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In a series of 4 highly technical papers, Shinichi Mochizuki claims to have proved the

• Only finitely many numbers of the form n(n+11)(n+111) are perfect cubes.

• Only finitely many numbers of the form n! + 1111 are perfect squares.

• The equation x^a + y^b = z^c has only finitely many solutions if x,y,z are relatively prime positive integers and a,b,c are positive integers with 1/a+1/b+1/c < 1.

Here two positive integers are

I have no idea whether Mochizuki's proof is correct, and I couldn't understand it without a couple years of very hard work. All I can say is that he's a serious mathematician, a Japanese algebraic geometer who has been developing Grothendieck's ideas on so-called 'anabelian geometry'. The relevant papers are the 4 entitled 'Inter-Universal Teichmuller Theory' at the bottom of his webapge here:

http://www.kurims.kyoto-u.ac.jp/~motizuki/papers-english.html

But what does the abc conjecture actually

The Wikipedia article lists many consequences of this conjecture - I've just skimmed the surface.

Thanks to +Alexander Kruel and +Scott Carnahan for pointing this out. If anyone knows the latest news on this work, let me know!

**abc conjecture**. Though you may have never heard of it, it's a powerful conjecture in number theory that would have millions of consequences, like:• Only finitely many numbers of the form n(n+11)(n+111) are perfect cubes.

• Only finitely many numbers of the form n! + 1111 are perfect squares.

• The equation x^a + y^b = z^c has only finitely many solutions if x,y,z are relatively prime positive integers and a,b,c are positive integers with 1/a+1/b+1/c < 1.

Here two positive integers are

**relatively prime**if the largest integer dividing both of them is 1. Also, n here is a positive integer. Also, we could replace the numbers 11, 111 and 1111 by any other integers if we wanted to!I have no idea whether Mochizuki's proof is correct, and I couldn't understand it without a couple years of very hard work. All I can say is that he's a serious mathematician, a Japanese algebraic geometer who has been developing Grothendieck's ideas on so-called 'anabelian geometry'. The relevant papers are the 4 entitled 'Inter-Universal Teichmuller Theory' at the bottom of his webapge here:

http://www.kurims.kyoto-u.ac.jp/~motizuki/papers-english.html

But what does the abc conjecture actually

*say*? It says that for every real number p > 1, there are only finite many triples (a,b,c) of relatively prime positive integers with a + b = c such that c > x^p, where x is the product of the prime factors of abc.The Wikipedia article lists many consequences of this conjecture - I've just skimmed the surface.

Thanks to +Alexander Kruel and +Scott Carnahan for pointing this out. If anyone knows the latest news on this work, let me know!

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- +Thomas R. Thanks for the link.Sep 9, 2012
- There's also an article in
*Nature*, pointed out by +Heather Vandagriff:

http://www.nature.com/news/proof-claimed-for-deep-connection-between-primes-1.11378

But you heard it here first, folks! :-) And the Mathoverflow page is the best for people who want technical details:

http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjectureSep 11, 2012 - I wonder if these results (assuming they pan out) will also advance or throw new light on the Langlands Program. Seems like they are in the same ball park.Sep 16, 2012
- What Mochizuki is claiming is a rather strong, completely effective form of ABC. (Perhaps
**too**strong to be conceivably true? - I have a feeling it might be in contradiction with the Stewart-Tijdeman "near-misses" of ABC). If true, it will certainly yield effective Mordell, and a new proof of FLT. (And yes, effective Mordell needs an**effective**ABC; they are actually, in a sense, equivalent. Also, effective ABC implies an effective Roth theorem).

For a raw, explicitly effective claim from Mochizuki's fourth paper, take a look at his Theorem 1.10, which is a preliminary, restricted version of the Szpiro inequality (one form of the ABC conjecture). There, the "log(q)" on the left-hand side of the asserted inequality is precisely the logarithmic minimal discriminant log D of the elliptic curve. On the right hand side, f is the conductor N of the elliptic curve; the other term is negligible if you restrict to the number field Q. So this assertion is, essentially, an explicit form of Szpiro's inequality 1/6 log(D) < (1+epsilon) log N + const.

@John Ramsden: As far as I know, the only hint of connection between ABC and Langlands is that ABC and Taniyama-Shimura (proved by Wiles; a tiny part of Langlands) can be unified into a single conjecture: every rational elliptic curve of conductor N admits a surjective map of degree << N^{2+epsilon} from the modular curve X_0(N) of level N. In this sense, they give estimates of complementary kind: Langlands optimizes the level N, ABC optimizes the degree of the covering.Sep 21, 2012 - Here's a really simple explanation of the ABC Conjecture:

(but not the proof, for obvious reasons)

http://youtu.be/RkBl7WKzzRwOct 12, 2012 - I'll share that one, Nic!Oct 12, 2012

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