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This week we celebrate the life of the most published mathematician in history, Paul Erdős (AIR-dosh), who was born 100 years ago on March 26. Dr. Erdős, who has been called the world’s greatest problem poser and solver, collaborated with over 500 mathematicians before his death in 1996.

“There are still a lot of Erdős’s vibrations going around,” says Ronald L. Graham, professor of computer science and engineering at the University of California, San Diego, and longtime Erdős friend and collaborator. “His impact will be felt for a long time.”

We’ll get a sense of this impact soon — Dr. Graham has suggested a terrific bonus puzzle for the week: an unsolved problem related to the work of Dr. Erdős. But before jumping in, let’s learn a bit more about the remarkable mathematician as a person, as captured by “The Boy Who Loved Math: The Improbable Life of Paul Erdős,” a soon-to-be-released book by Deborah Heiligman and LeUyen Pham.

The Boy Who Loved Math

There once was a boy who loved math. He grew up to be one of the greatest mathematicians who ever lived. He lived in Budapest, Hungary, with his mama.

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When Paul was 10, he fell in love. He fell in love with prime numbers. Prime numbers are special. They can’t be divided evenly. A prime number can be divided only by itself and 1.

Paul loved to think about prime numbers.

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Paul soon realized he didn’t fit into the world in a regular way. So he invented his own way to live. Paul would get on an airplane with everything he owned — a few clothes and some math notebooks. “I have no home,” he declared. “The world is my home.”

He helped people with their math problems and gave them more math problems to do. Plus, he was a math matchmaker. Paul knew that mathematician + mathematician + mathematician = more and better math.

All over the world mathematicians still talk about and love Uncle Paul. They talk about their “Erdős number.” If you did math with Paul you get an Erdős number 1. If you worked with someone who worked with Paul, your Erdős number is 2. People are so proud of their Erdős numbers.

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A long time ago there was a boy who loved math. Numbers were his best friends. He grew up to be the man who loved math. Numbers and people were his best friends. Paul Erdős had no problem with that.

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Paul Erdős made significant contributions to a number of fields, including number theory and computer science. Let’s take a look at his most lasting legacy: the pioneering of the probabilistic method, which can be used to prove the existence of something without pointing to a concrete example.

Here’s a brief introduction by Joel Spencer, professor of computer science and mathematics at the Courant Institute of Mathematical Sciences at New York University. Dr. Spencer, who co-wrote — with Dr. Erdős — “Probabilistic Methods in Combinatorics,” refers to the probabilistic method as Erdős Magic.

Suppose you have 500 sets and each set has exactly 10 elements. The sets may be disjoint, they may overlap, any pattern is possible, and the total number of elements can be anything up to 5,000. Now we want to split the elements into two classes Red and Blue (that is, two-color the elements) so that none of the 500 sets has all 10 elements the same color. How do we do this regardless of the pattern of the sets? We do it by coloring the elements at random. Each set has 2 chances in 1,024 or 1 in 512 of having all of its elements the same color. There are 500 possible sets, so the chance that any of them has all of its elements the same color is at most 500/512. (I use here that the probability that one of several things happen is at most the sum of the probabilities that they happen. This is usually an overestimate because the events can overlap, but here it’s O.K. as we only use it as an upper bound.) Thus the chance that none of the 500 sets have all the same color (which is what we want) is at least 12/512. The key is that this is strictly positive. Therefore, by Erdös Magic, such a coloring exists.

Let’s give it a try. We’ll start out with two-element sets. How many can we draw and be guaranteed we’ll find a way to color each element blue or red and avoid ending up with a set that’s all the same color?

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Erdős Magic would predict one: each set has a two in four, or 1/2, chance of being monochromatic, so we’re guaranteed that a single set of two elements (dots, in this case) will be two-colored. Erdős Magic won’t predict when we’ll actually reach the point where we can no longer avoid a monochromatic set, but we can play around a bit and find this happens when we reach three sets if they’re drawn in a triangle.

