Insights Fermat's Last Theorem

[fresh_42]

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 ...

 

[jedishrfu]

What was Fermat's level of mathematical knowledge time?

Was it comparable to what Andrew Wiles knew?

I’ve read that Fermat would write to other mathematicians of his time, presenting a problem without providing his own solution. Hence, there are few known proofs done by him.

Additionally, mathematicians during his time were less rigorous than those of today's proof-oriented mathematics.

 

[Office_Shredder]

What was Fermat's level of mathematical knowledge time?

Was it comparable to what Andrew Wiles knew?

It seems exceedingly unlikely because most of the techniques Wiles used were from branches of math invented well after Fermat died.

 

[bob012345]

Obviously Fermat thought he had a proof. Are there good candidate “proofs” people think may have been what he did and what are the typical flaws?

 

[Pikkugnome]

It is unfortunate that you picked a near miss by one, since it has an obvious parity problem.

 

[jedishrfu]

But the example does illustrate how close you can get and still its not close enough to disprove the theorem.

A recollection by G.H. Hardy while visiting Ramanujan at a local hospital via a taxicab with a rather dull number. Ramanujan quipped, "Oh no, Prof. Hardy, it is a most interesting number as being the sum of two cubes in two different ways: 9^3 + 10^3 and 12^3 + 1^3."

Thereafter, 1729 was known as a taxicab number.

The significance of this story lies in Ramanujan's investigation of sums of cubes to challenge Fermat's theorem. Several examples of near misses were discovered in his notebooks, along with a scheme to generate even more.

Prof Littlewood, Prof Hardy's associate, once said that all positive integers were Ramanujan's personal friends.

https://www.americanscientist.org/article/never-a-dull-number

https://plus.maths.org/content/ramanujan

 

[martinbn]

Obviously Fermat thought he had a proof. Are there good candidate “proofs” people think may have been what he did and what are the typical flaws?

He had a proof for $n=4$. For the general case one possibility, but this only a guess there is no hint of what his marvellous proof might have been, is that you factorize, for which you need $p$th roots of unity, and work within the ring $\mathbb Z[\xi]$, and expect that it has unique factorization just like $\mathbb Z$.

 

[bob012345]

He had a proof for $n=4$. For the general case one possibility, but this only a guess there is no hint of what his marvellous proof might have been, is that you factorize, for which you need $p$th roots of unity, and work within the ring $\mathbb Z[\xi]$, and expect that it has unique factorization just like $\mathbb Z$.

Did he have rings?

 

[martinbn]

Did he have rings?

Of course not! Is this a serious question?

 

[pines-demon]

Did he have rings?

Of course, yes. He married a cousin of his mother

 

[bob012345]

Of course not! Is this a serious question?

It seemed you were suggesting that because you were answering my question of what Fermat may have thought.

 

[fresh_42]

What was Fermat's level of mathematical knowledge time?

Was it comparable to what Andrew Wiles knew?

I’ve read that Fermat would write to other mathematicians of his time, presenting a problem without providing his own solution. Hence, there are few known proofs done by him.

Additionally, mathematicians during his time were less rigorous than those of today's proof-oriented mathematics.

Fermat was a smart guy, but not at the level of the great French mathematicians, but we must not forget that we speak about the 17th century, not the 18th or 19th. Even Newton's Principa hadn't been published yet. I wouldn't compare him with Wiles. It is believed that he had a proof for n=4 and possibly n=3, and presumably thought it would generalize to any power. He shared the fate with Kafka and Galois: many of his thoughts were published by his executor, his son.

Yes, those letters asking someone else to solve a problem while already having a solution were a mean game at these times. However, I am not 100% sure whether Fermat was one of them.

 

[fresh_42]

Obviously Fermat thought he had a proof. Are there good candidate “proofs” people think may have been what he did and what are the typical flaws?

I have presented the proof for n=4 in the article, and n=3 can be done in a similar way. But it will be hard without using complex numbers.

 

[fresh_42]

It is unfortunate that you picked a near miss by one, since it has an obvious parity problem.

