The abc conjecture

Recently, the Japanese mathematician Shinichi Mochizuki announced that he had a proof of the abc conjecture. This claim was significant enough to receive plenty of major news coverage, including from the New York Times. His strategy was to use entirely novel self-developed techniques — so-called inter-universal Teichmüller theory. In fact, practically no one else in the mathematics community is conversant in these techniques, so it will take quite a long time before the proof is sufficiently refereed. If it turns out to be accurate, however, it will undoubtedly a huge achievement!

So what is the abc conjecture, anyway? Essentially, it lies somewhere in the intersection of additive and multiplicative number theory. Curiously, it’s a statement about one of the simplest equations possible:

\( a + b = c \)

for natural numbers \( a \), \(b\), and \(c\). Naturally, there are infinitely many solutions to this equation. Now, it’s obvious that this equation is additive, but we can give it a multiplicative flavor if we consider the various prime divisors of \(a\), \(b\), and \(c\). In particular, for this conjecture, we’re interested in repeated divisors — cases where \(a\), \(b\), and \(c\) are divisible by perfect powers.

It’s probably helpful to use a concrete example to illustrate some of the ideas here. So suppose \( a = 81, b = 128, c = 209. \) We see that indeed \( a + b = c. \) Now, \( a = 3^4, b = 2^7, \) but \( c = 11 \cdot 19. \) Thus, although \( a \) and \( b \) are divisible by large prime powers, it seems that \( c \) is not. Of course this is simply one example, but in general this sort of pattern appears to be true. Try it out for yourself and see if you can find any counterexamples!

In order to describe the abc conjecture precisely, it will help to introduce a little terminology. Let’s define the radical of a natural number \( n \), denoted \( \mathrm{rad}(n), \)  to be the product of the distinct prime factors of \( n. \) For instance,

\( \mathrm{rad}(12) = \mathrm{rad}(72) = 2 \cdot 3 = 6, \mathrm{rad}(2^7) = 2, \mathrm{rad}(1) = 1, \) etc.

One formulation of the abc conjecture tries to compare \( c \) to \( \mathrm{rad} (abc) \). We can think of this in the following way: to what power must we raise the radical of \( abc \) in order to obtain \( c \)? In other words, in the form of an equation, what exponent \( q \) satisfies

\( \displaystyle (\mathrm{rad} (abc))^{q} = c \)?

The more that \( c \) is highly divisible by primes, the larger value of \( q \) we will need. Solving for \( q \), we obtain

\( \displaystyle q = \frac{\log c}{\log (\mathrm{rad} (abc))} \).

For convenience, let’s call this quantity \( q \) the quality of the abc-solution. We claim that large values of \( q \) arise when \( a\), \(b\), and \(c\) are divisible by prime powers. It’s worth verifying this claim in order to get some intuition for how this quantity \( q \) is a useful construct. So suppose for the sake of argument that \(a\), \(b\), and \(c\) are all perfect powers. For instance, let \( a = x^n, b = y^n, c = z^n \) for some natural numbers \( x, y, z, n. \) Then we get the infamous Fermat equation \( x^n + y^n = z^n \). Of course, it’s now known that no solutions exist for \( n \ge 3 \), but for the sake of argument, let’s suppose solutions do in fact exist for arbitrarily large values of \( n \). What is the quality of such an abc-solution? Well,

\(\begin{align*} \log c &= \log (\max(a, b, c))\\
&\ge \frac{1}{3}(\log a + \log b + \log c)\\
&= \frac{1}{3}\log abc\\
&= \frac{1}{3}\log (x^ny^nz^n)\\
&= \frac{1}{3}\log(xyz)^n\\
&= \frac{n}{3}\log(xyz)\\
&\ge \frac{n}{3}\log(\mathrm{rad}(abc)).


\( \displaystyle q =\frac{\log c}{\log (\mathrm{rad} (abc))} \ge \frac{n}{3}. \)

If \( n \) can be arbitrarily large, then so can \( q. \) However, the abc conjecture establishes a restriction on the size of \( q \) as follows.
The abc conjecture: For any real number \(C \) greater than \(1\), all but a finite number of \(abc\)-solutions have a quality \(q\) less than or equal to \(C\).
There are several alternative formulations as well — you can check them out at Wikipedia or MathWorld — but this one, adapted from Barry Mazur’s piece Questions about Number, appealed to me most. Let’s hope that Mochizuki’s proof is successful and opens up new doors in number theory! For further reading and research, here is a list of the consequences of the abc conjecture.