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Moving up to three-element sets — Erdős Magic says we’ll always find a way to two-color up to three sets of three dots: each set has a two out of eight, or 1/4, chance of being monochromatic; 3×1/4, or 3/4, is still less than one, so we’re guaranteed of finding at least one red and one blue dot in each set no matter how they’re drawn.

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Erdős Magic doesn’t tell us how many sets of three dots we’ll need (and in what configuration) before we’re no longer able to avoid creating a monochromatic set. We happen to find ourselves in this situation with seven sets, if they’re configured in a particular way.

Which brings us to our puzzle for the week. What would these seven sets look like? That is, how can you take a collection of dots and group them into seven sets with three dots in each set (dots can be in more than one set) in a way that makes it impossible, once you color each of the dots either red or blue, to avoid a monochromatic set?

We also have a bonus puzzle this week — an arithmetic progression challenge suggested by Dr. Graham, which he phrases as follows:

Call a set of 3 integers a < b < c “evenly spaced” if b-a = c-b. In other words, they form a three-term arithmetic progression. It is known that no matter how you two-color the first nine numbers 1,2, … ,9, you will always have a monochromatic set of 3 evenly spaced integers. It is believed that for any set of eight integers, it is always possible to two-color them so that no monochromatic evenly spaced set of 3 integers is formed but this has never been proved. You have some smart readers, and some readers who are good with computers. Maybe they can prove this, or at least create a convincing argument that it is true.

Solutions

Here’s Mike with the solution to the main problem:

I think this works for the main problem. Seven elements numbered 1-7, each element is part of three sets and each set shares at most one element with any other set. This also means that any two elements you choose will be part of exactly one of the seven sets:

1, 2, 3

1, 4, 5

1, 6, 7

2, 4, 6

2, 5, 7

3, 4, 7

3, 5, 6

In the words of Hans, “Beautiful!” Patrick C followed by noting the arrangement was something called a finite projective plane of order 2.

I asked Joel Spencer by e-mail to say a bit about this seven-set configuration, which clearly could not be 2-colored without creating a monochromatic set, and for a demonstration that six sets (one fewer) of three points could, in fact, always be 2-colored. Here’s his response:

This particular configuration is known as both the Fano plane and a S(2,3,7) Steiner triple system. The Fano plane is the nicest way to draw it. Take the seven sets in the Fano Plane. (Note that the six lines AND the circle are the seven sets.) Now suppose you have the same configuration but you don’t count the circle. Then color the three vertices of the triangle (very top, way left, way right) AND the center point red and all three other points blue. There is no monochromatic line. Similarly, suppose you didn’t count the bottom line but did count the circle. Then color the three points on the bottom red and the center point red and all the other points blue. Here is ONE of the cases for coloring six sets. SUPPOSE that (renumbering) there is one set with 12; another with 34; and another with 56. Now color 1,3,5 randomly but color 2,4,6 the opposite color of 1,3,5 respectively. Color all other points (if there are any) randomly. Now the first three sets are automatically OK. Indeed if there are any other sets that have 12 or 34 or 56 they are OK. All other sets are monochromatic 1/4 of the time. But there are only 3 of them so they are monochromatic at most 3/4 of the time. By Erdős Magic sometimes none of them are monochromatic and that is the coloring you want. (This is a nice variant of Erdős Magic in that the way you color, though it uses randomness, is not totally random.)

Let’s now turn to the bonus challenge proposed by Ronald Graham:

It is believed that for any set of eight integers, it is always possible to two-color them so that no monochromatic evenly spaced set of 3 integers is formed but this has never been proved.

Alex L., another Alex (from Budapest, Hungary) and David desJardins were among those who dug into this conjecture, which was ultimately proved by Noam D. Elkies:

OK, the 8-number conjecture is true. Of the 2²¹ = 2,097,152 combinatorially possible configurations of three-term arithmetic progressions (AP3’s), only 624 cannot be colored to avoid an AP3. But algebraically all but 8 of these 624 are impossible unless all 8 numbers are equal, and for the remaining 8 the only algebraic solution is to have all but one of the 8 numbers equal. (This is basically what David desJardins suggested; there’s no need for linear programming to exclude algebraic solutions that are not increasing.) Therefore any 8 *distinct* numbers can be 2-colored without producing a monochromatic AP3.

