Two Mathematicians Have Solved the First Part of the Weirdest Open Problem Ever
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A new approach has chipped away at a famously unsolved math problem.
The Erdos-Turan conjecture in additive combinatorics is one of the longest lasting unsolved problems.
The two mathematicians used density to prove that a certain set must behave a certain way.
Two mathematicians say they’ve untangled the first part of Paul Erdos’s famously thorny and unproven conjecture. In a new paper they’ve uploaded to arXiv and submitted to journals, mathematicians Thomas Bloom and Olof Sisask say they’ve jumped the first hurdle in the Erdos conjecture. If this is true, the next generation of researchers could start from that point with the first part finished and in hand.
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Let’s start at the very beginning (a very good place to start). Budapest-born mathematician Paul Erdos lived for most of the 20th century and made his name in discrete mathematics, specifically in combinatorics and number theory. Erdos was also a charismatic and Socratic figure not just in mathematics, but in the world at large, putting forward wild, thought-provoking ideas in such a prolific way that it’s not possible to list them all here. Here are his conjectures alone.
In fact, Erdos's work is so extensive that there’s an industry metric, the Erdos number, based on how many hops it takes to link your published work with the man himself.
Out of a lifetime of prolific mathematics, there’s a few most famous Erdos conjectures. One’s proper name is the Erdos-Turan conjecture—Erdos shares titles with dozens of mathematicians—and it deals with evenly spaced sets of numbers that extend infinitely in the right conditions.
Say you have 1 to infinity, the “natural numbers,” meaning the ones we use for counting. Within that set, there are in turn infinitely many progressions that themselves are infinite, including 2, 4, 6, 8 ... and 3, 6, 9, 12 ... and so on.
There are other small chips in the overall problem, like a “weaker” (more accessible) version put forward by Erdos and Turan themselves. These tweak factors like the size of the number set and which number sets qualify to be included. One measure is natural density, and another is whether the reciprocals of the set diverge, meaning the sum of all the fractions continues to move toward infinity rather than settling at a specific limit.
If all this overlapping infinity is makes your head spin, we get it. We gaze into the abyss and the abyss gazes back by 2s and 3s. But it gets weirder, because this conjecture is more specific. In your large-enough set, there are infinitely many sets of each finite size. And to begin proving part of the conjecture, Bloom and Sisask used infinity to make a fresh new argument.
They begin with the summing-to-infinity reciprocals we mentioned before, which Erdos himself laid out as a good measure of whether a set was large enough for his conjecture to apply. This implies the number list has density, which is required to have infinitely many evenly spaced sets. Think of it this way: counting 1, 2, 3 until infinity keeps going one at a time, but a set of, say, prime numbers or squares will keep growing further and further apart overall.
So less dense sets don’t hold up to a progression, for example, because there simply aren’t enough numbers close enough together. And in this proof, Bloom and Sisask show the opposite is also true: once a set is dense enough, it must have enough numbers close together.
The proof is over 70 pages long, and it must be combed through carefully by a variety of peers both for intellectual muster and because the unsolved problem it chips at is so huge. Making progress on progressions, too, happens just an increment at a time.
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