Mathematics is full of stories of geniuses toiling away in
isolation before presenting the world with a monumental result that
takes years for everyone else to understand. A famous example is Fermat's last theorem, a
350-year old problem that was solved in the 1990s by Andrew Wiles
after years of secretive work. Another is the Poincaré conjecture,
proved in the 2000s by the reclusive Grigory Perelman who later
refused all reward and recognition for his work.
Shinichi Mochizuki.
This summer the world of number theory made a renewed effort to
understand another such result, first published in 2012 by the mathematician
Shinichi Mochizuki. Mochizuki had spent many years working in
private on a famous problem known as the ABC conjecture. His
papers, which run to over 500 pages, purport to prove the conjecture,
but turned out to be impenetrable. There were
too many new formalisms and too much
unfamiliar terminology for other mathematicians to handle. Mochizuki didn't help by refusing to travel
outside of Japan to explain his work. A conference that took place in
Oxford last December, without Mochizuki, failed to elucidate the
proof, though it did give people a deeper insight into his methods. It
wasn't until this summer, nearly four years after publishing the
papers, that Mochizuki offered some more pentrable guidance at
a conference in Kyoto. And although his proof is far from understood,
attending mathematicians expressed some hope that they will eventually get
there.
The ABC conjecture cuts right to the heart of number theory by
linking its two most basic operations: addition and
multiplication. Multiplication is intimately linked to
those favourite friends of number theorists, the prime numbers. Primes
are natural numbers that are only divisible by themselves and
Examples are , , , and A famous result that was
already known to the ancient Greeks states that every natural
number can be written as a product of primes, for example
and
The result is the reason why multplication and primes go hand
in hand. For example, if you know the prime factors of two numbers
and , you immediately know the prime factors of their product
They are simply the prime factors of and the prime factors of taken
together. As an example, if and , you immmediately
know that the prime factors of the product () are (twice) and
(twice):
When it comes to addition, though, the primes aren't quite so
amenable. Suppose you have three numbers and such that
There's nothing obvious you can say about the prime factors
of the three numbers in this case. If and as above, then
which is itself prime. In this case the result of the
sum () involves a larger prime () than both and (which
involve primes and .) But this isn't always the case. If
and then the result of the sum
only involves a prime () that nestles in size
between those appearing in and ( and ). Is it possible
to say anything at all about the primes appearing in compared to
those appearing in and ?
One way of gauging the size of the primes that divide
three numbers and is to multiply all their prime factors
together, but only keeping one copy of each. For
and the prime factors are (twice), and
(twice). Multiplying them together but only using one copy of each
gives
Not quite as easy as ABC.
Even if you didn't know what and were to start with,
knowing that this product is would tell you immediately that the only prime
factors involved in the three numbers are and The
product is called the
radical of and written
as You can regard it as a sort of average. It doesn't tell
you what the three numbers involved are exactly, but it encodes
information about the size of their prime factors.
In our example with
and we have This
radical is bigger than the result of the sum. The same goes for
the example , and In this case the radical is
which is again bigger than
the result of the sum. Playing around with a few more examples will soon lead you to suspect that this is always the case.
This is what the ABC conjecture would like to say. That given a triple , and without common factors (because that would be boring), and with we always have
$$c
Further reading
You can read a lovely, and more detailed, article about the ABC conjecture and its purported proof in Quanta Magazine.
Comments
Great explanation, thank you!
Great explanation, thank you!