A Tantonalizing Problem

In times like these one craves distraction, or maybe anaesthesia. On the whole, mathematics is better for you than ethanol, and you can even do it while driving. So in spare moments I’ve been noodling away at the following problem, tweeted a week ago by James Tanton:

Tanton tweet

The answer to Tanton’s question is surely No: The series will never again land on an integer. I leaped to that conclusion immediately after reading the definition of the series and glancing at the first few terms. But what makes me so sure? Can I prove it?

I wrote a quick program to generate more terms:

1
2
5/2
17/6
37/12
197/60
69/20
503/140
1041/280
9649/2520
9901/2520
111431/27720
113741/27720
1506353/360360
1532093/360360
1556117/360360
3157279/720720
54394463/12252240
18358381/4084080
352893319/77597520

Overall, the trend visible in these results seemed to confirm my initial intuition. When the fractions are expressed in lowest terms, the denominator generally grows larger with each successive term. Looking at the terms more closely, it turns out that the denominators tend to be products of many small primes, whereas the numerators are either primes or products of a few comparatively large primes. For example:

\[\frac{9649}{2520} = \frac{9649}{2^3 \cdot 3^2 \cdot 5 \cdot 7} \qquad \textrm{and} \qquad \frac{18358381}{4084080} = \frac{59 \cdot 379 \cdot 821}{2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17}.\]

To produce an integer, we need to cancel all the primes in the factorization of the denominator by matching primes in the numerator; given the pattern of these numbers, that looks like an unlikely coincidence.

But there is reason for caution. Note the seventh term in the sequence, where the denominator has decreased from \(60\) to \(20\). To understand how that happens, we can run through the calculation of the term, which starts by summing the six previous terms.

\[\frac{60}{60} + \frac{120}{60} + \frac{150}{60} + \frac{170}{60} + \frac{185}{60} + \frac{197}{60} = \frac{882}{60}.\]

Then we calculate the mean, and add 1 to get the seventh term:

\[\require{cancel}\frac{882}{60} \cdot \frac{1}{6} = \frac{882}{360} = \frac{\cancel{2} \cdot \cancel{3} \cdot \cancel{3} \cdot 7 \cdot 7}{\cancel{2} \cdot 2 \cdot 2 \cdot \cancel{3} \cdot \cancel{3} \cdot 5} = \frac{49}{20} + 1 = \frac{69}{20}\]

Cancelations reduce the numerator and denominator of the mean by a factor of 18. It seems possible that somewhere farther out in the sequence there might be a term where all the factors in the denominator cancel, leaving an integer.

Another point to keep in mind: For large \(n\), the value of the Tanton function grows very slowly. Thus if integer values are not absent but merely rare, we might have to compute a huge number of terms to get to the next one. Reaching the neighborhood of 100 would take more than \(10^{40}\) terms.

value of tanton(n) for n from 1 to 300

So what do you think? Can we prove that no further integers appear in Tanton’s sequence? Or, on the contrary, might my instant conviction that no such integers exist turn out to be an alternative fact?


I’ve had my fun with this problem. I know the answer now, but I’m not going to reveal it yet. Others also deserve a chance to be distracted, or anaesthetized. I’ll be back in a few days to follow up—unless commenters explain what’s going on so thoroughly there’s nothing left for me to say.


Update 2017-01-30: Okay, pencils down. Not that anyone needs more time. As usual, my readers are way ahead of me. (See comments below, if you haven’t read them already.)

My own slow and roundabout voyage of discovery went like this. I had written a little piece of code for printing out n terms of the series, directly implementing the definition given in James Tanton’s tweet:

from fractions import Fraction as F
from statistics import mean

def tanton (n):
    seq = [F(1)]
    for i in range(n):
        print(seq[i])
        seq.append(mean(seq) + 1)

But this is criminally inefficient. On every pass through the loop we calculate the mean of the entire sequence, then throw that work away and do it all again the next time. Once you have the mean of \(n-1\) terms, isn’t there some way of updating it to incorporate the nth term? Well, yes, of course there is. You just have to appropriately weight the new term, dividing by n, before adding it to the mean. Here’s the improved code:

from fractions import Fraction as F

def faster_tanton (n):
    m = F(1)
    for i in range(1, n):
        print(m)
        m += F(1, i)

Tracing the execution of this function, we start out with 1, then add 1, then add 1/2, then 1/3, then 1/4, and so on. This is 1 plus the harmonic series. That series is defined as:

\[H_{n} = \sum_{i=1}^{n} \frac{1}{i} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\]

The first 10 partial sums are:

1
3/2
11/6
25/12
137/60
49/20
363/140
761/280
7129/2520
7381/2520

One fact about the harmonic series is very widely known: It diverges. Although \(H_{n}\) grows very slowly, that growth continues without bound as \(n\) goes to infinity. Another fact, not quite as well known but of prime importance here, is that no term of the series after the first is an integer. The simplest proof shows that when you factor the numerator and the denominator, the denominator always has more \(2\)s than the numerator; thus when the fraction is expressed in lowest terms, the numerator is odd and the denominator even. This proof can be found in various places on the internet, such as StackExchange. There’s also a good explanation in Julian Havil’s book Gamma: Exploring Euler’s Constant.

