Tuesday, April 16, 2019

The abc Conjecture: abc Triples

This is the second post in a series about the abc conjecture. For the first post, see here.

In the last post, we defined the radical of an integer n, namely the product of distinct prime factors of n. We suspected in the last post that for most equations a + b = c where a and b are relatively prime, rad(abc) > c. This is because this inequality expresses our hypothesis that there should not be too many high powers of primes in the factorizations of a, b, and c. As a result, we made the following conjecture:

Conjecture 1: For all but finitely many equations of the form a + b = c where a and b are relatively prime, rad(abc) > c.

However, as mentioned at the end of the previous post, this is in fact false. To prove this, we have to exhibit an infinite family of equations a + b = c with rad(abc) ≤ c. Any triple (a,b,c) of numbers satisfying this property is called an abc triple. The only example we've seen so far is (1,8,9), or in equation form, 1 + 8 = 9. In terms of this new definition, we are trying to show that there are infinitely many abc triples. The following claim gives the desired result.

Claim: For any prime number p grater than 2, the triple (a,b,c) = (1,2p(p-1) - 1,2p(p-1)) is an abc triple.

Proof: This family is infinite because there are infinitely many prime numbers p. The proof depends on a fact in elementary number theory known as the Euler-Fermat Theorem. This theorem may be used to show that b = 2p(p-1) - 1 is divisible by p2. This is significant because we now know that the radical of b cannot be greater than b divided by p; this is because taking the radical of b "forgets" about at least one of the factors of p. Of course, rad(1) = 1 and rad(2p(p-1)) = 2 so

Since p > 2, this last value is less than c, so that we do in fact have an infinite family of abc triples.

In fact the situation is even worse than this. Since the radical is less than 2c/p (as shown in the proof), it is not enough to replace the hypothesis rad(abc) > c with 2rad(abc) > c, or any higher multiple. We can make 2/p arbitrarily small by increasing p so that the radical is smaller than c by an arbitrarily large factor. For example, taking p = 5 gives the abc triple (1,1048575,1048576). Note that 52 = 25 divides b = 1048575, as claimed. Our proof guarantees that rad(abc) ≤ 2c/5. In fact the radical of this product is 419430. This is indeed less than 2/5 of c.

All of this shows that we cannot correct our conjecture 1 by adding a multiplicative factor to our inequality. The next reasonable thing one might try is a power law. Perhaps rad(abc)2 > c for all but finitely many equations, or something similar. This, in fact, is the correct idea. However, the choice of 2 as the exponent again seems arbitrary. We know already that the statement is false when the power is 1, so let's try increasing it just a little. This leads us to the actual abc conjecture.

The abc Conjecture: For any ε > 0, no matter how small, for all but finitely many equations of the form a + b = c where a and b are relatively prime, rad(abc)1 + ε > c.

The variable ε could for example be 1, and then we recover the rad(abc)2 > c inequality. However, ε could also be very close to 0, giving an exponent of 1 + ε very close to 1. Crucially, any function x1 + ε with ε > 0 eventually increases faster than any constant multiple of x, for example x1.1, x1.00001, etc. Therefore, this conjecture gets around the counterexample to conjecture 1. Nevertheless, the abc conjecture in some sense says that conjecture 1 is really close to being true. All we needed to do was increase the exponent by any positive amount. These concepts may become a little clearer with a new concept, called the quality of a triple (a,b,c). The formula for the quality, denoted q(a,b,c) is

This is another measure of how large c is compared to rad(a,b,c). In fact, rad(a,b,c)q(a,b,c) = c. For example, q(13,22,35) is about 0.386, and q(1,8,9) is close to 1.226. This allows a more succinct description of the conjecture: for most triples, q ≤ 1. It follows from our definition that abc triples are those for which q > 1. Finally, the abc conjecture is equivalent to the following.

The abc Conjecture (Second Formulation): For any ε > 0, no matter how small, for all but finitely many equations of the form a + b = c where a and b are relatively prime, q(a,b,c) < 1 + ε.

Let's see if our conjecture seems plausible from the numerical data. One possible way to do this is to come up with many triples and see how large the quality q is for each.

In the above diagram (click to enlarge), the x-axis is our variable c. For each c between 2 and 2000, the plot goes through all possible relatively prime values a and b adding to c, finds the triple among these with the highest quality, and plots a corresponding point there. Therefore, all points are already among the highest quality triples. Even among these, abc triples (those that lie above the horizontal q = 1 line) are rare. Furthermore, they seem to get even more rare as c increases. In terms of diagrams of this sort, the conjecture states that only finitely many dots lie above a given horizontal line q = 1 + ε for any ε > 0. The highest quality abc triple that appears on the plot is (3,125,128) = (3,53,27), with a quality of 1.426. Are there any higher quality triples out there?

In fact, there are well over a hundred known with higher quality, a list of which may be found here. Currently, the highest known quality belongs to the triple (2,6436341,6436343) = (2,310109,235), with q = 1.6299. Even assuming the abc conjecture does not answer the question of whether this triple is really the highest quality there is. All it says is that examples of this sort must eventually die out as we approach infinity. For instance, there may very well be no triples at all with q ≥ 2, meaning that crad(abc)2 may hold in all cases with no exceptions.

Of course, no matter how many examples we check, we are no closer to proving that the abc conjecture holds. In the last post, we will discuss attempts to prove it, as well as the applications of the statement, should it be true.

Sources: http://www.math.leidenuniv.nl/~desmit/abc/, Greg Martin and Winnie Miao: abc Triples; Arxiv:1409.2974v1 [math.NT] 10 sep 2014, Brian Conrad: The abc Conjecture, 12 sep 2013.