Is this Fermat’s Proof of the Last Theorem?
Dear Millenium Reader. Anyone with a passionate (or just passing) interest in discovering Fermat’s own proof to ‘Fermat’s Last Theorem’ may have been surprised that Steig Larsson included this subject in Part I Chapter 1 of Millennium II. What is more intriguing is that in Part IV Chapter 32, as Salander was approaching Zalachenko’s disued farm, and her possible death, she had a sudden incredible revelation of Fermat’s proof ? a revelation that appeared to make it simple and beautiful. (Was it on a par with Andrew Wiles’ 1994 revelation that helped him find a solution to the problem stopping him from proving FLT?). As a consequence I have looked through Millennium III to see if Salander had any recollection of this and found she didn’t. But for readers of the Millennium trilogy the following is what I believe is Fermat’s disarmingly simple proof.
Fermat’s Last Theorem
The questions that all mathematicians have asked is since his death is how did he start, and how did he prove it? Diophantus’Greek version of Arithmetica was published in 1621as the Diophanti Alexandrini Arithmeticorum Libri Sex et De Numeris Multanguls Liber Unus in both Greek and Latin by Claude-Gasper Bachet. In the early 1630’s Fermat became interested in the more subtle questions surrounding each of Diophantus’ questions, and around 1636/37 he thought that he could add something to the solution given to Question 8 of Bk II. This concerned the division of a given square into two squares, and it led him on to wonder if cubes and higher powers could be likewise divided. He played around with a few of the lower powers (as any modern mathematician would do) and found it was unlikely. Without any solution he eventually came to Q29 of Bk V which concerned the finding of three squares such that the sum of the squares is a square. Again looking to see what he could add, Fermat wrote in the margin that he could see a way of proving that a square could not be divided into two fourth powers, That is, in modern algebraic terminolgy x^4 y^4 ? z^2. His proof was never discovered, but an internet blogspot by Larry Freeman gives a conjectural reconstruction of it.
Fermat found that x^4 y^4 ? z^2 can be rewritten as x^4 y^4 ? z^4 because any z^4 can be written as z^2. Not only that, if x^4 y^4 ? z^4 then there are no solutions to any multiple (k) of the power 4. For instance, (x^4)^k (y^4)^k ? (z^4)^k is the convention for expressing a power raised to a power. Therefore, the multiples of power 4 (4k) that had no solutions are 4.1=4, 4.2=8, 4.3=12, 4.4=16, 4.5=20 …and so on. So how did Fermat go from here to proving his thoerem for all n > 2?
Although he may have searched for proofs to n =3, 5, 6, 7, etc, he simply went back to Q8 and asked himself the simple question: Why was it that squares be divided into two squares but not higher powers? He found that for any Pythagorean triple
z^2 – y^2 – x^2 always gave a zero difference. Applying this to higher powers he found that the nearest to a non-zero difference could only obtained with cubic triples (7,6,5), (9,8,6), (12,10,9), and the triple (3,2,1) for the fourth and higher powers. Therefore, there were no solutions for powers greater than squares. By applying the logic he used for power 4 he began to list all the multiples (nk) without solutions,
n (nk) multiples
3 3.1=3, 3.2=6, 3.3=9, 3.4=12, 3.5=15…
4 4.1=4, 4.2=8, 4.3=12, 4.4=16, 4.5=20 …
5 5.1=5, 5.2=10, 5.3=15, 5.4=20, 5.5=25…
6 6.1=6, 6.2=12, 6.3=18, 6.4=24, 6.5=30…etc..
But one thing stood out immediately. The fourth multiple of n = 3, 5, 6, etc is always numerically equal to a multiple of n = 4! Therefore, if the multiples of n = 4 had no solutions then the fourth multiples of any other power had no solutions. Using the logic of mathematical proof introduced by the ancient Greeks Fermat realized that he could now vigorously prove that all n > 2 had no solutions (indicated by sign ?);
Let A = (x^4)^1 (y^4)^1 ? (z^4)^1
Let B = (x^4)^k>2 (y^4)^k>2 ? (z^4)^k>2
Let C = (x^k>2)^4 (y^k>2)^4 ? (z^k>2)^4
Let D = (x^k>2)^1 (y^k>2)^1 ? (z^k>2)^1
As A is proved to have no solutions,
Then A implies B when k > 2, and B has no solutions.
As B is equal to C, then C has no solutions.
As D implies C, then D has no solutions.
Proof that no power greater than squares can be divided into two powers can be demonstrated by inserting any k in the equations. For instance let k > 2 = 3,5,6,7…etc. Fermat was obviously very pleased with this simple proof and he annotated the long-hand theorem in the margin of Q8 Bk II. The ‘marvellous’ aspect about it was that he didn’t have to prove the infinitude of powers individually.
With reference to Bachet’s question about the area of Pythagorean triangles not having square areas, this is conjecturally reconstructed in Larry Freeman’s blogspot on ‘Fermat’s One Proof’ and shows that P^4 – Q^4 ? z^2 can be expressed as x^4 y^4 ? z^4. This was achieved sometime in the mid to late 1650’s, when he began corresponding with Christiaan Huygens. Not only did he describe in general terms his proof of this, but also indicated that he had proved the case for n = 3. This proof has never been found, and I believe that it was just an intellectual exercise to show that he could. In these and all his proofs, it seems that the way he used his method of infinite descent depended on whether questions were ‘affirmative’ or ‘negative’.