Neither of those sources mentions anything about the origin or author of the proof. When I scouted around for more information, I found more than a dozen sources that attribute the proof to “Taeisinger 1915,” but with no reference to an original publication. For example, a recent paper by Carlo Sanna (Journal of Number Theory, Vol. 166, September 2016, pp. 41–46) mentions Taeisinger and cites Eric Weisstein’s Concise Encyclopedia of Mathematics; consulting the online version of that work, Taeisinger is indeed credited with the theorem, but the only reference is to another secondary source, Paul Hoffman’s biography of Erdős, The Man Who Loved Only Numbers; there, on page 157, Hoffman writes, “In 1915, a man named Taeisinger proved. . .” and gives no reference or further identification. So who was this mysterious and oddly named Taeisinger? I have never heard of him, and neither has MathSciNet or the Zentralblatt or the MacTutor math biography pages. In Number Theory: A Historical Approach John J. Watkins gives a slender further clue: The first initial “L.”

After some further rummaging through bookshelves and online material, I finally stumbled on a reference to a 1915 publication I could actually track down. In the Comptes Rendus Mathematique (Vol. 349, February 2011, pp. 115–117) Rachid Aït Amranea and Hacène Belbachir include this item in their list of references:

L. Taeisinger, Bemerkung über die harmonische Reihe, Monatsch. Math. Phys. 26 (1915) 132–134.

When I got ahold of that paper, here’s what I found:

Opening lines of the Theisinger 1915 paper

Not Taeisinger but Theisinger!

I still don’t know much of anything about Theisinger. His first name was Leopold; he came from Stockerau, a small town in Austria that doesn’t seem to have a university; he wrote on geometry as well as number theory.

What I do know is that a lot of authors have been copying each other’s references, going back more than 20 years, without ever bothering to look at the original publication.

Posted in mathematics, problems and puzzles | 7 Comments

The Polite Apocalypse

On the Beach cover of 1957 editionI’ve been rereading On the Beach, Nevil Shute’s novel about humanity’s last gasp in the aftermath of a nuclear war. The book was published 60 years ago, in 1957. I first read it in 1963, which is roughly when the events in the story are supposed to take place. It was also just a few months after the Cuban missile crisis, when annihilation was in the air.

Spoiler alert: Everybody dies.

The setting is Melbourne, Australia, the southernmost major city on the planet. The entire population of the Northern Hemisphere was wiped out in the war, and airborne radioactivity is slowly creeping across the Equator. Darwin and Cairns, on Australia’s north coast, are already ghost towns, and the people of Melbourne are told they have less than a year to go.

A U.S. submarine takes refuge in Melbourne’s harbor. Over a period of weeks the captain of that vessel, Dwight, forms an attachment to a young woman named Moira. There’s affection on both sides, and maybe passion, but Dwight is determined to remain faithful to his wife and children back in Connecticut. He buys them presents: a diamond bracelet, a fishing rod, a pogo stick. He speaks of them in the present tense. Is Dwight delusional? Not exactly. He knows perfectly well that his family are all dead, and that he’ll never rejoin them except in the sense that he too will soon be dead. But those deaths are abtractions.

He had seen nothing of the destruction of the war . . . ; in thinking of his wife and of his home it was impossible for him to visualize them in any other circumstances than those in which he had left them. He had little imagination, and that formed a solid core for his contentment in Australia.

It’s not just Dwight who lacks imagination—or chooses to ignore the truths it reveals. Moira studies shorthand and typing for a future job that will never exist. Her father harrows fields for crops that will never grow. Another couple plant hundreds of daffodils whose blooms they will never see, and they invest in a lawn mower for grass they’ll never cut.

The author himself seems to share this selective connection to reality. Everyone in his doomed society is unfailingly polite, and usually cheerful. Civilization may be ending, but not civility. There’s not a single act of violence or malice or even selfishness in the entire story. Shute mentions no hoarding or profiteering, much less rape and pillage. No marauding bandits or desperate refugees from the contaminated north descend on this last haven. On the Beach is the antithesis of that other Australian vision of apocalypse: the Mad Max movies. (The first of the series, in 1979, was filmed near Melbourne.)

It’s also worth noting that no one in Shute’s world takes any steps to prolong life. The government is not hollowing out mountains to keep the human germ line going until the atmosphere clears. Families are not digging fallout shelters in the back yard. These last few representatives of Homo sapiens may indulge in a variety of follies, but hope isn’t one of them.


Why am I writing about this sad book just now? Well, obviously, it’s inauguration day. Which feels more like termination day.

The threat of nuclear disaster has continued to shadow us through all the years since Shute wrote his novel. The danger of a planet-scouring war seemed particularly urgent when I was 13 and reading On the Beach for the first time. I stood up in front of my eighth-grade English class to give an oral report on the book. My performance was not interrupted by a duck-and-cover drill, but it could have been.

Now we have handed control of 4,700 nuclear warheads to a petulant brat, and the danger seems greater than ever.

Revisiting that sense of menace is why I picked up the book, but it’s not what has made the strongest impression on me this second time around. I am both drawn to and appalled by the stoic acceptance of Shute’s fictional Melbournites. Given their circumstances, their reaction is not inappropriate. The worst has already happened, there’s nothing they can do to change it, they may as well make the best of it. In the face of certain extinction, what can you do but shrug your shoulders? Maybe the best way of muddling through is just to plant some daffodils.

Given the current mood of the nation and the world, I suddenly find it easier to understand Dwight’s behavior. The urge to pretend is powerful. I too want to believe that life can go on as normal, that I can continue to enjoy the private pleasures of family and friends, that I can retreat to a cozy office or library and lose myself in the world of ideas, in the “less fretful cosmos” of mathematics and science, or art and literature for that matter.

