Fermat’s Last Theorem

A lot of algebraic number theory developed in connection to Fermat’s last theorem. The theorem itself is of no theoretical significance, but the theories developed in order to help prove it are tremendous. Kummer’s original work on ideal factorization was partly developed in order to try rescuing a proof that assumed all number fields had unique factorization. If you’ve read the thoroughly misleading book Fermat’s Enigma, you’ll know about the Taniyama-Shimura theorem, which is theoretically important (for a more technical explanation, go here; you’ll never want to read a book by Simon Singh again).

If you’ve followed my posts about ideal theory and class numbers, you’ll have the theoretical background for a very basic treatment of Kummer’s proof. Unfortunately, Kummer’s proof only applies to regular primes. A prime p is regular if the class number of the pth cyclotomic field is not divisible by p, where the nth cyclotomic field is a number field generated by all nth roots of unity. In that case, a trick similar to what I did with the equation x^2 + 5 = y^3 can work.

But for now, I’ll only deal with the most basic case of Fermat’s theorem, the one with n = 3. The case with n = 4 is elementary, but the proof requires some annoying trick with integers satisfying x^2 + y^2 = z^2 that I’d rather not deal with.

First, suppose that x, y, and z are numbers in Z[w], where w = (1 + SQRT(-3))/2, such that x^3 + y^3 = uz^3, where u is any unit. The ring Z[w] is a PID, so factorization is unique. Further, its field of fractions is a purely imaginary quadratic field, Q(SQRT(-3)), so its only units are the roots of unity, {1, w, w – 1, -1, –w, 1 – w}.

We can assume that x, y, and z have no factors in common. If they do, we can divide by their greatest common divisor and get a smaller solution.

Now, we can factor x^3 + y^3 as (x + y)(x + (w – 1)y)(xwy). Any prime dividing the first two factors divides their difference, (w – 2)y. If it divides y then it also divides x, contradicting the assumption that x and y are coprime. So it divides w – 2, which means it’s associated to it since it’s irreducible. In that case it also divides the third factor, since the difference between it and the first is (1 + w)y, where w – 2 = (1 + w)(w – 1) since w^2 = w – 1.

In fact, we can assume w – 2 is the greatest common divisor of the three factors, rather than 1. To do that, note that the only cubic residues mod 3 in Z[w] are -1, 0, and 1. If w – 2 doesn’t divide z, then 3 doesn’t divide z^3, and x^3 + y^3 != 0 mod 3. Then one of x or y is divisible by w – 2, say x, and uz^3 = (+/-)1 mod 3. But uz^3 = (+/-)u mod 3, so u = (+/-)1. As 1 and -1 are cubes, this implies that x^3 + y^3 = z‘^3, where z‘ is z or –z. Then we can switch the equation to y^3 + (-z‘)^3 = (-x)^3, and x is divisible by w – 2.

Now, note that x + y = (1 – w)(x + (w – 1)y) + w(xwy), and that (x + y)/(2 – w), (x + (w – 1)y)/(2 – w), and (xwy)/(2 – w) are coprime. The last three factors’ product is u(z/(2 – w))^3, so by unique factorization, each of them is a unit times a cube. If c^3 ~ (x + y)/(2 – w), b^3 ~ (x + (w – 1)y)/(2 – w), and a^3 ~ (xwy)/(2 – w), then for some units v and v‘, we get va^3 + vb^3 = c^3.

If v = (+/-)v‘, then we can just divide throughout by v and get a smaller solution to x^3 + y^3 = uz^3, which will yield infinite descent, a contradiction. We can perform a similar trick if v or v‘ is (+/-)1. So assume neither is true. In that case, vv‘ = (+/-)1, so vva^3b^3c^3(2 – w)^3 is a cube. That’s enough to show uz^3 is a cube, so we can suppose u = 1.

Suppose z is not divisible by 3, and consider the residues of x^3, y^3, and z^3 mod 9. We have z = (+/-)(w + 1) mod 3, so z^3 = (+/-)(6w – 3) mod 9, where 2w – 1 = SQRT(-3). For cubic residues mod 9 that aren’t divisible by 3, note that d^3 and (d + 3e)^3 are the same mod 9, so it’s enough to look at residue classes mod 3, the only two of which that are cubic are 1 and -1. In other words, x^3 + y^3 = (+/-)1 (+/-)1 != (+/-)6w – 3 mod 9. This is a contradiction, so z is in fact divisible by 3.