Much has been made of Fermat’s reluctance to provide any solutions when he corresponded with other mathematicians. We must remember that in the period of Fermat’s mathematical education the Francois Viete algebraic notation (introduced between 1584 and 1589) was used in France, and he continued using it throughout his life. However, in 1637 Rene Descartes published a work which severed the dependence of all mathematics on geometry and became known as analytical geometry. This eventually superseded the Viete notation. It isn’t surprising that when Fermat did correspond with mathematicians from several European countries, his questions were posed in Latin, the language of the educated classes. Not only did he not want to give away the secrets of his method of infinite descent, but his solutions existed in the out-of-date Viete notation, and nobody would have understood them. This obviously made things difficult for his son Clement-Samuel when he came to republish the Arithmeticorum with his fathers notes and theorems.
So Millenium Reader, if you are interested, there several outstanding mysteries:
– After his death, what became of Fermat’s copy of the Arithmeticorum and the prolific amount of notes and jottings made during his life?
– Knowing that Fermat proved FLT for only one power, n = 4, why didn’t all
19 comments on “Is this Fermat’s Proof of the Last Theorem?”
So, would you agree that Salander realized this solution as she was circling the farm b/c she is “k>2? in this proof? The Section & Zalachenko had their multitude of coverups “all squared away” until the “power” of Salander kept interfering with the perfectly solved equations.
Posted by CMM in MI ,
Thank you CMM in MI for the punny/funny observation, but I don’t think there are any other examples where the metaphorical replaces the literal sense in Stieg Larsson’s writing style.
It should be noted that html replaced the ‘not equal to’ mathematical symbol with 8#8808.
Fermat’s theorem was only found when his son included it in his Arithmeticorum in 1670, five years after Fermat’s death. The proof was probably amongst the mass of notes and jottings which are now ‘lost’ to posterity. Fermat never put this problem to any of his correspondents which is why mathematicians have always believed he never had a proof, or if he did it was flawed. This discussion board is probably the first opportunity for anyone since Fermat to have seen a mid seventeenth-century proof that works. The question is: How did all the most brilliant mathematicians over the last 340 years fail to see how Fermat simply extended the case for n=4?
Posted by Phil Cutmore in Bristol UK ,
Hocus Pocus Dominocus
Posted by Mary Malone in Detrkoit, MI ,
This is why she said a philosopher should be looking at it. She was referring to how Fermatt tried to get people to work together through his life. The solution to this, was by uniting people to solve a problem that he thought unsolvable, I believe. So the solution he referred to in the margin, was a philosophical solution, not a mathematical one.
Posted by Kaylie in Cambridge ,
I’m not a mathematician but I was convinced that FLT was only proved math’ true by Wiles in ’94 based on the T S conjecture (as you refer)… Am I wrong? was it not?
Posted by Pete in Lx ,
Read your analysis with interest. However, due to some resolution problems, the equations you have mentioned don’t display properly on my system. Can you repost the same please?
Posted by Ranga in Oxford, UK ,
I have never been so perplexed by one page of a novel before. I read it over and over and couldn’t figure out what was so simple. Kayle’s answer makes sense. I didnt know what the philisophical meaning could be but then again, I don’t know of fermat’s reputation that well
Posted by Katie in St. Cloud ,
As requested by Ranga in Oxford UK, 12/12/10 I have added an easier to read version of what I believe to be Fermat’s final arguments.
Fermat had a “Eureka” moment when he noticed that n.k:12.1 is the lowest common multiple (LCM) of n.k:4.3 and nk:3.4. This demonstrated that as the first multiple (k = 1) of of n = 12 and the third multiple (k = 3) of n = 4 had no solutions, then the fourth multiple (k = 4) of n = 3 also had no solutions. By implication therefore, the first multiple (k = 1) of n = 3 had no solutions. Fermat had gone from proving that (x^4)^1 (y^4)^1 = (z^4)^1 had no solutions to implying that (x^3)^1 (y^3)^1 = (z^3)^1 had no solutions.
Having proved the case for n = 4 and its n.k multiples he could now prove the case for all the n.k multiples of any n > 2 using the logic of mathematical proof introduced by the ancient Greeks. But to overcome a paradox that occurs when k > 1 = 2 minimum when D implies Pythagoras theorem, the value of any multiple has to be k > 2 = 3 minimum.
Let A = (x^4)^1 (y^4)^1 = (z^4)^1
Let B = (x^4)^k >2 (y^4)^k >2 = (z^4)^k >2
Let C = (x^k >2)^4 (y^k > 2)^4 = (z^k >2)^4
Let D = (x^k >2)^1 (y^k > 2)^1 = (z^k >2)^1
As A is proved to have no solutions,
and A implies B, then B has no solutions.
As B = C, then C has no solutions.
As D implies C, then D has no solutions.
In order to prove that no power greater than squares can be separated into two like powers simply replace (k > 2) with any integer k > 2, i.e. 3 minimum, in equations B, C, and D, and D is a re-statement of Fermat’s Last Theorem .
With reference to Pete in Lx 10/12/10
Yes I agree. The only generally accepted version of a published proof to FLT is that derived by Andrew Wiles in 1994, in which he proved an equality known as the Tantiyama-Sahimura conjecture. But through the centuries mathematicians beginning with Euler, then Lejeune-Dirichlet, Legendre, Lame and Cauchy, and finally Kummer all found proofs of individual prime values of n, but could not extend their arguments for all n. What I have demonstrated is that as Fermat had only proved the case for n = 4 (twice), he could easily have gone on to extend his arguments for all n > 2. The important aspect here is that this particular proof works.
The historical record shows there is no indication that he could have done it this way, or in any way. The reason for this blog is to show that he could have.
Posted by Phil Cutmore in Bristol UK ,
I’m maijoring in math. The proof is 100 pgs! Larsson refers to a phylo. CMM is the closest.
Posted by Faith in Leb ,
“as D implies C”…
that is totally gibberish, D DOES NOT implies C. think about it: 3 4 is not 5, but 3^2 4^2 IT IS 5^2.
also, even if that would be true, you just proved it for even powers bigger then 2, but you still need to prove it for ALL odd primes…
amazing how all cranks believe Fermat was so stupid… “this is his proof”… etc.
Posted by signature in town ,
Okay, here’s my take.
Someone added that page after Larsson’s death. Fermat’s riddle was left unsolved just like Larsson’s books. Fermat’s riddle is that only the one or two dimensional can fit on a two dimensional margin of paper. A cube cannot. Four dimensions cannot, etc. That’s the joke. Larsson made a joke on us too with his family, etc. God’s Revenge is really about us wanting a book for the ending, but we won’t get it b/c Larsson died. That’s my take. We’ll see if I’m right.
Posted by Megan Van Zelfden in Plano, Texas ,
congratulations. u have appended yourself to the ever burgeoning flt troll-list; otherwise you are an ordinary idiot.
grrr. isn’t maths hard enough w/o people like u muddying the water?
Posted by fermat in bordeaux ,
Here is my solution – I was reading it in the wee small hours because I couldn’t put it down – which may help to explain a few things!
Can’t remember exact wording: She giggles and then there is something to the effect that it is more a question for philosophers or something like that.
“She GIGGLES” – it is a joke – it is funny! – get it?
… as in 1 1=”a window” all kids learn that – get it yet?
Well, fine, ok: what do horrible, unsolvable maths questions do? they put you to sleep, right? and z3 is in fact “zzz” which any child can tell you means “sleep” so x3 y3 = “zzz”.
x2 y2 = z2 is pythagorus theorum for a right angled triangle and completely solvable: apparantly x3 y3 = z3 is not.
I don’t know the maths of trying to solve it but late at night reading the book, she is a maths genious who could not solve the problem, until she “gets the joke and giggles” – I actually just came on line to see if someone else had “discovered” the answer and reached the same conclusion I did – I laughed out loud reading it because my sleep deprived brain actually made the connection between being tired and “zzz”.
x3 y3 = z3 is a parody and the answer is funny.
there is a Simpsons episode that had a maths joke with the solution being rdr2 (which is rdrr, or – say it – r d r r – “ha de ha ha”)
ps just in case: ( 1 1=window: if you write 1 with touching the 1 then write another 1 touching the other side of the 1 then write the = as one line above the and one below, you have drawn a window )
of course I could have it completely wrong but I don’t think so and I like it! She is a maths genious and recognises it can’t be solved and that the mathematician was having a laugh, all these serious people trying with great seriousness to solve the problem and they don’t realise it is a joke – she has just got the joke!
“The answer was so disarmingly simple. A game
with numbers that lined up and then fell into place in a simple formula that was most
similar to a rebus.
Fermat had no computer, of course, and Wiles’ solution was based on mathematics
that had not been invented when Fermat formulated his theorem. Fermat would never
have been able to produce the proof that Wiles had presented. Fermat’s solution was
She was so stunned that she had to sit down on a tree stump. She gazed straight
ahead as she checked the equation.
So that’s what he meant. No wonder mathematicians were tearing out their
Then she giggled.
A philosopher would have had a better chance of solving this riddle.
She wished she could have known Fermat.
He was a cocky devil.”
Posted by Jess in Wellington, NZ ,
I believe Fermat’s proof related to a parametric solution of the Fermat equation of the theorem.I appreciate your proof, however.
Pl.read .A simple and short analytical proof of Fermat’s last theorem’ Vol.2,No.3, CMNSEM,March 2011,pg 57-63
Posted by radpiyadasa in Kelaniya,Sri Lanka ,
Jess in Wellington. There is probably a very high probability that you will never read these words…but I think you are on the money. PS. I loved the books. Almost timeless classics.
Posted by Fermat in Melbourne ,
Yes, but what is the actual point? Of math like this? Like, what... is the point?
thanks, interesting read
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