But we are not yet huddled on the beach, the last of the doomed. It’s late, but not yet too late. This is not the moment for resignation and acquiescence. Tomorrow we march!

Posted in off-topic | 2 Comments

Joint Mathematics Morsels

Atlanta. I’m at the Joint Mathematics Meetings, the annual smorgasbord where I never have time to fully digest my helping of algebraic geometry before I move on to a desert of cohomology. Here are a few easily swallowed morsels.


Carey’s Equality. Has everyone but me known all about this for ages and ages?

In a stationary population—where births equal deaths—the number of individuals who have lived a years is the same as the number who still have a years left to live. Here’s a more precise statement from James W. Vaupel of the Max-Planck-Institute for Demographic Research:

If an individual is chosen at random from a stationary population with a positive force of mortality at all ages, then the probability the individual is one who has lived a years equals the probability the individual is one who has that number of years left to live. For example, it is as likely the individual is age 80 as it is the individual has 80 years to live—not 80 years of remaining life expectancy but a remaining lifetime of precisely 80 years.

Carey's Equality for 80 years

Is this fact obvious, a trivial consequence of symmetry? Or is it deep and mysterious? Apparently it was not clearly recognized until about 10 years ago, by James R. Carey, a biological demographer at UC Davis and UC Berkeley who was studying the age structure of fruitfly populations. The equality was proved in 2009 by Vaupel. A more general statement of the theorem and a more mathematically oriented proof were published in 2014 by Carey and Arni S. R. Srinivasa Rao of Augusta University.

I learned all this from a wide-ranging talk by Rao: “From Fibonacci to Alfred Lotka and beyond: Modeling the dynamics of population and age-structures.”


Go with the Green. Every weekday you walk from your home at the corner of 1st Avenue and 1st Street to your office at 9th Avenue and 9th Street. Since your city is laid out with a perfectly rectilinear grid, you have to go eight blocks east and eight blocks north. Assuming you never waste steps by turning south or west, or by straying outside the bounding rectangle, how many routes can you choose from?

8 block street grid with one possible sw to ne path

It would be quite a chore to count the paths one by one, but combinatorics comes to the rescue. The answer is \(\binom{16}{8}\), the number of ways of choosing eight items (such as eastbound or northbound blocks) from a set with 16 members:

\[\binom{16}{8} = \frac{16!}{(8!)(8!)} = 12{,}870{.}\]

You could walk to work for 50 years without ever taking the same route twice. Which of those 12,870 paths is the shortest? That’s the beauty of the Manhattan metric: They all are. Every such path is exactly 16 blocks long.

But just because the routes are equally long doesn’t mean they are equally fast. Suppose there’s a traffic light at every intersection. Depending on the state of the signal, you can proceed either north or east without interruption, but you’ll have to wait for the light to change if you want to cross the other way. A sensible strategy, it seems, is always to go with the green if you can. Following this rule, you will never have to wait for a light unless you are on the north or the east boundary edge of the square.

The street grid with traffic lights came up in a talk by Ivan Corwin of Columbia University, titled “A Drunk Walk in a Drunk World.” The more conventional term for this subject is “random walks in random environments.” In an ordinary random walk (with a nonrandom environment), the walker chooses a direction at each step according to a fixed probability distribution—the same at all sites and at all times. With a random environment, the probabilities vary both with position and with time. In a brief aside, Corwin offered the street grid with traffic lights as an example of a random environment. If the lights are uncorrelated on the time scale of a pedestrian’s progress through the grid, the favored direction at any intersection is an independent random variable. Then the following question arises: If the walker always takes the green-light direction when that’s possible, which paths are the most heavily traveled?

Corwin’s answer is that the walker will likely follow a stairstep path, never venturing very far from the diagonal drawn between home and office. Thus even though the distance metric says all routes are equal, the walker winds up approximating the Euclidean shortest path.

Corwin gave no proof of his assertion, although he did show the result of a computer simulation. After ruminating on the problem for a while, I think I understand what’s going on. One way of thinking about it is to break the 16-block walk into two eight-block segments, then consider the single vertex that the two segments have in common. Suppose the common point is the central intersection at 5th Avenue and 5th Street. There are 70 ways of getting from home to this point, and for each of those paths there are another 70 ways to continuing on to the office. Thus 4,900 paths pass through the center of the grid. In contrast, only one path goes through the corner of 9th Avenue and 1st Street. The same kind of analysis can be applied recursively to show that the initial eight-block segment of the walk is more likely to pass through 3rd Avenue and 3rd Street than through 5th Avenue and 1st Street.

Another way to look at it is that it’s all about the binomial theorem and Pascal’s triangle. The binomial coefficient \(\binom{n}{m}\) is largest when \(m = n/2\), making the “middle-way” paths the likeliest.

This argument says that always going with the green will give you the fastest route across town (at least in terms of expectation value), and the route you follow is likely to lie near the diagonal. What the argument doesn’t say is that deliberately biasing your choices so that you stay near the diagonal will get you to work sooner; that’s clearly not true.

When I mentioned Corwin’s example to my friends Dan Silver and Susan Williams, Susan immediately pointed out that the model fails to capture some important features of walking in an urban environment. Streets have two sides, and generally two sidewalks. To get from the southwest corner of an intersection to the northeast corner, you need two green lights. I’m not sure whether the conclusions hold up when these complications are taken into account.

I should add that solving this citified problem was not the main point of Corwin’s talk. Instead, he was addressing the problem of a bartender who wants to build a tavern in rough and ever-changing terrain near the rim of the Grand Canyon. The bartender needs to know how close he can come to the edge without endangering inebriated customers who might wander over the cliff.


