Sums with too few primes

Whole numbers have two structures, one additive and one multiplicative. Adding and multiplying are each entirely understood. What is not understood is how they interact, and two neighbours sharing out their small primes was one view of that. Consecutive numbers cannot both be built only from small primes, except for finitely many pairs. Take any two numbers and add them. Knowing the prime factors of $a$ and $b$ tells almost nothing about the prime factors of $a + b$. That single sentence states most of the hard problems about integers.

In 1985 David Masser and Joseph Oesterlé put it into a formula. For a coprime sum $a + b = c$, let the radical $\operatorname{rad}(abc)$ be the product of the distinct primes that divide $abc$. It is what is left when all repetition is stripped from the factorisation. Typically, it is a large number, typically much larger than $c$. The conjecture says that it cannot be much smaller, in a precise sense. Dorian Goldfeld called it “the most important unsolved problem in Diophantine analysis”, and it has been at the centre of a long controversy.

What the conjecture says #

The simplest case is $1 + 8 = 9$. Here the primes are 2 and 3, so the radical is 6, and that is less than 9. Measuring how much less gives the quality:

$$q(a, b, c) = \frac{\log c}{\log \operatorname{rad}(abc)}.$$

For $1 + 8 = 9$ it is $\log 9 / \log 6 = 1.2263$. For the great majority of sums, the quality is well below one: $1 + 2 = 3$ has radical 6 and quality 0.613. When the quality is above one the sum is called an abc triple. The conjecture is not that these do not exist, as there are infinitely many. It says that for each $\varepsilon > 0$, only finitely many triples have $q > 1 + \varepsilon$. Equivalently, $c < K_\varepsilon \operatorname{rad}(abc)^{1 + \varepsilon}$, where $K_\varepsilon$ is a constant depending only on $\varepsilon$.

So what is being bounded is a sum against repetition. A number with very little repetition can be big only if it has many different primes. If $a$, $b$ and $c$ are all built from high powers of few primes, then the conjecture says that they cannot add up exactly very often. The picture of a large power of 23 being almost exactly 3 to the 10th times 109 shows how unusual that is.

Every triple up to 200,000 #

A table of radicals makes this easy to enumerate. Since $\operatorname{rad}(a)\operatorname{rad}(b) < c/\operatorname{rad}(c)$ is necessary, only values of $a$ with small radicals have to be tried for each $c$, and that makes the search fast.

The counts below $10^3$, $10^4$ and $10^5$ are 31, 120 and 418, and they agree with the published tables. Between ten thousand and 200,000 the counts grow on a line of slope about 0.54. Stewart and Tijdeman showed in 1986 that infinitely many triples exist, and that their excess can even be large: there are triples with $c > \operatorname{rad}(abc) \exp(4\sqrt{\log c}/\log\log c)$, roughly, and this rules out replacing $\varepsilon$ by nothing. This is why there has to be a margin. The quality can exceed one, by an amount shrinking slowly, infinitely often.

A simple construction shows that much. Take $a = 1$, $c = 2^{2 \cdot 3^{k-1}}$ and $b = c - 1$. Then $3^k$ divides $b$, so that $b$ has a large power in it, and that makes the radical at most about $2b/3^{k-1}$, much smaller than $c$. It is a one-line way of producing as many triples as needed, and every one of them has a quality only slightly above 1.

Of those 606, 206 are above quality 1.1, 65 above 1.2, 17 above 1.3, and 7 above 1.4. The largest in this range is $1 + 4374 = 4375$, or $1 + 2 \cdot 3^7 = 5^4 \cdot 7$, with quality 1.5679. So it is not a big number that holds the record here, and the best of the range sits at a modest size.

Taking the range a decade at a time shows why no trend can be read from it. The best quality among the 6 triples up to 100 is 1.2920, at $1 + 80 = 81$. Among the 25 from there to a thousand it is 1.4266, at $3 + 125 = 128$. The 90 from there to ten thousand reach 1.5679. The 298 in the next decade reach 1.5471, at $343 + 59049 = 59392$, or $7^3 + 3^{10} = 2^{11} \cdot 29$, whose radical is just 1,218. The 187 from a hundred thousand to two hundred thousand reach only 1.4127. So the best value rises, then falls, then rises again further out to Reyssat’s. It behaves much as any maximum of a thin set should behave, and that is what any picture of this conjecture must look like, since it is a statement about what happens at the extreme tail, at all sizes.