Was this truly the solution? I asked Ronald Graham what he thought of the proof, and if he happened to know Dr. Elkies.

I know Noam quite well. Actually, I introduced him on the stage of the National Academy of Sciences when he won a mathematics prize some years ago (and I used a juggling ball to do it, since his work involved sphere packing). If Noam said he did it, then I believe it!

The 44-year-old Dr. Elkies is truly brilliant: a music, chess, and math prodigy who, at age 26, became Harvard’s youngest tenured professor ever. Why did it take Dr. Elkies to solve this problem? Was the problem that difficult? I sent an e-mail to David desJardins and received this reply:

It’s not the hardest problem ever (when I saw Ron Graham say this was an unsolved problem, I immediately thought, “OK, I’m sure I can solve that; it’s only unsolved because no one has really tried yet). But it’s trickier than you suggest. The hardest part is getting “all possibilities” to be a small enough number that you can write a program to check them all. If you take the point of view that each triple (i,j,k) might be either an arithmetic sequence or not, then there are 56 = B(8,3) such triples, and therefore 256 possible configurations (each triple is either a sequence or it isn’t). 256 is 72 million billion possibilities, it’s too many to exhaust over. The main insight here is that some of these configurations are inconsistent, e.g., if (x1,x2,x3) is an arithmetic sequence then it’s not possible for (x1,x2,x4) to also be an arithmetic sequence. Or, if (x1,x2,x5) is an arithmetic sequence then it’s not possible for (x1,x3,x4) to also be an arithmetic sequence. If you write down the rule for what makes two triples inconsistent with one another, and then you write a program to generate all configurations that have no inconsistent pair, then you get only 2²¹ configurations that have no inconsistencies. This is a small enough number to exhaust over, so the rest is pretty easy. (Noam acknowledged me only because I posted the method first and he didn’t want to take credit. But what he did is pretty straightforward and obvious and I’m sure he would have done exactly the same thing if I hadn’t mentioned it.) The *most* interesting thing here is to figure out *why* there are *exactly* 221 configurations. Neither Noam or I can figure it out, and he’s one of the world’s top mathematicians. That is a really interesting and (apparently) hard problem, much more interesting and much harder than Ron’s original problem. If we haven’t figured this out by Monday, I’m going to set a bunch of other people to work on it.

I followed up with Dr. Elkies as well: what did he think when he saw the problem? And what were the keys to solving it? Here’s Dr. Elkies’ e-mail response.

When I saw the problem I was intrigued because it was a new (to me) and natural variation of a bit of mathematics that I’d heard of before but hadn’t thought that much about: one can bicolor {1,2,3,…,8} to avoid any monochromatic AP3 (arithmetic progression of length 3), but not {1,2,3,…,9}. I was surprised that such a question could still be open. Since Ron Graham asked it, I figured that the usual analytical techniques had been tried, but my first thought was that 8 is small enough that modern technology should bring this easily within reach — computers these days can routinely do several billion (109) operations per second, and if the problem can be parallelized on several CPU’s then one can get at least an order of magnitude more. My second thought was that 8 is small but the number 56 of triples is still too large: as David desJ. already explained, 256 is much too big a number for a mindless exhaustive computer search. The third thought was that most of those 256 possibilities can never arise due to the consistency condition: if a,b,c and A,B,C are different triples that both index an AP3 then we cannot have a≤A, b≥B, and c≤C. But it was not obvious whether this would suffice to bring the computation down from the extravagant 256 to something feasible. Here I took a bit of a detour: rather than try to impose this condition immediately on all triples, I tried to get a sense of how stringent the condition is by imposing it one number at a time: assuming I’ve already done all the triples not involving the largest of n numbers, how many possibilities are there for the set of triples involving the n-th? Using the consistency condition just on the triples of the form (i,j,n) for n=3, 4, 5, 6 yields the counts 2, 5, 14, 42. This is the start of the famous sequence of Catalan numbers, and it turns out to be a nice exercise to prove that this pattern persists for all n. So, just the consistency of all triples with the same third entry already brings the count from 256 down by more than 8 orders of magnitude to 2 * 5 * 14 * 42 * 132 * 429 = 332972640. This already makes the problem feasible: remember a computer can do 332972640 elementary operations in a fraction of a second, and even though testing a given collection of triples is not elementary I figured it would be straightforward enough to check them all within an hour, and then do linear algebra to screen the others and either prove or disprove the conjecture. First I figured I might as well impose the full consistency condition (and now we re-join the path that David DesJ. outlined). Again trying small n, I found the counts 2, 8, 64, 1024 for n=3,4,5,6; the appearance of 64=26 already aroused some attention, and 1024=210 could not be ignored. The next count of 32768 = 215 for n=7 confirmed the pattern “beyond reasonable doubt”, though I still didn’t have a proof. [BTW I wouldn’t read that much into the fact that I personally can’t prove it — David desJardins is too kind to call me “one of the world’s top mathematicians”, and even if I were, it wouldn’t be for my work in this kind of enumerative combinatorics.] The next power of 2 in this pattern is 221 or about 2.1 million, which was small enough that I felt sure that a simpleminded exhaustive calculation was all that is necessary. (There were further refinements available, such as exploiting the symmetry between (say) 0,1,2,4 and -4,-2,-1,0, which are clearly equivalent but have different AP3 patterns; this might roughly halve the computing time, but the extra effort to code this correctly would not be worth the time saving.) At this point I went back to your blog and saw that all the ingredients, as well as some others that I didn’t expect to be necessary (like linear programming), had already been posted — which as David notes wasn’t that surprising. But by now I already had all the pieces to finish the computation. I rigged up some C code to find all the consistent configurations by depth-first search, and try all 27=128 colorings for each one (not 28, because switching red for blue yields an equivalent coloring, so I can assume that the largest number is blue — *this* twofold symmetry was easy), and then outputting those that had no solution. Testing this first for n=7 found a unique consistent configuration of triples that could not be 2-colored, namely 125, 136, 147, 234, 246, 257, 345, 367, 456 (note that this is symmetric, as it must be given that it’s unique: pairs 125/367, 136/257, 234/456, and the symmetrical 147,246,345). But the only solutions to the linear equations b-a=e-b, c-a=f-c, etc. are those for which a,b,c,d,e,f,g are all the same, so this confirms that any set of 7 numbers can be 2-colored without producing a monochromatic AP3. So then I ran the n=8 search and get a list of 624 configurations to try. (This took less than 20 seconds on my laptop.) Fortunately modern computers can also do 624 linear algebra problems of this size faster than we can set up the computation. As I reported on your blog, it turned out that only 8 of those 624 are possible without all 8 numbers being the same — and in those 8 all but one of the 8 is the same; indeed they’re just the unique n=7 configuration with an idle 8th number in any of the 8 possible positions! So this completed the proof — and (as I’d hoped) with no need for linear programming. Having gone this far I thought it might be interesting to check for configurations that can be colored in only one way, and found the two that I posted on your blog (plus their mirror images). If there’s a computer-free solution of Ron Graham’s question, it must somehow find these unique solutions; this might be one of those theorems that are true but have no conceptually simple proof. This still leaves the power-of-2 mystery. By now I’ve checked it also for n=9 and n=10, getting the expected counts of 228 and 236 configurations respectively. The n=10 configuration finally approached the limits of this technique, taking just under 3 hours on my laptop, so I’m not going to try n=11. For n=3 through n=10 I also recorded in each case the number of configurations of each size (there’s 1 empty configuration, n(n-1)(n-2)/6 consisting of a single triple, etc.), and found a few more patterns in the size and number of the largest configurations: n 3 4 5 6 7 8 9 10 size 1 2 4 6 9 12 16 20 # 1 3 4 28 64 960 4096 126976 the maximal size is easy to account for by counting the number of times each number arises as the middle of an AP3: the smallest can’t occur at all, the second-smallest at most once, the third at most twice, etc., and likewise for the largest, second-largest, etc.; so for instance the counts for 6 and 7 are at most 0+1+2+2+1+0 = 6 and 0+1+2+3+2+1+0 = 9 — and these are actually attained by the AP3’s in {1,2,3,…,n}. But once n>3 there are other configurations that attain this too. Up to n=10, every other count 1, 4, 64, 4096 is a power of 2, with exponents 0, 2, 6, 12 or 0*1, 1*2, 2*3, 3*4; and each of the remaining counts 3, 28, 960, 126976 is obtained by multiplying the previous one by 3, 7, 15, 31, these being one less than 4, 8, 16, 32 (powers of two). That could be much easier to prove than the power-of-two formulas for the numeration of all consistent configurations. [NB most of these maximal combinatorial configurations don’t arise arithmetically.] A final curious observation is that the consistent configurations always split evenly between even and odd size; e.g. for n=7 the counts are: size 0 1 2 3 4 5 6 7 8 9 # 1 35 420 2310 6538 10076 8593 3899 832 64 so the even configurations total 1 + 420 + 6538 + 8593 + 832 = 16384, same as the count of odd configurations 35 + 2310 + 10076 + 3899 + 64.