TASEP. I’m a sucker for simple models of complex behavior. This week I learned of a new one—new to me, anyway. Jinho Baik of the University of Michigan talked about TASEP, a “totally asymmetric simple exclusion process” (admittedly not the most vividly descriptive name). Here’s what little I understand of the model so far.

The setting is a one-dimensional lattice, which could be either an infinite line or a closed loop of finite size. Some lattice sites are vacant and some are occupied by a particle. (No site can ever host multiple particles.) At random intervals—random with an exponential distribution—a particle “wakes up” and tries to move one space to the right (on a line) or one space clockwise (on a loop). The move succeeds if the adjacent site is vacant; otherwise the particle goes back to sleep until the next time the exponential alarm clock rings. Given some initial distribution of the particles, how does that distribution evolve over time.

Baik TASEP

When I see a model like this one, my impulse is to write some code and see what it looks like in action. I haven’t yet done that, but this is my current understanding of what I should expect to see. If you start with the smoothest possible particle distribution (alternating occupied and vacant sites), the particles will tend to clump together. If you start with a maximally clumpy state (one area solidly filled, another empty), the particles will tend to spread out. Baik and his colleagues seek a more precise description of how the density fluctuations evolve over time. And they have found one! Unfortunately, I’m not yet prepared to explain it, even in my hand-waviest way. The best I can do is refer you to the most recent paper by Baik and Zhipeng Liu.


Debunking Guy. If you ever have an opportunity to hear Doron Zeilberger speak, don’t pass it up. At this meeting he gave a spirited and inspiring defense of experimental mathematics, under the title “Debunking Richard Guy’s Law of Small Numbers.” Sitting in the front row was 100-year-old Richard Guy. Neither one of them was in any way daunted by this confrontation. In any case, Doron’s talk was more homage than attack. Later, I had a chance to ask Guy what he thought of it. “His heart is in the right place,” he said.

Guy’s Strong Law of Small Numbers says:

There aren’t enough small numbers to meet the many demands made of them.

As a consequence, if you discover that \(f(n)\) yields the same value as \(g(n)\) for several small values of \(n\), it’s not always safe to assume that \(f(n) = g(n)\) for all \(n\). Euler discovered a cautionary example that’s now well known: The equation \(n^2 + n + 41\) evaluates to a prime for all \(n\) from \(-40\) to \(+39\), but not outside that range.

Zeilberger doesn’t deny the risk of mistaking such accidents for mathematical truths. As a matter of fact, he discusses some of the most dramatic examples: the Pisot numbers, some of which produce coincidences that persist for thousands of terms, and yet ultimately break down. But such pathologies are not a sign that “empirical” mathematics is useless, he says; rather, they suggest the need to refine our proof techniques to distinguish true identities from false coincidences. In the case of the Pisot numbers, he offers just such a mechanism.

A paper by Zeilberger, Neil J. A. Sloane, and Shalosh B. Ekhad (Zeilberger’s computer/collaborator) outlines the main ideas of the JMM talk, though sadly it cannot capture the theatrics.


Soundararajan on Tao on Erdős. Take a sequence of +1s and –1s, and add them up. Can you design the sequence so that the absolute value of the sum is never greater than 1? That’s easy: Just write down the alternating sequence, +1, –1, +1, –1, +1, –1, . . . . But what if, after you’ve selected your sequence, an adversary applies a rule that selects some subset of the entries. Can you still count on keeping the absolute value of the sum below a specified bound? This is a version of the Erdős discrepancy problem, which Paul Erdős first formulated in the 1930s.

The question was finally given a definitive answer in 2015 by Terry Tao of UCLA. In the “Current Events” session of the JMM, Kannan Soundararajan of Stanford gave a lucid account of thre proof. You can read it for yourself, along with three other Current Events talks, by downloading the Bulletin.


Proust’s Powdered-Wig Party. Finally, a personal note. In the closing pages of Marcel Proust’s immense novel A la Recherche du Temps Perdu, the narrator attends a party where he runs into many old friends from Parisian high and not-so-high society. He is annoyed that no one told him the party was a costume ball: All of the guests are wearing white powdered wigs, as if they were gathering at the court of Louis XIV. Then the narrator catches sight of himself in a mirror and realizes that he too is coiffed in white.

At these annual math gatherings I run into people I have known for 30 years or more. For some time I’ve been aware that the members of this cohort, including me, are no longer in the first blush of youth. This year, however, the powdered wigs have seemed particularly conspicuous. Everyone I talk to, it seems, is planning for imminent retirement.

But of course this geriatric impression owes more to selection effects than to the aging of the mathematical population overall. Indeed, the corridors here are full of youngsters attending their first or third or fifth JMM. Which brings us back to Carey’s Equality. If we can safely assume that the population of meeting attendees is stationary, then the proportion of people who have been coming to these affairs for 30 years should be equal to the proportion who will attend 30 more meetings.

Posted in mathematics | 1 Comment

Truth, Trump, and Trisectors

What is truth? said jesting Pilate, and did not stay for an answer.

Lately there’s been a lot of news about fake news (some of it, for all I know, fake). Critics are urging Facebook, Google, and Twitter to filter out the fraudulent nonsense. This seems like a fine idea, but it presupposes that the employees—or algorithms—doing the filtering can reliably distinguish fact from fiction. Even if they can tell the difference, can we count on the companies to stand up to the prevaricators? Sure, Facebook can block traffic from a clickbait website run by a teenager in Macedonia. But what if the lies were to come from an account registered in the .gov domain?

