Mathematicians Name 50 of the Highest-Stakes Problems in Math

Mathematicians love to collect elegant, profound questions that defy proof: the longer the holdout, the better. And one day in June, groups from around the world gathered to do just that. The goal was a new list of 50 high-stakes questions with an important caveat: there had to be a way to automatically check solutions to the problems. Workshops took place in six cities: London, Toronto, Los Angeles, New York City, Berkeley, Calif., and Cambridge, Mass.That morning Yang-Hui He, a mathematician at the London Institute for Mathematical Sciences, was accompanied by a very young guest. His son was on school break, so He brought him along to the workshop. The two entered a room full of mathematicians who were well supplied with lunch, snacks, and coffee and got to work. Under the watchful eye of his cake-munching son, He collaborated with others to submit three problems that touched on fields from knot theory to algebra, topology and number theory. “There’s a big culture in mathematics, which is solving open conjectures,” He says. “Many people get Fields Medals because they solve open conjectures. This is very big.”The event was held at the request of Epoch AI, an organization that benchmarks the progress of artificial intelligence systems. Epoch AI senior researcher Greg Burnham says the problems cover a broad spectrum of topics. Lately AI models from various companies have been cutting through a large swath of math’s biggest challenges. Earlier this month, for instance, OpenAI claimed a solution to the Navier-Stokes problem, one of the towering Millennium Prize Problems.On supporting science journalismIf you're enjoying this article, consider supporting our award-winning journalism by subscribing. By purchasing a subscription you are helping to ensure the future of impactful stories about the discoveries and ideas shaping our world today.As of late September, a few of the problems from the new list have been solved, including some landmark ones. The official list still remains peppered with unsolved problems, although OpenAI recently announced that its unreleased internal AI model has solved more than 100 open questions. “My guess is that it likely contains at least a few [of these],” Burnham says.Here are a few of the critical problems the mathematicians identified.The Sum of Three CubesStatus: Open Mathematicians would like to find integer solutions to polynomial equations that contain three different cubes. “It’s strongly suspected that there are integer solutions to the equation x3 + y3 + z3 = 114,” Burnham says. “But it’s suspected that the smallest solutions have something like 30 digits in them.”In 2020, Andrew Booker of the University of Bristol and Andrew Sutherland of the Massachusetts Institute of Technology settled the case for when the sum of x3 + y3 + z3 totals 42. Their efforts consumed 1.3 million computing hours, distributed across volunteer home computers. The next total is 114. Burnham estimated that the same strategy for 114 would cost perhaps $100 million or more, making such a blunt approach unrealistic. Checking the solution simply requires evaluating the equation with the new values.“Apéry-Style” Irrationality ProofStatus: Rumors swirl that this has been solvedIrrational numbers are decimals that can’t be rewritten as a fraction of two integers. Pi and √2 are familiar examples. Although the idea is simple, proving that a number is irrational is notoriously difficult—it took two millennia to prove that pi was irrational.In 1978 the eccentric French mathematician Roger Apéry proved that a number known as zeta(3) was irrational. The zeta numbers originate from Riemann’s famed zeta function, which describes intriguing statistical patterns in the arrangement of the prime numbers. Apéry’s discovery has shaped modern number theory.Evaluating the zeta function at 3 gives the sum of the series 1⁄n3: 13 + 1⁄23 + 1⁄33 + 1⁄43+ .... The idea that this series should sum to an irrational number might be come as a surprise, but centuries ago Euler proved that all the even values of the zeta function are tied to powers of pi, making them irrational. This led mathematicians to suspect that the odd values were also likely irrational.To find the proof for zeta(3), Apéry sandwiched the function between two infinite series that converged at just the right speed to ensure zeta(3) must be irrational. Mathematicians were baffled: How did he find the two series? Frustratingly, Apéry replied that he found them in a flower pot. The mystery of how he located them remains outstanding today. If anyone can reconstruct Apéry’s reasoning, the odd values of zeta may finally be proven irrational.“A credible solution to Apéry-style irrationality—for one of the relevant numbers, the one called zeta(5) —has been circulating,” Burnham says.The Lonely Runner ConjectureStatus: OpenLike the previous two problems, the “lonely runner conjecture” seems simple but has deep ties to many fields. On a circular track, place a group of runners, all of whom run at unique, constant speeds. The lonely runner conjecture predicts that, at some point, a runner will be maximally distant from all the other runners—the saddest, loneliest phase of the race. Mathematicians want to rescue this runner with a counterexample: some number and combination of speeds that will break the pattern. The problem has stood since 1967.The Jones Unknot ConjectureStatus: OpenTake a tangled loop of string. Can you tell whether it’s truly knotted or just a circle in disguise?In 1984 the late mathematician Vaughan Jones introduced a new way to assign polynomials to knots, thereby tangling the two fields together. He later won a Fields Medal for the discovery. If two knots have different Jones polynomials, they’re different knots. But the reverse doesn’t hold—different knots might be linked to the same polynomial. In fact, mathematicians already know that knots of two or more links have ambiguous polynomials. But what about the simplest knot, a circle? In knot theory, a circle is known as an “unknot” and has the trivial Jones polynomial 1. Mathematicians strongly suspect that there is a genuinely knotty loop that also has a Jones polynomial 1. It would only take a single example to show that the Jones polynomials can’t reliably detect whether a knot is there at all.The Core in Approval-Based Committee ElectionsStatus: Recently solved by humans and AIImagine a town of 1,000 people elects a 10-member council. Each citizen can vote for as many candidates as they like. What would make the election fair? One natural answer is proportional representation of the town’s people. For instance, a group of 300 people should wind up with three seats’ worth of influence.How do you tell whether a particular voter is well represented? Count how many of the candidates they voted for that end up winning seats.Now suppose the group of 300 proposes its own three-person slate. Take a resident who voted for only one of the 10 winners but for two of the slate’s three candidates. The council gives her only one representative. But the slate would give her two. If every member of the group likewise ends up with more of their picks on the slate than on the council, the group has a legitimate reason to complain. With just the three seats the group is owed, it could serve every one of its members better than the entire council.A council that no group can object to is said to be in the “core,” a term borrowed from game theory. In a 2016 preprint paper, a team of computer scientists asked whether every election has at least one council in the core.For nearly a decade, progress was slow. Certifying a council as fair means comparing it against every possible group of voters and every slate of candidates that group can claim. As election sizes grow, the number of such combinations explodes. Researchers managed only smaller cases, for instance, 15 candidates in 2025.On September 10 three researchers based in Germany, France and England posted a preprint on arXiv.org that settled the question: every election has a fair council, and there’s an efficient method for finding one. The method doesn’t need to check every group because the proof guarantees that the council it produces is fair. The researchers employed AI as an assistant, alongside their own mathematical intuition, and verified their result using the proof assistant Lean.