The ABC’s of Arithmetic

abc123Some of the hardest problems in arithmetic are those that relate multiplication to addition. For example: Is every even number the sum of two primes? This is most assuredly a hard problem—mathematicians have been tackling it for centuries and so far nobody’s solved it. And it relates multiplication to addition. As soon as you talk about primes, you’re (implicitly) talking about multiplication, and of course when you talk about sums, you’re talking about addition.

Or: How many ways can you write the number 2 as the difference of two primes? You can write 2 = 5-3, or 2 = 7-5, or 2 = 13-11. That’s three so far. How many more are there? The betting is that the answer is “infinitely many”, but nobody knows for sure. This problem has stumped some of the best and the brightest not just for centuries but for millennia. And again it relates multiplication to addition. (Well, it relates multiplication to subtraction, but of course subtraction is just addition in reverse.)

The ABC problem has only been around for a few decades, but it’s in many ways the most interesting and important of the bunch. Tomorrow I’ll explain how you can help solve this problem. Today I’ll explain what the problem is.

Start with any three numbers—call them A, B and C—that satisfy the equation A+B=C. Well, not any three numbers—stick to numbers that have no common factors. So you could start, with, say, 4+11=15. Now list all the primes that divide any of these numbers. In this case we have 2 (which divides 4), 11 (which divides 11) and 3 and 5 (which divide 15). Multiply all these primes together and call the result D. In this case, D = 2 x 11 x 3 x 5 = 330. Notice that C (that is, 15) is considerably smaller than D (that is, 330).

Try playing with some examples, and you’ll find that C is almost always smaller than D—often by quite a lot. But not always. If you start with 2+243=245, the primes are 2 (which divides 2), 3 (which divides 243), and 5 and 7 (which divide 245). So D = 2 x 3 x 5 x 7 = 210. In this case, C (namely 245) is bigger than D. In fact, it’s bigger by a pretty respectable amount.

The ABC problem is, roughly, to figure out how often this happens. Nobody knows for sure, but the smart money says that C is rarely much bigger than D. Alright, that’s a little vague. In fact it’s doubly vague—I still have to tell you both what “rarely” means and what “much bigger” means.

When does C count as “much bigger” than D? We could arbitrarily say that C counts as “much bigger” when it’s bigger than, say, D3. (This would rule out our example where 245 is bigger than 210, but not bigger than 2103). The betting is that this happens rarely in the sense that out of all the infinitely many A’s, B’s and C’s you could start with, it happens only a finite number of times.

Now a critic could come along and say that you’ve chosen far too demanding a definition of “much bigger”. Maybe you should be a little more liberal and count C as “much bigger” than D whenever it’s bigger than, say, D2. Now there will be more examples—but the betting is that the number of examples is still finite.

Now let’s loosen up a little further and count C as “much bigger” whenever it’s bigger than D1.5, or D1.1 or D1.01 or D1.001 or D1.000000000001. (Under the last few of these criteria, our example with C=245 and D=210 qualifies.) The ABC conjecture says that no matter how many 0′s you throw into that exponent, there will still be only finitely many examples where C counts as much bigger than D.

The ABC problem is to determine whether the ABC conjecture is true, and our inability to solve this problem goes to the heart of our limited understanding of arithmetic. It turns out that this problem has deep connections to many of the hardest and most ancient problems about numbers. An affirmative solution to the ABC problem would yield, as side benefits, completely new proofs of some of the most spectacular recent results in number theory, including the Mordell Conjecture and (most of) Fermat’s Last Theorem. But it would also take us much farther. While Fermat’s Last Theorem deals with the equation xm+ym=zm, a solution to the ABC problem would allow us to enumerate all solutions to the far more general equation xm+yn=zk.

One of the nice things about the ABC problem is that you don’t have to solve it to deepen our understanding of arithmetic. Forget about the whole “C is much bigger than D” business, and just concentrate on finding examples where C is bigger than D. Compiling a more extensive list of examples can already be a substantial contribution. And you can help. Tomorrow I’ll tell you how.

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5 Responses to “The ABC’s of Arithmetic”


  1. 1 1 dave

    *rolls his sleeves up*
    at first glance i bet that the abc conjecture is false.
    its a series problem, right? like the search for primes themselves?

  2. 2 2 Tim Roberts

    This is the clearest exposition of the ABC conjecture I have come across – so much so that I hope to plagiarise it profusely when I add it to the Unsolved Problems web site at http://unsolvedproblems.org/

    Hope this is OK…..

    Tim

  3. 3 3 Steve Landsburg

    Tim Roberts: Thanks for the kind words. It’s an honor to be plagiarized.

  4. 4 4 Neil

    I must be missing something. Why cannot a computer program be written that systematically determines and compiles a list of numbers for which C is bigger than D?

  5. 5 5 Steve Landsburg

    Neil: One can. But there are two problems with this.

    First, the goal is to find numbers for which C is much bigger than D, for infinitely many different definitions of “much”. So you need not one computer program, but infinitely many.

    Second, no matter how long you run one of those programs, it will (after a finite time) have found only a finite number of examples. That throws exactly zero light on whether the total number of examples is finite or infinite.

    Nevertheless, it can be extremely useful to compile lists of examples. By studying these lists, mathematicians can begin to guess at the underlying patterns, and there’s a lot to be learned from that.

    So what we need is lots of people willing to put their computers to work on this problem. Which is what I’ll talk about tomorrow.

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