Besides that, I have no idea what you are talking about. Please write down the case ##n=4## step by step, and you will see where the problems are; or I can tell you. But this general handwavy form is almost impossible to deal with.
 
  • #64
noelbcornerstone said:
a^2 + b^2 = c^2 does have solutions, infinite solutions, that's my point. You would be able to divide both sides of any solution by c, so in any solution dividing by c must remove an indice from the left hand side. In the two cases there are only solutions to 1, and 2.

a + b = c

let's rewrite this as a^2 + b^2 = c^2 + k, where k = 0

I want to raise the indice of c by one and a and b by 1.

We know a + b cannot equal c, since the square root of a^2 + b^2 cannot be a + b

We know a + b = c + k must have a solution then if any solution exists and we state that there is a d^2 + e^2 - k = c^2

If we divide by c, we need to remove -k, and d^2 + e^2 must be divisible by c to give a + b. Infinite solutions to that, especially if k = 0 :)

I think because I didn't write out the examples, now you can see how d and e are employed, and you don't think about the initial a and b. In other words, you are not reading or understanding that the equations:

1. a^2 + b^2 = c^2

and

2) a^2 + b^2 = c^2 + k

represent all integers and any solution that can exist for any powers for FLT

What was confusing is I made it look like I was keep the original integers for a and b and this was a bit lazy, yet depends how you do it, if you understand those two equations represent all integers and all possible solutions, it's kinda obvious really why looking at ALL the indices is a pretty dumb approach when they are all represented by two simple equations. ALL possible solutions are represented above.
The REALLY important part above should be clarified as:

Any solution that is not a^2 + b^2 = c^2 can be written as

a^n + b^n = c^n

and that can be rewritten as:

d^2 + e^2 = c^2 + k

So ANY solution to Fermat's Last Theorem HAS to be of the form:

d^2 + e^2 - k = c^2

If we add k back, we get a^n + b^n = c^n

For ANY solution.

k must be a multiple of c

k must raise both d and e by the same indice by multiplication BUT d and e cannot be the same number. Why?

That's very last and most obvious question to be the winner, take this and be the first to trivialise this problem. I'm not going to post, it can be yours, all you have to do is show why d and e cannot be the same number and you'll be a maths God, yawn.
 
  • #65
fresh_42 said:
Because ##k=c^2-a^2-b^2## is so arbitrary that it is useless, furthermore, division by ##c## leaves the realm of integers. And what does it have to do with ##a^3+b^3=c^3##?

Besides that, I have no idea what you are talking about. Please write down the case ##n=4## step by step, and you will see where the problems are; or I can tell you. But this general handwavy form is almost impossible to deal with.
 
  • #66
Ok, here's the bit to understand. a^2 + b^2 = c^2, we can take any integer d and write it in this form as we are for c and for some it is an integer, others it is not a whole number. This is indisputable.

Every other integer d, can be written in the form:
a^2 + b^2 - k = c^2 where is d is again presented by c^2

Do you understand that d is c^2 in this discussion?

It is the exact same statement as: a number is the sum of two squares or it is not the sum of two squares.

Then, for ANY solution we know it is of the form:

a^2 + b^2 - k = c^2 since this is the form any integer d takes that is not the sum of two squares, so c^2 in this instance is not the sum of two squares and represents any integer d that is not the sum of two squares.

ALL integers d are covered. If there is a d that is c^n and that is the sum of a^n plus b^n it is covered by the above. This is also indisputable. A number is either the sum of two squares or it isn't.

So, in the second equation, the variable k stores the additional indices of a and b for the solution if it is added, and this must also be a multiple of c, since the right hand side is a multiple of c, so if we add it, c's indice increase, as do a and b's
 
  • #67
Repetition does not add value. Please write down the case ##n=4.##
 
  • #68
For n = 4

a^n + b^n = c^4, a doesn't b doesn't c, all are integers

This can be rewrittens as:

d^2 + e^2 = c^2 where a^n and b^n are equal to the sum of two squares that is itself a square.

For ALL other values for c^4, we can rewrite as follows:

a^4 + b^4 = c^4

d^2 + e^2 - k = c^2, d != e

We took off the amount needed to make c^4 a square, we didn't divide by c

We took off the amount to drop of a and b by two indices.

k must be divisible by c or no solution for c^4 exists.

k must equal c^2 to exist, and d and e would have to be equal for a solution.
 