Training games

I’ve played a couple training games with a friend of mine from undergraduate recently. Here is one of them, which features an instructive rook and pawn endgame.

[Event "ICC"]
[Site "Internet Chess Club"]
[Date "2012.09.14"]
[Round "?"]
[White "Aaron"]
[Black "Jason"]
[Result "1-0"]
[WhiteELO "?"]
[BlackELO "?"]

1.d4 Nf6 2.c4 e6 3.Nf3 Bb4+
{The Bogo-Indian. A solid choice.}
4.Bd2 Qe7 5.g3 Nc6 6.Nc3
( 6.Bg2 Bxd2+ 7.Qxd2 $6 Ne4 8.Qc2 Qb4+ {and Black equalizes.} )
6...O-O 7.Bg2 d6
( 7...Bxc3 8.Bxc3 Ne4 {and the bishop is snagged.} )
8.O-O e5
( 8...Bxc3 9.Bxc3 Ne4 10.Be1 {is now possible.} )
9.Nd5 $1
{Emphasizing Black's failure to capture on c3.}
9...Nxd5 10.cxd5 Bxd2
{Forced. But now White has an interesting choice of captures.}
11.Qxd2 $6
( 11.dxc6 Bh6 12.dxe5 dxe5 13.Qd5 {was the most precise.} )
11...Nxd4 12.Nxd4 exd4 13.Qxd4 $5
( 13.Rfe1 {is simpler, but the game continuation is more combative.} )
{Facing the prospects of a passive defense of the weak c7-pawn, Black
chooses instead to grab some material and start a fight. But now White
obtains a massive lead in rook development.}
14.Rfe1 Qb5
( 14...Qg4 15.Re4 )
15.Rac1 Re8
{The idea of Black's Queen placement. Now Black's back rank hangs by a
thread! }
{Finally, I decided on this simple continuation. Trading off a pair of
rooks highlights Black's pathetic rook on a8.}
( 16.Qc4
( 16.Bf1 Qd7 17.Qg4 Qd8 18.Qg5 Bd7 )
( 16...Qxc4 $4 17.Rxe8# )
16...Qxe8 17.h4
( 17.Rxc7 $2 Qe1+ 18.Bf1 Bh3 )
{Black tries to defend c7, but he cannot hold it for long.}
18.Qc3 Bf5
{Black finally mobilizes, but now the material balance is
reestablished, with White maintaining the active Rook.}
19.Qxc7 Qxc7 20.Rxc7 Rb8
{With perfect play, the position is likely a draw. However, Black's
pieces are so passive that White has excellent practical chances.}
21.Bf3 $1
{Starting a good plan of kingside expansion. In order to create
winning chances, White must activate his King and probe for more
{Removing the a-pawn from the vulnerable 7th rank.}
22.g4 Bc8
{Not ideal, but if the Bishop abandons the c8-h3 diagonal, White can
continue with Rc7-d7, picking off the d6-pawn. }
23.Kg2 Kf8 24.Kg3 Ke8 25.Be4 $1
{Probing for weaknesses in order to try to gain entry for the King
into Black's position.}
{Regardless of whether Black plays h7-h6 or g7-g6, he creates holes in
his sixth rank.}
26.Kf4 Bd7 27.Bf5 Bxf5 $2
( 27...Bb5 {and White has trouble coordinating his King and Bishop,
since both of them would like to occupy the f5-square.} )
28.Kxf5 b5
{Starting the plan of b5-b4 and Rb8-b5. But it's too slow.}
29.g5 $1 hxg5 30.hxg5 b4 31.g6 $1
{The point. White gets into the sixth rank. }
31...fxg6+ 32.Ke6
{White is now 'up a King' and winning.}
32...Kf8 $6
( 32...Rd8 33.Rxg7 Kf8 34.Rxg6 {and White is still winning, but not as
quickly.} )
33.Kxd6 Rb6+
{Basically a spite check, as Black cannot realistically stop White's
34.Rc6 Rb7
{This move does lose immediately, but it's hard to criticize since
Black is losing regardless.}
35.Rc8+ Kf7 36.Rc7+
{and Black resigned.}
( 36.Rc7+ Rxc7 37.Kxc7 {and the d-pawn is unstoppable.} )