When misinformation is stamped with the imprimatur of the president or other high government officials, there’s not much hope of shutting it down at the source or breaking the chain of transmission. This problem was not created by the new communication technologies of the internet age, and it is not unique to the incoming Trump administration. I have probably been lied to by every president who has served during my lifetime, and I could name seven of those presidents whose fibs are well documented. But Trump is different. He is not a devious liar, careful not to be caught in a contradiction. He is simply indifferent to truth. When challenged to support a dubious claim, he shrugs or changes the subject. The question of veracity seems not to interest him. And his election suggests that some part of the voting public feels the same way.

What to do? The only practical remedy I can suggest is to work diligently to uncover the truth, to publish it widely, and to help the public reach sound judgments about what to believe. All three of these tasks are difficult, but the last one, in my view, is the real stumper. As the signal-to-noise ratio in public discourse dives toward zero, we would all do well to sharpen our powers of discrimination. But I worry most about that subpopulation for whom strict factual accuracy is not the primary criterion when they choose stories to pass on to their friends and to embrace as the basis of important decisions. I don’t know how to change this, but I feel it’s important to try.

I’d like to begin with a more personal and less political anecdote. Some years ago, when the internet was young, a friend began sending me emails with subject lines like “Save 7 y.o. Jessica Mydek from cancer” or “Fw: Fw: FW: Fw: Bill No. 602P 5-cent tax on every email.” I would reply with a link to the debunking report at Snopes. My friend would thank me and sheepishly apologize, then the next month she would forward a message warning me not to blink my high beams if I saw a car with the headlights off—I’d be attacked by gang members conducting a rite of initiation. Email exchanges like these continued for a year or so, then they tapered off. Had my friend developed a measure of skepticism? Yes, but not in the way I had hoped. She had become skeptical of snopes.com. After all, it’s a website with a funny name, run by smug, self-appointed know-it-alls who make fun of gullible people. Why should she trust them?

Instead of an ad hoc watchdog like Snopes, maybe we should have an official arbiter of factuality, a certified and sanctified public agency. Call it the Ministry of Truth. And let’s give it enforcement powers: No social network or news outlet is allowed to publish anything unless the ministry attests to its accuracy.

Okay, that’s not such a hot idea after all.

In any case, no amount of scrupulous fact-checking would have cured my friend’s addiction to hoax email. There was something in those messages she wanted to believe. Even if 7 y.o. Jessica Mydek doesn’t exist, a world where chain letters can cure cancer is more appealing and empowering than the Snopesian world of grim facts, where you can only watch helplessly while a child dies. When you see a car driving without headlights, it’s more exciting to imagine a murderer at the wheel than a forgetful old fool. I’m sure my friend had her own doubts about some of these breathless pleas and warnings, but she was willing to overlook dodgy evidence or flawed logic for the sake of a good story.

As far as I know, my friend’s lax attitude toward factuality never caused grievous harm to herself or anyone else. But sometimes credulity can be disastrous.


Those who can make you believe absurdities can make you commit atrocities.

—Voltaire (paraphrase)

Let’s talk about Edgar Maddison Welch, the young man who showed up at the Comet Ping Pong pizzeria with a rifle and a handgun. By his own account, he sincerely believed he was going to rescue children being held captive in a basement room and subjected to unspeakable acts by Hillary Clinton and her associates. Where did that idea come from? Apparently it began with leaked emails from the hacked account of John Podesta, Clinton’s campaign chairman. According to an outline in the New York Times, eager sleuths on Reddit and 4chan discovered the phrase “cheese pizza” in the email texts, and recognized it as a code word for “child pornography.” Connecting the rest of the dots was easy and obvious: Podesta had corresponded with the owner of Comet Ping Pong, and Barack Obama had been photographed playing ping pong with a small boy, and so the basement of the restaurant must be where the Democrats slaughter their child sex slaves. However, the would-be rescuer with the AR-15 found no basement kill room—in fact, no basement at all. “The intel on this wasn’t 100 percent,” he told a Times reporter.

In case there’s even the slightest doubt, let me say plainly that I don’t believe a word of that grotesque tale about child abuse in the pizza parlor. Indeed, I can make sense of it only as a stupid joke, a parody, a deliberately preposterous confection. If I were fabricating such a malicious fiction, and if I wanted people to believe it, I would come up with something that’s not such a total affront to plausibility. Yet at least one reader of these fantasies took them in deadly earnest. We’ll never know how many more believe there might be a “grain of truth” in the story, even if specific details are wrong. And the purveyors of the myth are not backing down. In an AP story that ran in the Times on December 9 they propose that the Comet Ping Ping event was a “false flag,” yet another twist in the larger plot:

James Fetzer, a longtime conspiracy theorist who also believes the Sandy Hook school shooting was a hoax, told The Associated Press that Welch’s visit to the pizzeria was staged to distract the public from the truth of the “pizzagate” allegations. . . .

Fetzer and other conspiracy theorists seized on the fact that Welch had dabbled in movie acting as a giveaway that his visit to the restaurant was staged. . . . Blogger Joachim Hagopian, a false-flag proponent, told the AP that conspirators look for “a patsy or stooge” to pose as a lone gunman with an assault rifle. Welch, he said, “fits the pattern” with his acting background.

“He’s got an IMDB (Internet Movie Database) profile,” Hagopian said.

It’s easy to heap ridicule on these ideas. Indeed, by quoting them at length that’s exactly what I’m doing. How could anyone possibly believe in such contrived and convoluted schemes, such teetering towers of improbabilities? But it’s useful to keep in mind that the incredulity goes both ways. The conspiracy theorists would snigger at my naiveté for believing what I read in New York Times. Anyone who’s paying attention knows that all the big papers and TV networks are parties to the conspiracy. (Snopes is surely in on it too.)


Mathematics alone proves, and its proofs are held to be of universal and absolute validity, independent of position, temperature or pressure. You may be a Communist or a Whig or a lapsed Muggletonian, but if you are also a mathematician, you will recognize a correct proof when you see one.

—Philip J. Davis, American Mathematical Monthly, 79(3):254 (March 1972)

A high-stakes presidential election and accusations of child rape and murder certainly add force and immediacy to a discourse on the nature of truth, but they also distract. I would like to retreat from these incendiary themes, at least for a few paragraphs, and look at the calmer universe of mathematics, where we have well-developed mechanisms for distinguishing between truth and falsehood.

Take the case of angle trisectors—people who claim they can divide an arbitrary angle into equal thirds with the standard Euclidean toolkit of straightedge and compass. In some respects, trisectors are like peddlers of pizza parlor pedophilia, but when a trisector comes before you, you can give a stronger response than: “What you claim is contrary to common sense.” You can offer an absolute refutation: “What you claim is impossible. Pierre Laurent Wantzel proved it 180 years ago.” But I wouldn’t count on the trisector meekly accepting this answer and going away.

A few years ago, writing in American Scientist, I made an earnest effort to explain the Wantzel proof in some detail and in plain words, and I provided an English translation of Wantzel’s own paper from 1837. Soon after the article appeared, I began receiving letters festooned with elaborate geometric diagrams, some of them quite pretty, which the authors presented as proper straightedge-and-compass trisections. I wasn’t surprised at this development, but I was at a loss for how to respond. If a mathematical proof fails to persuade the reader of the truth of a mathematical proposition, what other kind of argument could possibly be more effective?

In the past few weeks I’ve given this incident further thought, and I’ve come to see it in a different light. The task of “persuading the reader,” even in mathematics, is not just about truth; it’s also about trust, or rapport, or social solidarity. The quip by Philip Davis that I reproduce above has long been a favorite of mine, but at this point I am tempted to turn it inside out. What I would say is not “If you’re a mathematician, you’ll recognize a proof” but “If you recognize a proof, you’re a mathematician.” The ability and willingness to engage in a certain style of reasoning, and to accept the consequences of that mental process no matter what the outcome, marks you as a member of the mathematical tribe. And, conversely, if you respond to a proof by saying “It may be impossible but I can do it anyway,” then you are not a member of this particular affinity group.

I am not arguing here that mathematical truth is some kind of socially determined quantity, and no more fundamental than religious or political doctrines. Quite the contrary, I am one of those stubborn prepostmodernists who believes in a reality that’s not just my private daydream. I’m convinced we all share one universe, where certain things are true and others aren’t, where certain events happened and others didn’t. The interior angles of a plane triangle will always sum to 180 degrees no matter what I say. Nevertheless, the process by which we recognize such truths and reach consensus about them is a social one, and it’s not infallible.

The same essay in which I discussed Wantzel’s proof also mentioned the infamous Monty Hall problem.

In 1990 Marilyn vos Savant, a columnist for Parade magazine, discussed a hypothetical situation on the television game show “Let’s Make a Deal,” hosted by Monty Hall. A prize is hidden behind one of three doors. When a contestant chooses door 1, Hall opens door 3, showing that the prize is not there, and offers the player the option of switching to door 2. Vos Savant argued . . . that switching improves the odds from 1/3 to 2/3.

In the following weeks thousands of letter writers berated vos Savant for her blatant error, insisting that the two remaining closed doors were equally likely to conceal the prize. Quite a few of those critics identified themselves as mathematicians or mathematics teachers. Even Paul Erdős took this side of the controversy (although he didn’t write a letter to Parade). But of course vos Savant was right all along.

This story was already well known when I told it in my American Scientist essay, but I have a reason for retelling it yet again now. Along with the mail from angle trisectors I also received irate messages from Monty Hall deniers, who insisted that the probabilities really are 1/2 and 1/2. But this time it wasn’t professional mathematicians who raised objections; they had long since resolved their differences and settled on the correct answer. Now it was outsiders, dissidents, who attacked what they perceived to be an ignorant, entrenched orthodoxy enforced by the professoriat. In other words, the same two factions continued to fight over the same question, but they had switched positions.

The point I’m making here is the unsurprising one that social factors influence judgment. We are all predisposed to go along with the views of those we know and trust, and we are skeptical, at least initially, of ideas that come from outsiders. We listen more attentively and sympathetically when the speaker is a trusted colleague. The scrawled manuscript from an unknown author claiming a simple proof of the Riemann hypothesis gets a cursory reading or none at all. There’s nothing wrong with making such distinctions. The alternative—equal treatment for the competent and the crackpot—would certainly not help advance the cause of truth. But it has to be acknowledged that these practices further alienate outsiders. By pushing them away and closing off the channel of communication—treating them as irredeemables and deplorables—we diminish the chance that they will ever find a path into the community.


Why do the nations so furiously rage together, and why do the people imagine a vain thing?

—Psalms, 2:1, via George Frideric Handel

Do these skirmishes over minor mathematical questions have anything to do with “Fakebook” news that might have turned the tide of a presidential election? I submit there is a connection. In both cases the nub of the problem is not discovering the truth but persuading people to recognize and own it. The mathematical examples show that even the most irrefutable kind of evidence—a deductive proof—is not always enough to win over skeptics or opponents.