Thank you, Dr. Elkies, and for everyone who participated this week. In alphabetical order: Alex L, Alex, David desJardins, Gary, Gary Antonick, Giovanni Ciriani, Golden Dragon, Hans, Maya, Mike, Mr Bill, NDE (Noam D. Elkies), Patrick C, Ravi, Ricardo Ech and WC.

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I’d like to thank Ronald Graham for suggesting the arithmetic puzzle and for other contributions to this post. Dr. Graham is chief scientist at the California Institute for Telecommunications and Information Technology and a professor in computer science and engineering at the University of California, San Diego. He is credited by the American Mathematical Society as being “one of the principal architects of the rapid development worldwide of discrete mathematics in recent years,” and has published about 320 papers, including nearly 30 with Paul Erdős, and five books, including “The Mathematics of Paul Erdős Volumes I & II.” I’d also like to thank Joel Spencer for the explanation of Erdős Magic. Dr. Spencer is a professor at the Courant Institute of Mathematical Sciences of New York University. He has worked on probabilistic methods in combinatorics and as well as on Ramsey theory, and is the author of a number of books including “The Probabilistic Method,” with Noga Alon, and “The Art of Counting” with Paul Erdős and others. Quoted text and images from the delightful “The Boy Who Loved Math: The Improbable Life of Paul Erdős” by Deborah Heiligman and illustrated by LeUyen Pham provided courtesy of Roaring Brook Press, a division of the Macmillan Children’s Publishing Group. What’s your Erdős number? Find out by visiting Jerry Grossman’s Erdős Number Project, where you can learn more about research that qualifies for an Erdős number and related concepts like the Kevin Bacon game. To number comments, render TeX and display comment images, try Gary Hewitt’s Numberplay Comment Enhancer. You may also use the Enhancer to test your TeX before posting. Do you have a favorite puzzle? Send it to Numberplay@NYTimes.com. You can also reach me directly at gary.antonick@NYTimes.com.

LeUyen Pham was gracious enough to share with me her preliminary sketches for “The Boy Who Loved Math.” Nearly all of Ms. Pham’s illustrations in the book contain mathematical concepts important to Paul Erdős; the image below shows the types of primes assigned to several buildings in the city of Budapest, where Paul Erdős grew up: Mersenne primes, prime triplets, unique primes and palindromic primes. (This level of detail can be found throughout the book, creating an enduring adventure for kids of any age: from 3 to 34 and beyond.)