Neighbours and near-powers #

Seventy-seven of the 606 triples have $a = 1$, so that $b$ and $c$ are consecutive numbers with very few primes between them. The 1.4557 of $1 + 2400 = 2401$ is one of these, where $2401 = 7^4$ and $2400$ has only the primes 2, 3 and 5. Such pairs are where this essay meets two neighbours sharing out their small primes. Størmer’s theorem says that for any finite set of primes only finitely many neighbours are built from them, and it is a theorem one can compute with. Abc goes much further, as it would limit the size of a pair by the size of the product of its primes, whatever those primes are, and that is exactly the piece of information that Størmer’s method does not give.

The same inequality speaks to powers close together. Eugène Catalan conjectured that 8 and 9 are the only consecutive perfect powers, and Preda Mihăilescu proved it in 2002 by quite different means. But $x^p - y^q$ being small, without being one, is still open: Hall’s conjecture says that $|x^3 - y^2|$ cannot be much smaller than $\sqrt x$ unless it is nought, and a form of abc implies a weak version of that. Pillai asked whether each difference occurs only finitely often. Abc would answer all of these in one stroke, in the form “only finitely many, for each fixed gap”. In each case the mechanism is the same. A sum of perfect powers has a radical that is small, because repetition has been stripped from it, and that is only possible a limited number of times. The problem of three cubes making thirty-three looks related and is not, since there the sum has three terms, and a sum of three numbers with few primes has no comparable restriction. The abc statement really is about exactly two summands.

The records #

Eric Reyssat’s $2 + 3^{10} \cdot 109 = 23^5$ has quality 1.6299. Benne de Weger found $11^2 + 3^2 \cdot 5^6 \cdot 7^3 = 2^{21} \cdot 23$, at 1.6260. Jerzy Browkin and Juliusz Brzeziński found $19 \cdot 1307 + 7 \cdot 29^2 \cdot 31^8 = 2^8 \cdot 3^{22} \cdot 5^4$, at 1.6235. Those are the only three known above 1.6, in spite of a distributed search in the Netherlands, and of very large computations by others, which have listed all triples up to $10^{18}$ and looked for more.

How does one find such things? Dividing Reyssat’s equation by $23^5$ shows that $3^{10} \cdot 109/23^5$ is within $3 \times 10^{-7}$ of one. So a triple of high quality is a place where a product of powers of primes is extremely close to one. In terms of logarithms it is a near-relation among numbers such as $\log 2$, $\log 3$ and $\log 23$. Lattice basis reduction, with the shortest vector in a lattice as its goal, finds such relations, and so does a good use of continued fractions as in a fraction that never closes. Alan Baker’s theorems on linear forms in logarithms set a lower limit on how close such a thing can come, and they supply the only known unconditional bound on $c$. Stewart and Kunrui Yu’s best version says that $c$ is less than about $\exp(\operatorname{rad}(abc)^{1/3 + \varepsilon})$. That is exponentially larger than what is conjectured. It is the gap between what is proved, an exponential, and what is believed, nearly linear in the radical.

A line for Fermat #

The reason the conjecture attracts so much attention is how much follows from it. The clearest example takes three lines.

Suppose $x^n + y^n = z^n$ with $x$, $y$, $z$ coprime. That is an abc sum with $c = z^n$. The radical of $x^n y^n z^n$ is the radical of $xyz$, which is at most $xyz$, and that is less than $z^3$. So the quality is greater than $\log z^n / \log z^3 = n/3$. An effective form of the conjecture, which gave an explicit limit on quality, would dispose of Fermat’s last theorem for all large $n$ in this one line. Even the weak statement that the quality is always below 2 would settle it for $n \ge 6$. It was historically an extraordinary effort to prove what this would do in an instant, as well as the reduction modulo primes that had preceded it. In fact, the relation is closer than an analogy. Gerhard Frey’s curve, built from a would-be solution of Fermat’s equation, is central to Andrew Wiles’s proof, and Lucien Szpiro’s conjecture about elliptic curves’ discriminants and conductors is essentially equivalent to abc, using the same curve.

There are many more consequences. Noam Elkies showed in 1991 that an effective form of abc gives Mordell’s conjecture effectively, so that the rational points on a curve of genus at least two could actually be found. Andrew Granville showed that it implies strong results on squarefree values of polynomials. Joseph Silverman showed that it implies that infinitely many primes are not Wieferich primes, which is a statement nobody has proved otherwise, even though nearly all primes are believed to be of that kind. Roth’s theorem on rational approximation of algebraic numbers follows too, and that is a Fields-Medal result of 1955. The range of the implications comes from how general the statement is. It is about addition alone against multiplication alone.

The proof for polynomials #

There is a version where everything is known, and it shows that the idea is right. For polynomials over the complex numbers, replace the size of an integer with degree, and replace the radical with the product of distinct linear factors. Then the theorem of Walter Stothers in 1981 and Richard Mason in 1984 says: if $a + b = c$ are coprime polynomials, not all constant, then

$$\max(\deg a, \deg b, \deg c) \le \text{(number of distinct roots of } abc) - 1.$$

There is no $\varepsilon$ and no constant, and the proof takes less than a page. Divide by $c$ and differentiate. That gives $a/c + b/c = 1$, so that $(a/c)' + (b/c)' = 0$, which is a Wronskian identity relating $a$ and $b$. A root of $a$ of multiplicity $m$ is a root of the Wronskian of multiplicity at least $m - 1$, and this bounds the degrees. Derivatives do all the work. That is what integers lack. There is nothing that acts as $d/dx$ on whole numbers, and all efforts to find an “arithmetic derivative” that behaves in this way have not produced one that is useful. In a sense this absence is what is difficult about the integer case. It is also why the polynomial version of Fermat’s theorem is a one-paragraph exercise.

How small a radical usually is #

Squarefree numbers are their own radical, and there are $6/\pi^2$ of them, a number connected to the sum of reciprocal squares. The count up to 100,000 gives 60.793%, in agreement with the limit to five digits. Numbers with very small radicals are rare: of the 100,000, just 161 have a radical at most 10, and 28,255 have one at most 10,000. These are built from small primes, as in the typical factorisation, and a heuristic for abc is built on that rarity. Choose $a$ and $b$ independently of size about $N$. The chance that all of them have small radicals is small, and summing the probabilities predicts the number of triples of each quality. Oliver Robert, Cameron Stewart and Gérald Tenenbaum refined such a count in 2014, and proposed a precise form of $K_\varepsilon$. It predicts that the largest quality should indeed be bounded, and that the best known values might be close to the best there are, although there is no evidence for that beyond its arithmetic plausibility.

How far these figures go #

The figures cannot see the conjecture, which concerns what happens at all sizes. Every triple up to 200,000 is a very small sample, and the record itself is off the scale of the scatter plot, at 6.4 million. Enumerations reaching $10^{18}$ have shown the qualities thinning at each scale without a clean law. And a picture of qualities tells nothing about the constant $K_\varepsilon$, and it cannot distinguish the conjecture from a slightly different statement with a slowly growing correction, of the kind suggested by the Stewart–Tijdeman lower bound. No finite data could. So nothing here is evidence in the sense of a test the conjecture could have failed, because every triple found has a finite quality, and the claim concerns only whether finitely many exceed each level.

What was computed and what was quoted #

All triples with $c$ up to 200,000 were found by exact integer arithmetic from a sieve of radicals, and the counts below $10^3$, $10^4$ and $10^5$ were checked against the published values. Quality is computed in double precision from exact integers. The named triples were all recomputed: the identities were checked exactly and the radicals recomputed from factorisations. The squarefree share and the small-radical counts are exact for $n$ up to 100,000. The Fermat inequality is given in full.

The conjecture itself is unproved. The theorems of Stewart and Tijdeman, Stewart and Yu, Stothers and Mason, Elkies, Granville and Silverman, and the statements on large computations and the records, are quoted. The heuristic of Robert, Stewart and Tenenbaum is quoted, and it was not compared numerically with these counts.

Still open: the status of the proof #

Is the abc conjecture true? In August 2012 Shinichi Mochizuki posted a series of four long papers on what he calls inter-universal Teichmüller theory, claiming a proof. They were accepted and published in a journal of his own institute in 2021. In 2018 Peter Scholze and Jakob Stix examined them, and they reported what they consider to be a serious and unfixable gap in a key corollary, a statement that is not accepted by Mochizuki and his colleagues. The result is an unusual situation in mathematics: a published claim that most number theorists outside one school do not treat as established. Efforts to formalise parts of the argument have been proposed, but none has settled it. The practical test is simple, as a working proof ought to yield effective bounds and many corollaries, and nobody has yet derived those from the papers in a way widely accepted.

Beyond that, there are questions at a smaller scale: what is the supremum of quality over all triples? It may be exactly Reyssat’s 1.6299, or something a little larger, and there is no way to tell. The weakest conjecture, that quality is bounded by a constant, is itself open. So it is not known that a triple with quality 100 does not exist.

Addition against repetition #

A sum of two coprime numbers can be built from only a few primes, but rarely, and the abc conjecture says that the size of the sum cannot much exceed the product of its distinct primes. Enumerating every triple up to 200,000 shows how quickly high qualities thin out: from 606 triples to 7 above 1.4, with a maximum of 1.5679, and with Reyssat’s 1.6299 not far beyond.

The appeal of the statement is that it makes a clean claim of independence between adding and multiplying, while being simple enough to test with a sieve. Fermat’s last theorem, for large exponents, would follow in one line. For polynomials, it takes one derivative. And for whole numbers, after forty years and one disputed proof, nobody knows.