Proof is said to “compel belief”: You embrace the result even against your will. Once you grant the premises, and you work through the chain of implications, accepting the validity of each step in turn, you have no choice but to accept the ultimate conclusion. Or so one might think. But this view of proof as an irresistible engine of reason underestimates the flexibility and creativity of the human mind. In fact we are all capable of believing impossible things before breakfast, and denying certainties after dinner, if we choose to. Mathematicians—members of the tribe—promise not to do so, but that pledge is not binding on anyone else.

When I look back over my various encounters with angle trisectors and other mathematical mavericks, I can’t recall a single instance where I successfuly persuaded someone to give up an erroneous belief and accept the truth. Not one soul saved. This record of failure does not give me great confidence when I think of venturing forth to combat fake political news, where we don’t even have the secret weapon of deductive proof.

I’m left with the thought that compelling people to acknowledge a truth may be the wrong approach, the wrong attitude. Voltaire was a great hero of free-thinking, but his motto “Écrassez l’infâme!” is a bit too militaristic for my taste. However you choose to translate that phrase, he meant it as a call to arms. Let us crush superstition, wipe out error and ignorance, put an end to fanaticism and irrationality. I’m for all that, but I don’t want to be bludgeoning people into accepting the truth. It doesn’t really change their minds, and at some point they bludgeon you back.

Rather than force the people to give up their false notions and vain things, I would let the truth seduce them. Let them fall in love with it. Doesn’t that sound grand? If only I had the slightest clue about how to make it happen.


I like mathematics largely because it is not human and has nothing particular to do with this planet or with the whole accidental universe—because, like Spinoza’s God, it won’t love us in return.

At this point my only consolation is a cold and severe one. Trump may be indifferent to truth, but the universe, in the long run, is utterly indifferent to him and his foibles. Our new president can declare that climate change is a hoax, and purge government agencies of all those who disagree, but those acts will not lower the concentration of carbon dioxide in the atmosphere.

Mathematical truths are even more aloof from human interference. In Orwell’s 1984 the Thought Police boast of making citizens believe that two plus two equals five. But all the sophistry of the Ministry of Truth and all the torture chambers of the Ministry of Love cannot alter the equation itself. They cannot make two and two equal five.

These are very small islands of certainty in a vast maelstrom of confusion, but they offer refuge, and maybe a place to build from.

Posted in mathematics, modern life, off-topic | 6 Comments

ABC and FLT

This weekend I’ll be attending a workshop at the University of Vermont, “Kummer Classes and Anabelian Geometry: An introduction to concepts involved in Mochizuki’s work on the ABC conjecture, intended for non-experts.” It was the last phrase in this title that gave me the courage to sign up for the meeting, but I have no illusions. There are degrees of non-expertness, and mine is really quite advanced. The best I can hope for is to go home a little less ignorant than I arrived.

I’m glad to be here all the same. The ABC conjecture is one of those beguiling artifacts in number theory that seem utterly simple one moment and utterly baffling the next. As for Shinichi Mochizuki’s 500-page treatise on the conjecture, that’s baffling from start to finish, and not just for me. Four years after the manuscript was released, it remains a proof-on-probation because even the non-non-experts have yet to fully digest it. The workshop here in Burlington is part of the community’s digestive effort. I’m grateful for an opportunity to see the process at work, and naturally I’m curious about the eventual outcome. Will we finally have a proof, or will a gap be discovered? Or, as has happened in other sad cases, will the question remain unresolved? For me the best result will be not just a proof certified by a committee of experts but a proof I can understand, at least in outline, if I make the effort—a proof I might be able to explain to a broader audience.

The drive up to Burlington gave me a few hours of solitude to think about the conjecture and how it fits in with more familiar ideas in number theory. One notable connection is summed up as “ABC implies FLT.” Proving the ABC conjecture will bring us a new proof of Fermat’s Last Theorem, independent of the celebrated Andrew Wiles–Richard Taylor proof published 20 years ago. Interesting. But how are the two problems linked? As I cruised through the chlorophyll-soaked hills of the Green Mountain state, I noodled away at this question.

[Correction: ABC actually implies only that FLT has no more than a finite number of counterexamples, and only for exponent \(n \ge 4.\)]


I have written about the ABC conjecture twice before (2007 and 2012). Here’s a third attempt to explain what it’s all about.

The basic ingredients are three distinct positive integers, \(a\), \(b\), and \(c\), that satisfy the equation \(a + b = c\). Given this statement alone, the problem is so simple it’s silly. Even I can solve that equation. Pick any \(a\) and \(b\) you please, and I’ll give you the value of \(c\).

To make the exercise worth bothering with, we need to put some constraints on the values of \(a\), \(b\), and \(c\). One such constraint is that the three integers should have no factors in common. In other words they are relatively prime, or in still other words their greatest common divisor is 1. Excluding common factors doesn’t really make it any harder to find solutions to the equation, but it eliminates redundant solutions. Suppose that \(a\), \(b\), and \(c\) are all multiples of 7; then dividing out this common factor yields a set of smaller integers that also satisfy the equation: \(a\,/\,7 + b\,/\,7 = c\,/\,7\).