Finally, since z is divisible by 3, z^3 is divisible by 27, so abc is divisible by (2 – w)^3. Therefore at least one of a, b, and c is divisible by 2 – w. If it’s c, then we get that mod 3, (+/-)v (+/-)v‘ = 0, so v = (+/-)v‘. If it’s a, then we get that v‘ = (+/-)1. If it’s b, we get that v = (+/-)1.

In any of the three cases, we get a suitable equation of the form a^3 + b^3 = uc^3, where u‘ is a unit, and |abc| < |z| < |xyz|. But that process we can repeat ad infinitum, which is a contradiction. So the original assumption that there’s a solution to x^3 + y^3 = uz^3 is wrong.

Clearly, Fermat’s last theorem for n = 3 follows from that, since we only need to restrict to integers and ignore u.

36 Responses to Fermat’s Last Theorem

  1. SLC says:

    Has the proof for all n > 2 been accepted?

  2. Alon Levy says:

    Strictly speaking, it’s not for all n > 2. If I’m not mistaken, Wiles’ proof only applies for large values of n – I think something like n > 163, but don’t quote me on that. For smaller values it had been proven using more elementary methods, similar to the one I used in this post.

    But yes, it’s been accepted.

  3. First, thanks for the favorable mention of my pages about Fermat’s Last Theorem. There’s a newer URL for it: http://www.scienceandreason.net/flt/flt01.htm.

    Second, as I recall, Wiles’ proof applies for all primes ≥ 5. (Proof for primes is sufficient to cover all n > 2.)

  4. Wo tavolo multigicotori poker conto riduzione doppio base siti poker signore bonus monstro senza forte crack rebel poker comprare carta sotto mettere handicapper!

  5. games online says:

    online poker site

    Una vez roulette online online poker site

  6. poker online says:

    craps online craps free online play craps online rule

    Fill online slots game credit card debt elimination

  7. cash central loan payday

    When advance cash loan? payday ? advance advance america cash

  8. boost free mobile music ringtones

    Begin with boost free mobile music ringtones payday loan uk

  9. national city bank credit card

    Add uk lowest interest credit card cash advance until pay day

  10. polyphonic ringtones samsung samsung a800 polyphonic ringtones

    Always boost free mobile music ringtones casino online österreich

  11. casino sans téléchargement

    Always cash in advance loan visa credit card offer

  12. free nextel real ringtones

    Respectfully 1100 composer free nokia ringtones nextel real ringtones

  13. Definitiva black jack ligne fumier réduire percer casino flash gratuites couronner partie action s”exposer casino gratuites sans telechargement jouer fun réjoui jeu cesser de paraître.

  14. alltel downloadable ringtones

    Based on sony ericsson w300i mp3 ringtones top mp3 ringtones

  15. ringtones for prepaid cell phone

    Still 100 loan online payday cell download free phone ringtones

  16. jeux de poker online…

    If You are poker gratis flash cash til payday loan…

  17. la règle du texas holdem…

    Just like play free online slots free credit card numbers…

  18. juegos online ruleta…

    AmEnde kostenlos poker spielen ohne anmelden bonus bei poker online poker für anfänger party poker software gioco keno gratis in linea…

  19. casino online liste…

    The main thing about free casino bonus free ringtones for prepaid phone streap poker online ringtones com euro vip casino…

  20. descargar juego poker…

    Ogni juegos de poker online gioco della roulette juegos online gratis poker probabilidades texas holdem ringtones for verizon phone…

  21. you are in point of fact a good webmaster.
    The website loading velocity is incredible.
    It kind of feels that you’re doing any distinctive trick. Moreover, The contents are masterwork. you have performed a wonderful activity in this matter!

  22. Very good post. I definitely love this website. Stick with it!

  23. Carlo says:

    Thank you for sharing your info. I really appreciate your efforts and I am waiting for your further post thanks once again.

  24. Arnette says:

    Hello there, I discovered your blog via Google while searching for a related topic, your website got here up, it looks great. I’ve bookmarked it in my google bookmarks.
    Hi there, just changed into aware of your weblog through Google, and located that it is really informative. I am going to be careful for brussels. I will appreciate if you continue this in future. Numerous other folks will probably be benefited from your writing. Cheers!

  25. Hey there! I’m at work browsing your blog from my new iphone 3gs!
    Just wanted to say I love reading your blog and look forward to
    all your posts! Carry on the fantastic work!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: