The conjecture is set in the world of clock arithmetic. You start by placing the whole numbers on a number line, then you wrap the line around the face of a clock so that the numbers repeat after some prime number, p. Say p is 7, for instance. In this setting, 0, 7, 14, and all other multiples of 7 are equivalent — meaning that you can add two positive numbers (like 3 and 4) and get zero.
Graham asked the following: If you pick any set of nonzero numbers off this number line (for any p), can you always rearrange them so that the partial sums you get are all different?
The challenge depends on how big your set is compared to p. The more numbers you pick, the more sums there are to manage. But if you choose fewer numbers, there will be fewer ways to rearrange them. These different cases inspire different approaches.
Müyesser, along with his former adviser, Alexey Pokrovskiy of University College London, tackled the case where your set includes almost every possible number up to p. With sets this large, it can be extremely hard to construct a valid ordering. But it turned out that starting with a random ordering can bring you most of the way there.
“Computer scientists often call this a ‘finding the hay in the haystack’ problem,” Müyesser said. You might know that lots of good orderings are out there, but actually finding one is hard. “If you do it randomly, it’s likely going to work, but it’s hard to explicitly describe what the solution is supposed to look like.”
Müyesser and Pokrovskiy needed to ensure that no sequence of numbers anywhere in the ordering added up to zero. Otherwise, adding those numbers to the previous partial sum would repeat that sum.
A completely random ordering might have a few of these troublesome sequences. So Müyesser and Pokrovskiy first set aside a few specially chosen numbers from the set, then randomly scrambled the rest. They scanned their random ordering for any problems; if they came across an interval that added up to zero, they could insert one of the spare numbers to change it. In 2022, they posted their solution, though it was hidden in a paper that focused on applying the same technique to a more general problem.
A couple of years later, Noah Kravitz of Oxford, unaware of Müyesser and Pokrovskiy’s solution, stumbled on Graham’s conjecture in an online archive of unsolved problems. “I saw there was an open problem, and I was like, it’s embarrassing for humanity that we don’t know this,” Kravitz said. “This situation just had to be rectified.”
He decided to approach the conjecture from the opposite end. Together with Benjamin Bedert of Oxford, he considered the case where the set of numbers is tiny compared to p — for instance, Alon said, if you have a set of 100 numbers where p is 1 billion.
Kravitz and Bedert solved Graham’s conjecture for those cases and posted their proof in September 2024. Müyesser saw it and reached out, sharing his own work; the three of them (plus two other colleagues) then teamed up to extend Müyesser’s original approach.
“It was a pretty unlikely combination of people,” Kravitz said. He and Müyesser come from two areas of combinatorics that don’t typically collaborate. “Different sections have completely different techniques,” he said.
Their paper, which they posted in August 2025, handled more cases where the set of numbers is relatively large compared to p. But between those cases and the small-set cases that Kravitz and Bedert had covered, a gap remained. No one could figure out what to do about medium-size sets, such as those that include roughly half as many numbers as p. “Our methods didn’t work there, and there were clear reasons that they would not have worked,” Müyesser said.
It seemed as though research on the problem might enter another long hiatus.
Then, in February 2026, a surprise appeared online.
The Fountain
Lisa Sauermann and Huy Tuan Pham were old friends. The two mathematicians had met in 2015 at Stanford University, where Sauermann was a graduate student and Pham an undergraduate. Today they live on different continents — Sauermann in Bonn, Germany, and Pham in Chicago. But a conference in Germany in September 2025 provided a rare chance for them to share a chalkboard again, and afterward Pham followed Sauermann to Bonn for a short visit. All they needed was a problem to work on.
At the conference, they heard two talks on Graham’s conjecture by mathematicians who had attempted but failed to bridge the gap. They were intrigued. And as it later turned out, Sauermann had encountered a closely related problem in the International Mathematical Olympiad as a high school student. She solved it correctly, and by the time she finished high school, she’d won a gold medal in the prestigious competition four times. (Most likely, it was Chung who placed the problem on that year’s exam, as she was on the committee that wrote the questions, and she frequently took inspiration from Graham’s many puzzles.)
By the end of their three-day visit, Sauermann and Pham had a plan for how to crack the case.
It hinged on a technically demanding method called anti-concentration. Here, an anti-concentration statement asserts that some event has a particularly low chance of happening. But the mechanics of proving these kinds of statements are so intricate that, though Kravitz and others were aware that such an anti-concentration approach might succeed, “we just hadn’t had the guts to actually try it,” he said.
First, though, Sauermann and Pham began the way their predecessors had. They randomly reordered their set of numbers and came up with a procedure to fix any problems — that is, any sequences that add up to zero. Any time they found a zero-sum sequence, they swapped out the last number in the sequence with another one.
This procedure often went without a hitch. But three types of “bad events” would cause it to fail. One: A zero-sum sequence might occur toward the end of the entire arrangement; then there would be no other numbers to swap in. Two: Many zero-sum sequences might appear too close together, making it impossible to fix them all. And three: Fixing one bad sequence might create another zero-sum sequence down the line.
Sauermann and Pham hoped to prove, using anti-concentration, that each of these bad events was sufficiently unlikely. Then there would have to be a way to rearrange the set of numbers to satisfy the conjecture.
To do this, the duo used Fourier analysis — an area of math that lets you rewrite functions as sums of simple waves — to show that in general, when you add up random sets of numbers, no one sum is especially likely to appear. They then used this insight to carefully estimate the probability that each bad event would occur, ultimately showing that the total chance of getting a bad event was less than 100%. That was enough to settle the conjecture.
A few months after their stint in Germany, Sauermann and Pham posted their 27-page proof online. They had shown not only that a satisfactory rearrangement was always possible, but that a random ordering could be rearranged to eliminate bad events at least 90% of the time — a massive success rate.
The mathematicians who had previously worked on the problem were surprised to see the remaining case closed so quickly. “Their approach is just completely different,” Müyesser said.
Together, the four papers prove Graham’s conjecture for sets of all sizes. But they all assume that p is very large; though no one has calculated its exact value, think along the lines of 10 raised to the 100th power. To mathematicians, that’s fine — the salient point is that you’re working in the setting of clock arithmetic. But another aspect of the problem technically remains unsolved — you might still try to resolve the conjecture for all p. And if you want to use the result to choreograph a real juggling routine, you’re out of luck: To correspond to such a large p, the routine would have to be much too long.
The proof confirms that even within these strange, limited number settings, “there are some nice structures that always exist,” Alon said. You can always achieve some degree of flexibility, shuffling the numbers in your set around to avoid revisiting the same partial sums.
“To pose a good problem is really an art,” Chung said. “I think Ron would be extremely happy to see the problem solved.”