The ABC conjecture imposes a further constraint, and this is where the arithmetic finally gets interesting. We are to restrict our attention to triples of distinct positive integers that pass a certain test. First we find the prime factors of \(a\), \(b\), and \(c\), and cast out any duplicates. For example, given the triple \(a = 4, b = 45, c = 49\), the prime factors are \(2, 2, 3, 3, 5, 7, 7\); eliminating duplicates leaves us with the set \(\{2, 3, 5, 7\}\). Now we multiply all the distinct primes in the set, and call the product the radical, \(R\), of \(abc\). Here’s the punchline: The solution is admissible—it is an “\(abc\)-hit”—only if \(R \lt c\). For the example of \(a = 4, b = 45, c = 49\), this condition is not met: \(2 \times 3 \times 5 \times 7 = 210\), which of course is not less than 49.

The ABC conjecture holds that \(abc\)-hits are rare, in some special sense. Hits do exist; try working out the radical of \(a = 5, b = 27, c = 32\) to see an example. In fact, there are infinitely many \(abc\)-hits, with constructive algorithms for generating endless sequences of them. Yet, it’s part of the maddening charm of modern mathematics that objects can be both infinitely abundant and vanishingly rare at the same time. The particular kind of rareness at issue here says that \(R\) can be less than \(c\), but seldom by very much. As a measure of how much, define the power \(P\) of an \(abc\)-hit as \(\log(c) \,/\, \log(R)\). Then one version of the ABC conjecture states that there are only finitely many \(abc\)-hits with \(P \gt (1 + \epsilon)\) for any \(\epsilon \gt 0\).


On first acquaintance, all this rigmarole about radicals seems arbitrary and baroque. Who came up with that, and why? The answer to the who question is Joseph Oesterlé of the University of Paris and David W. Masser of the University of Basel, in 1985. If you play around long enough with integers and their prime factors, you can find all sorts of curious relations, so what’s so special about this one? Whether or not the conjecture is true—and whether or not it’s provable—why should we care?

As I tooled along the Interstate, I tried to answer this question to my own satisfaction. I made a little progress by thinking about what kinds of numbers we might expect to produce \(abc\)-hits. Are they big or small, nearly equal or of very different magnitudes? Are they primes or composites? Do we find squares or other perfect powers among them?

Primes are always a good place to start. Can we have an \(abc\)-hit with \(a\), \(b\), and \(c\) all prime? One complication is that either \(a\) or \(b\) will have to be equal to 2, because 2 is the only even prime, and we can’t have \(a + b = c\) with all three numbers odd. But that’s all right; there are still lots of triples (and conjecturally infinitely many) of the form \(2 + b = c\) with \(b\) and \(c\) prime; they are called twin primes. However, none of them are \(abc\)-hits. It’s easy to see why: If \(a\), \(b\), and \(c\) are all prime, then their radical is simply the product \(abc\), which for numbers larger than 1 is always going to be greater than \(a + b\).

We can extend this reasoning from the primes to all squarefree numbers, that is, numbers that have no repeated prime factors. (They are called squarefree because they are not divisible by any square.) For example, \(10, 21,\) and \(31\) form a squarefree \(abc\) triple, with prime factorizations \(2 \times 5\), \(3 \times 7\), and \(31\). But they do not produce an \(abc\)-hit, because their radical \(2 \times 3 \times 5 \times 7 \times 31 = 6510\) is clearly larger than \(a + b = c = 31\). And the same argument that rules out all-prime \(abc\)-hits applies here to exclude all-squarefree hits.

These results suggest that we look in the opposite direction, at squarefull numbers—and even cubefull numbers, and so on. We want lots of repeated factors. This strategy immediately pays off in the search for \(abc\)-hits. It maximizes the sum \(a + b = c\) without overly enlarging the radical—the product of the distinct primes. The very first of all \(abc\)-hits (in any reasonable ordering) offers an example. It is \(1 + 8 = 9\), or in factored form \(1 + 2^3 = 3^2\). This is a high-power hit, with \(\log(9) \,/\, \log(6) = 1.23\). The triple with highest known power is \(2 + (3^{10} \times 109) = 23^5\), yielding \(\log(6436342) \,/\, \log(15042) = 1.63.\)

Let’s look more closely at that first \(abc\)-hit, \(1 + 8 = 9\). Note that 8 and 9 are not just squarefull numbers; they are perfect powers. This triple is the subject of another famous conjecture. Eugène Catalan asked if \(8\) and \(9\) are the only consecutive perfect powers. Preda Mihailescu answered affirmatively in 2002. Thus we know that the equation \(1 + x^m = y^n\) has only this single solution. However, if we relax the rules just a little bit, we can find solutions to \(a + x^m = y^n\) where \(a\) has some value greater than 1. For example, there’s 3 + 125 = 128 (or 3 + 5^3 = 2^7), which is another high-power \(abc\)-hit.

Suppose we tighten the rules instead of relaxing them and ask for solutions to \(x^n + y^n = z^n\), where the three members of the triple are all nth powers of integers. If we could find solutions of this equation with large values of n, they would surely be a rich ore for high-power \(abc\)-hits. But alas, that’s the equation that Fermat’s Last Theorem tells us has no solutions in integers for \(n \gt 2\). The ABC conjecture turns this implication on its head. It says (if it can be proved) that the rarity of \(abc\)-hits implies there are no solutions to the Fermat equation.


That’s as far as I was able to get while musing behind the wheel—a vague intuition about the balance between addition and multiplication, a tradeoff between increasing the sum and reducing the radical, a hint of a connection between ABC and FLT. Not much, but a better sense of why it’s worth focusing some attention on this particular relation among numbers.

Now morning breaks over Burlington. Time to go learn something from those who are less non-expert.

Posted in mathematics | 5 Comments