Easy as ABC?

Submitted by Marianne on October 5, 2016

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.

Watch Andrew Wiles talk about his work!

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 $1.$ Examples are $2$ , $3$ , $5$ , $7$ and $11.$ 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

$4 = 2 \times 2,$

$9 = 3 \times 3,$

and

$12 = 2 \times 2 \times 3.$

The result is the reason why multplication and primes go hand in hand. For example, if you know the prime factors of two numbers $a$ and $b$ , you immediately know the prime factors of their product

$c=a \times b.$

They are simply the prime factors of $a$ and the prime factors of $b$ taken together. As an example, if $a=4=2^2$ and $b=9=3^2$ , you immmediately know that the prime factors of the product ( $36$ ) are $2$ (twice) and $3$ (twice):

$4 \times 9 = 2^2 \times 3^2 = 36.$

When it comes to addition, though, the primes aren’t quite so amenable. Suppose you have three numbers $a,$ $b,$ and $c$ such that

$a+b=c.$

There’s nothing obvious you can say about the prime factors of the three numbers in this case. If $a=4$ and $b=9$ as above, then

$c=a+b=13,$

which is itself prime. In this case the result of the sum ( $c$ ) involves a larger prime ( $13$ ) than both $a$ and $b$ (which involve primes $2$ and $3$ .) But this isn’t always the case. If $a=4$ and $b=5$ then the result of the sum

$c=4+5=9=3^2,$

only involves a prime ( $3$ ) that nestles in size between those appearing in $a$ and $b$ ( $2$ and $5$ ). Is it possible to say anything at all about the primes appearing in $c$ compared to those appearing in $a$ and $b$ ?

One way of gauging the size of the primes that divide three numbers $a,$ $b,$ and $c$ is to multiply all their prime factors together, but only keeping one copy of each. For $a=4=2^2,$ $b=5$ and $c=9=3^2$ the prime factors are $2$ (twice), $5,$ and $3$ (twice). Multiplying them together but only using one copy of each gives

$2\times 3\times 5 = 30.$

Not quite as easy as ABC.

Even if you didn’t know what $a,$ $b$ and $c$ were to start with, knowing that this product is $30$ would tell you immediately that the only prime factors involved in the three numbers are $2,$ $3$ and $5.$ The product is called the radical of $abc$ and written as $rad(abc).$ 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 $a=4,$ $b=5$ and $c=9$ we have

$rad(abc)=2 \times 3 \times 5 = 30.$

This radical is bigger than the result $c=9$ of the sum. The same goes for the example $a=4$ , $b=9$ and $c=13.$ In this case the radical is

$rad(abc) = 2 \times 3\times 13 = 78,$

which is again bigger than the result $c=13$ 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 $a$ , $b$ and $c$ without common factors (because that would be boring), and with $a+b=c,$ we always have

$c<rad(abc).$

Unfortunately though, this isn’t true. A counter example is $a=2$ , $b=243$ and $c=2+243=245$ . The only prime factor of $a$ is $2$ , the only prime factor of $243$ is $3$ (since $243=3^5$ ) and the prime factors of $245$ are $5$ and $7$ (since $245=5\times 7^2$ ). The radical is therefore

$rad(abc) = 2 \times 3\times 5\times 7 = 210,$

which is smaller than $c=245.$

So instead, the ABC conjecture says that $c<rad(abc)$ is usually true, where the meaning of "usually" is made up of two components. The first component relates to the fact that there are infinitely many triples $a$ , $b$ and $c$ with no common factors (co-prime triples) and with $a+b=c.$ You can easily make as many as you like yourself: simply pick two co-prime numbers $a$ and $b$ and add them together to get $c$ . The "usually" in this context is interpreted as meaning that only a finite number of such triples don’t comply with the requirement that

$c<rad(abc).$ A finite number can still be large of course (think of 1,000,000) but it would still pale into insignificance compared to the infinity of cases that do comply. So is it true that only finitely many of the infinite number of co-prime triples $a$, $b$ and $c$ with $a+b=c$ do not have $c<rad(abc)$? Unfortunately not — it’s possible to find infinitely many such triples — which leads to the second meaning of the word "usually". The other way to create a little leeway is to allow $c$ to be a tiny little bit bigger than $rad(abc)$. Given any positive number $r$, the number \[ rad(abc)^{1+r} \] is bigger than $rad(abc)$. For example, if $r=1$ we get \[ rad(abc)^{2}, \] which is quite a lot bigger than $rad(abc).$ If you take $r$ to be really small, for example $r=0.000001,$ then \[ rad(abc)^{1+r} \] is only a tiny little bit bigger than $rad(abc).$Applying this meaning of "usually" and combining it with the first meaning gives us the ABC conjecture in its full glory. Given any number positive $r$, which can be as small as you like, there are only finitely many co-prime triples $a$, $b$ and $c$ with $a+b=c$ that do not have \[ c<rad(abc)^{1+r}. \] The triples that do not comply depend on the value you choose for $r$. There’ll be more such triples for a smaller $r$ than there are for a larger one. But however small or large your $r$, if the conjecture is true, you can rest assured that only finitely many triples don’t comply. The ABC conjecture might not be as easy to explain as Fermat’s last theorem, but a proof would be no less exciting. Many other important results in number theory would follow from it, including Fermat’s last theorem itself. It’s an exciting prospect — but it seems we need to wait for a little while longer to see it fulfilled. $$

Comments

Great explanation, thank you!

Permalink Submitted by Anonymous on November 11, 2016

Great explanation, thank you!

Easy as ABC?

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Comments

Great explanation, thank you!