Near misses are extremely rare. I have found near misses of the order $\pm 10^{33}.$ Although they are still relatively close since the summands were of order $10^{39},$ I thought the parity problem would be less of a thing than the big gaps otherwise.

 

[fresh_42]

Did he have rings?

Rings, unique factorization in non-integer rings, elliptic curves etc. were all unknown. However, n=3 requires dealing with zeros of $x^2+x+1$ for which we would use complex numbers, but the people at the time were pretty resourceful and used their own environments surprisingly effectively.

 

[martinbn]

It seemed you were suggesting that because you were answering my question of what Fermat may have thought.

Yes, but I said rings because it was easier and shorter. Clearly you don't have to use this terminology.

 

[TensorCalculus]

you know, last year in school we were learning Pythag theorem, and I had just read Simon Singh's book on Fermat's last theorem. Seeing the similarity between pythagoras and what I had read about, I went up to my maths teacher after class and asked if he would be able to prove Fermat's last theorem since it wasn't in the book in detail and I was curious.
Poor guy didn't know what to say haha - it's only a few months later on that I eventually realise what a stupid question it was to ask... imagine being a maths teacher and one of your students just asks you to prove Fermat's last theorem on the spot, after class when they have another lesson to go to, and they're probably expecting like a 5 minute answer... whoops.

 

[jackjack2025]

you know, last year in school we were learning Pythag theorem, and I had just read Simon Singh's book on Fermat's last theorem. Seeing the similarity between pythagoras and what I had read about, I went up to my maths teacher after class and asked if he would be able to prove Fermat's last theorem since it wasn't in the book in detail and I was curious.
Poor guy didn't know what to say haha - it's only a few months later on that I eventually realise what a stupid question it was to ask... imagine being a maths teacher and one of your students just asks you to prove Fermat's last theorem on the spot, after class when they have another lesson to go to, and they're probably expecting like a 5 minute answer... whoops.

There are worse experiences as a professor than that :)

I have met Simon Singh. You may like the Code book, the Enigma stuff is interesting.

 

[TensorCalculus]

There are worse experiences as a professor than that :)

he teaches secondary school children (so 11-18 year olds) I can imagine there are probably worse things yes

I have met Simon Singh. You may like the Code book, the Enigma stuff is interesting.

whoa meeting Simon Singh is really cool! I have read the code book, personally I prefer his book on Fermat's last theorem but that one is really good too :)

 

[jackjack2025]

he teaches secondary school children (so 11-18 year olds) I can imagine there are probably worse things yes

whoa meeting Simon Singh is really cool! I have read the code book, personally I prefer his book on Fermat's last theorem but that one is really good too :)

What the teacher should have said is: Yes of course I can prove it, but the margin is too narrow to contain the proof (or some variation of that get out clause)

 

[martinbn]

you know, last year in school we were learning Pythag theorem, and I had just read Simon Singh's book on Fermat's last theorem. Seeing the similarity between pythagoras and what I had read about, I went up to my maths teacher after class and asked if he would be able to prove Fermat's last theorem since it wasn't in the book in detail and I was curious.
Poor guy didn't know what to say haha - it's only a few months later on that I eventually realise what a stupid question it was to ask... imagine being a maths teacher and one of your students just asks you to prove Fermat's last theorem on the spot, after class when they have another lesson to go to, and they're probably expecting like a 5 minute answer... whoops.

I don't see what the problem is.

 

[jackjack2025]

I don't see what the problem is.

digging out Andrew Wiles' proof and explaining it on the spot would be a bit of a problem, no?

 

[martinbn]

digging out Andrew Wiles' proof and explaining it on the spot would be a bit of a problem, no?

I meant, i don't see the problem if a student asks the teacher how to prove Fermat and the teacher diesn't know.

 

[nomadreid]

I meant, i don't see the problem if a student asks the teacher how to prove Fermat and the teacher diesn't know.

As a longtime mathematics secondary school teacher, I know the solution to this situation if one doesn't want to admit that one doesn't know: "Excellent question! That's your homework assignment for tomorrow."