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  • #69
noelbcornerstone said:
For n = 4

a^n + b^n = c^4, a doesn't b doesn't c, all are integers

This can be rewrittens as:

d^2 + e^2 = c^2 where a^n and b^n are equal to the sum of two squares that is itself a square.

For ALL other values for c^4, we can rewrite as follows:

a^4 + b^4 = c^4

d^2 + e^2 - k = c^2, d != e

We took off the amount needed to make c^4 a square, we didn't divide by c

We took off the amount to drop of a and b by two indices.

k must be divisible by c or no solution for c^4 exists.

k must equal c^2 to exist, and d and e would have to be equal for a solution.
This is nonsense.
 
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  • #70
Getting back to the first point, wouldn't it be amazing fun to discover a proof that Fermat could have found using the tools he knew? He had a good reputation at the time, and it would have been out of character for him to claim to have a proof that he did not have. Maybe there is a proof out there, waiting to be discovered by someone with a grasp of rather simple math and a lot of persistence!
 
  • #71
Al_ said:
Getting back to the first point, wouldn't it be amazing ...
It would be amazing!
 
  • #72
fib(n)^2 + Fib(n+1)^2 =Fib(2n+1) Fib = fibonacci number

fib(n)^3 +fib(n+1)^3 = fib(n-1)^3 + fib(3n+1)
Wouldn't it be great if this relationship between the fibonacci sequence and Fermat's theorem could be somehow used in a proof.
 
  • #73
Al_ said:
Getting back to the first point, wouldn't it be amazing fun to discover a proof that Fermat could have found using the tools he knew? He had a good reputation at the time, and it would have been out of character for him to claim to have a proof that he did not have. Maybe there is a proof out there, waiting to be discovered by someone with a grasp of rather simple math and a lot of persistence!

But he didn't really claim a proof. It was just a quick note in a book.
 
  • #74
fresh_42 said:
There are inifinitely many Pythagorean triplets (##n=2##) and no solution for ##n=4##, why? Euler needed two articles to cover the case ##n=3,## which is a sure indicator that it is not trivial.

But even if we accept the case ##n=3## as given, there is no such thing as an induction. Any induction step fails.
fib(n)^2 + Fib(n+1)^2 =Fib(2n+1) Fib = fibonacci number
fib(n)^3 +fib(n+1)^3 = fib(n-1)^3 + fib(3n+1) Wouldn't it be great if this relationship between the fibonacci sequence and Fermat's theorem could be somehow used in a proof.
 
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  • #75
Hornbein said:
But he didn't really claim a proof. It was just a quick note in a book.
I thought he did? But it was too big to fit in the margin? (I read Simon Singh's book so I'm an expert.....)
 
  • #76
"I have discovered a truly marvelous proof of this, which this margin is too small to contain."
 
  • #77
pinball1970 said:
I thought he did? But it was too big to fit in the margin? (I read Simon Singh's book so I'm an expert.....)
My article cites the exact wording (Latin) and its English translation:

“Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duas ejusdem nominis fas est dividere: cujus rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.”

“It is not possible, however, to decompose a cube into two cubes, or a biquadrate into two biquadrates, and in general, to decompose a power higher than the second into two powers with the same exponent: I have discovered a truly wonderful proof for this, but this margin is too narrow to grasp here.”

Source: https://www.physicsforums.com/insights/fermats-last-theorem/
 
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  • #78
fresh_42 said:
My article cites the exact wording (Latin) and its English translation:

“Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duas ejusdem nominis fas est dividere: cujus rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.”

“It is not possible, however, to decompose a cube into two cubes, or a biquadrate into two biquadrates, and in general, to decompose a power higher than the second into two powers with the same exponent: I have discovered a truly wonderful proof for this, but this margin is too narrow to grasp here.”

Source: https://www.physicsforums.com/insights/fermats-last-theorem/
Could we say his ‘proof’ wasn’t even marginal?
 
  • #79
bob012345 said:
Could we say his ‘proof’ wasn’t even marginal?
It is hard to tell, since he didn't even wrote that in any letter as it was usual at his time. He only scribbled this into Diophant's book, and it was his son who published it after his death, sorting his remains.

I think he had a proof for ##n=4, ## which relies on the argument for ##n=2## which he certainly knew, and possibly some ideas for ##n=3,## and he thought that these could be extended to any number. But this is an opinion. We do not have reliable information.
 

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