 

[TensorCalculus]

@martinbn - I ended up searching up the proof a little while later and it's pretty long, can't imagine someone would have it memorised... or if they did be able to explain it to a 12 year old kid (who, sure, is a huge nerd, but also 12 so uhm yeah said kid's maths knowledge is really limited) in less than 5 minutes... I mean it's just a bit awkward I guess, and looking back on it was probably not the best question to ask haha

 

[fresh_42]

I once received a very unsatisfactory answer to a question I asked my math teacher, too. He basically said it was difficult and beyond my horizon. I don't think this was true. He could have said: I currently don't know, but I will look it up and tell you tomorrow, for example.

In my opinion, the best answer about FLT would have been: It uses a standard proof technique by transforming the problem into another, more general one that can then be solved. However, this more general problem dates back to a conjecture made in 1958 that itself requires higher mathematics to prove. The fact that FLT has been unsolved for 350 years and the other one for 35 years can be viewed as evidence of their complexity.

 

[TensorCalculus]

Yeah that's probably quite a good way of putting it - in the end he just said that it would take a while to explain so I should look it up and then ask him if I had any questions - spoiler alert I didn't have any questions because unsurprisingly I didn't understand a thing . Not as bad as just going "it's difficult" though I guess :)

 

[fresh_42]

Yeah that's probably quite a good way of putting it - in the end he just said that it would take a while to explain so I should look it up and then ask him if I had any questions - spoiler alert I didn't have any questions because unsurprisingly I didn't understand a thing . Not as bad as just going "it's difficult" though I guess :)

There is also the possibility of saying: Solve it for $n=2$ and use this result on $a^4+b^4=c^2$ to solve it for $n=4.$ Still a bit tricky, but doable for a 12-year-old. This risks frustration for the promise of the feeling of success. I'm not sure whether I would risk this in the role of a teacher. On the pro list is that the case $n=3$ then paves the way to think about complex numbers.

 

[TensorCalculus]

There is also the possibility of saying: Solve it for $n=2$ and use this result on $a^4+b^4=c^2$ to solve it for $n=4.$ Still a bit tricky, but doable for a 12-year-old. This risks frustration for the promise of the feeling of success. I'm not sure whether I would risk this in the role of a teacher. On the pro list is that the case $n=3$ then paves the way to think about complex numbers.

hmm, 1 year older and 1 year wiser, I might try that and see what I get it sounds fun :)
and try n = 3 too of course

 

[fresh_42]

hmm, 1 year older and 1 year wiser, I might try that and see what I get it sounds fun :)
and try n = 3 too of course

Ask for help / hints here before frustration takes over.

 

[TensorCalculus]

ah - but I'm the sort of person who loves learning by trying again and again... and failing again and again, and again... but if I start to get annoyed them I'll ask here, sure :)
the feeling when you get it right after like an uncannily long time (hours... and hours... and hours... sometimes tens of hours... spent on problems... and getting stuck so many times before finally finding my way out...) is one of the best in the world

 

[fresh_42]

ah - but I'm the sort of person who loves learning by trying again and again... and failing again and again, and again... but if I start to get annoyed them I'll ask here, sure :)
the feeling when you get it right after like an uncannily long time (hours... and hours... and hours... sometimes tens of hours... spent on problems... and getting stuck so many times before finally finding my way out...) is one of the best in the world

There are a few key points that have to be found. Like crossings, where all depend on whether you take the right way. Here is where practice and experience come into play. But, sure, this is a nice problem to gain it.

 

[martinbn]

it has an obvious parity problem.

What has?

 

[fresh_42]

it has an obvious parity problem.

See post #14 for an explanation. If you like it better without parity problems, then consider
$$
3987^{12}+4365^{12} =(4472.0000000070592907382135292414 \ldots)^{12} \sim 4472^{12},
$$
in which case, you should read the meaning of "almost" as relative error.

 

[TensorCalculus]

@fresh_42 I got n=4 a while ago but I've been stuck for too long on n=3
do you mind giving me a little hint at what you meant when you said to use complex numbers pleeeease ?

 

[fresh_42]

@fresh_42 I got n=4 a while ago but I've been stuck for too long on n=3
do you mind giving me a little hint at what you meant when you said to use complex numbers pleeeease ?

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.

 

[Hornbein]

I just would have said to the student, "the proof is five hundred pages long."

 

[TensorCalculus]

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, $1-\omega^2=2-\omega,$ 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.

Wow....
I think this took me a couple of reads (okay... the ring bit... much more than a couple) over to figure out what was going on, but when I did - I was pretty astonished (and you did a good job explaining it in a very understandable way, thank you)
I mean, I wouldn't have been able to come up with something like this on my own for sure... but it's a pretty elegant solution once you get going. Once you realise the trick is to express it in the form
$$ c^p=a^p+b^p=(a+b)\cdot(a+\zeta b)\cdot (a+ \zeta^2 b)\cdot \ldots \cdot (a+\zeta^{p-1}y) $$
then everything just seems to... flow from there. I guess I just never decided to factorise these things because I have never come across a situation in which factorising and getting a complex number in the brackets... if there is a complex root to, say, a quadratic I'm trying to solve, I would just use the quadratic formula and not factorise simply because I am not used to it...

These properties are not obvious, so either you trust me or you have to prove them.

I choose option 2, more fun :)

I'm going to try to do some research on Eisenstein Numbers then prove the properties of rings... then prove the n=3 case as a final challenge XD
ahh this is such a nice problem - thank you so so much for the very understandable guidance because I genuinely think I would have been stuck forever without it . Now time to do some reading on Eisenstein Numbers...

 

[TensorCalculus]

aagh the TeX is not rendering

 

[fresh_42]

I just would have said to the student, "the proof is five hundred pages long."

One hundred, plus correction.

 

[martinbn]

One hundred, plus correction.

He includes some of the prerequisites. For a high school student, five hundred is too optimistic. May be one more zero at the end.

 

[bob012345]

$\mathbb{Z}[\zeta]$ 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.

You said 3 is not prime above? I don’t follow your logic. You said Fermat had a proof for n=4 and possibly 3. What would that have looked like, i.e. without modern techniques? Thanks.

 

[martinbn]

You said 3 is not prime above?

Not prime in that ring. It is a prime in the ring of the usual intergers.

I don’t follow your logic. You said Fermat had a proof for n=4 and possibly 3. What would that have looked like, i.e. without modern techniques? Thanks.

The proof for four was given above. I doubt he had one for three.

 

[fresh_42]

You said 3 is not prime above? I don’t follow your logic.

I had a typo ($\omega$ instead of $\zeta$), which I now corrected. I also admit that I took the information about the non-primality of $3\in \mathbb{Z}[\zeta]$ from the Wikipedia page without checking. We have with $\zeta=\dfrac{1}{2}+\dfrac{i}{2}\sqrt{3}$
$$
3=\zeta \cdot (2-\zeta)^2,
$$
so all we have to do is to prove that $2-\zeta\in \mathbb{Z}[\zeta]$ isn't a unit since $3\,\nmid\,(2-\zeta).$ From
$$
1=(2-\zeta)(a+b\zeta)=2a+(2b-a)\zeta-b\zeta^2=(2a+b)+(b-a)\zeta
$$
we get $b=a$ and $3a=1$ which isn't solvable in $\mathbb{Z}[\zeta].$ Hence $2-\zeta$ isn't a unit, and $3$ divides a product without dividing one of its factors, so it cannot be prime.

That $2-\zeta$ is actually a prime is a bit more work.

You said Fermat had a proof for n=4 and possibly 3. What would that have looked like, i.e. without modern techniques? Thanks.

Well, I don't know what it would look like without using modern terms. One would have to study Fermat's inheritance, since this is where his remarks were found. But I do know that mathematicians at this time have been very ingenious in creating workarounds for modern terms. One could as well operate with an unknown $x$ instead of the explicit formula for $\zeta,$ simply as a non-trivial solution for $x^3=-1$ and then consider divisibilities. Zachows (1st entry in my list of sources) states the following timetable:

Fermat (1601/08–1665) ca. $1630$ problem statement; proof for $n = 4$ and later in hints for $n=3$

Euler (1707–1783) proof for $n = 3$ using complex numbers, distributed in two different papers.

Gauß (1777–1850) complete proof for $n = 3$ in one piece.

etc.

So it's a) unclear whether Fermat really had a proof, and b) probably easier to search in Euler's or Gauß' papers.

 

[martinbn]

There is a book that might be of interest.

"Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory" - Harold M. Edwards​

This book is an introduction to algebraic number theory via the famous problem of "Fermat's Last Theorem." The exposition follows the historical development of the problem, beginning with the work of Fermat and ending with Kummer's theory of "ideal" factorization, by means of which the theorem is proved for all prime exponents less than 37. The more elementary topics, such as Euler's proof of the impossibilty of x+y=z, are treated in an elementary way, and new concepts and techniques are introduced only after having been motivated by specific problems. The book also covers in detail the application of Kummer's ideal theory to quadratic integers and relates this theory to Gauss' theory of binary quadratic forms, an interesting and important connection that is not explored in any other book.​

 

[fresh_42]

The exposition follows the historical development of the problem, beginning with the work of Fermat and ending with Kummer's theory of "ideal" factorization, by means ...

I am not 100% certain, but it could well be that Kummer's ideal numbers led to the term ideal. I have found in Dieudonné's book about the history of mathematics between 1700 and 1900 a very nice description of Kummer's idea. He also mentions ideals in the realm of Lie theory, where they have been called "hypercomplex systems". Dieudonné writes about Kummer

As already mentioned, Kummer's goal was to prove Fermat's Conjecture. This is not all; inspired by the work of Gauss, Dirichlet, Jacobi, and Eisenstein on the use of complex numbers in the investigation of the cubic and biquadratic reciprocity laws, he wanted to formulate and prove the general reciprocity laws for the p-th powers. Kummer devoted about twenty of his particularly creative years to these two problems, and the ideas he developed have had a tremendous influence on modern mathematics, the significance of which far exceeds the results he actually achieved. In the course of his research, Kummer was inspired to introduce "ideal numbers." The clarification and generalization of this concept took several decades. Thanks to the work of Kronecker, Dedekind, and E. Noether on this problem, the face of algebra and algebraic geometry has changed completely.

 

[martinbn]

I am not 100% certain, but it could well be that Kummer's ideal numbers led to the term ideal.

This is 100% the case.

 

[TensorCalculus]

Ring theory is really cool
but I think I might need to read up a bit more on it before trying to apply it, @martinbn, the book you mentioned, would it cover ring theory? And/or Eisenstein Integers? (And do you think that it would be understandable for a child - amazon said that it was elementary but I know absolutely nothing about maths )

 

[Petek]

I have a copy of Edwards' book on Fermat's Last Theorem. Here are some comments:

• The book doesn't cover ring theory, per se, but uses the properties of rings. Edwards' genetic approach is historical. He uses the terminology at the time the theory was developed.
• The book also doesn't use the term Eisenstein integers. However, the discussion of the case n = 3 uses what are now called by that name.
• The book goes far beyond the individual cases of FLT. It does not cover Wiles' proof.
• It's a good read just for the historical comments.
• The book is part of Springer's Graduate Texts in Mathematics series. It's written at a higher level than an undergraduate text.
• You can read the Introduction and Table of Contents here (in the Read sample link).

I'd be glad to answer any questions that you might have about the book.

 

[TensorCalculus]

I have a copy of Edwards' book on Fermat's Last Theorem. Here are some comments:

• The book doesn't cover ring theory, per se, but uses the properties of rings. Edwards' genetic approach is historical. He uses the terminology at the time the theory was developed.
• The book also doesn't use the term Eisenstein integers. However, the discussion of the case n = 3 uses what are now called by that name.
• The book goes far beyond the individual cases of FLT. It does not cover Wiles' proof.
• It's a good read just for the historical comments.
• The book is part of Springer's Graduate Texts in Mathematics series. It's written at a higher level than an undergraduate text.
• You can read the Introduction and Table of Contents here (in the Read sample link).

I'd be glad to answer any questions that you might have about the book.

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...