Mathematicians sometimes think of their research as a garden and unsolved problems as seeds waiting to sprout.
Some problems are analogous to tulip bulbs. As mathematicians work to solve them, they may appear stagnant and stuck underground, leaving onlookers questioning whether they will ever produce a dazzling result. If they eventually grow into flowers, however, their glow brings the whole garden to life.
Other unsolved mathematical mysteries are akin to the branches of trees. The trees themselves—fields within the wider subject of math —are strong and towering, rooted firmly in established findings. The branches represent chances to grow the trees—expand the fields—and solving these problems one by one pushes the trees toward the sky.
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Still other open questions are like soil—mathematical material that appears ordinary but connects seemingly disparate plants—areas of math—and helps to nourish the whole garden.
We asked mathematicians which open questions intrigued them the most at the moment and what the implications of solving those problems might be. Their answers are below.
Are There Odd Perfect Numbers?
My favorite problem is also the oldest-known problem in math: Are there odd perfect numbers? A perfect number is the sum of its proper factors such as 6 = 3 + 2 + 1 or 28 = 14 + 7 + 4 + 2 + 1. All known perfect numbers are even. The even perfect numbers are also interesting because they are related to the largest prime numbers that have been constructed.
The problem is compelling because one does not even know what to expect. I believe there are odd perfect numbers but that they are very, very large and that a clever search will find one within the next 100 years. The endeavor is not completely hopeless. There are methods to hunt cleverly in a large set of numbers. —Oliver Knill, Harvard University
How Efficiently Can we Factor Integers?
For the sake of argument, suppose an integer n is a product of two prime numbers, p and q. If I write out n for you (in decimal notation, say), how can you recover p and q? What is an efficient general algorithm for this task that you could program in a computer? (By “efficient,” we mean that the run time of the algorithm should grow modestly with the size of n, say linearly or quadratically with the number of digits.) We know of some inefficient algorithms, such as trial division (just checking primes one after another), but the run times of these grow exponentially (or almost so).
In practice, we simply can’t factor random integers with many hundreds of digits. Is it impossible? Or are we just missing one big new idea? This problem has so many fascinating aspects: It’s incredibly simple and as ancient as the integers themselves. It seems to be very difficult. And solving it now would upend the world. To that last point: many of our modern cryptographic protocols are based on the assumed difficulty of factorization. When your computer connects to a secure website, the system sends or receives secret information in a disguised form. To reveal it seems to require factoring a very large integer—which no one knows how to do (without a quantum computer). Finding an efficient algorithm now would probably immediately wreak havoc on one’s own life, as well as the global economy. So do we really want to solve this problem?
Any appealing unsolved problem can serve as a touchstone for research, and this one is no different. One can’t poke and prod a problem like this without discovering connections to many fundamental aspects of number theory. There’s something deliciously unambiguous about algorithmic problems: either you can factor that big number, or you can’t. Perhaps we will never solve it, but the joy is in the exploration. —Katherine Stange, University of Colorado Boulder
The Kummer-Vandiver Conjecture
One of the unsolved math problems that fascinates me is the Kummer-Vandiver conjecture in number theory. It concerns divisibility of class numbers, which in turn reflects the failure of unique factorization into primes.
Early on, when we learn about numbers, we learn how to break them into building blocks by factoring them into primes, and we discover the beautiful fact that this decomposition is unique for any integer. For example, 18 = 2 × 3 × 3, and that is the one and only way to break it up into prime factors.
As we get to more abstract number systems, which might include imaginary numbers, unique factorization might fail. For example, if we include the imaginary number √–5 in our number system, we can see that 6 = 2 × 3 but also that 6 = (1 + √–5)(1 – √–5), and this is another decomposition into irreducible factors, or factors that cannot be broken down any further in our number system. The “class number” of a number system measures the failure of unique factorization. Class number 1 means factorization into primes is unique, whereas a higher class number indicates multiple ways a number can be broken up into irreducible factors.
Cyclotomic fields are number systems that are similarly obtained by including imaginary numbers that are roots of 1—numbers that, when raised to a certain power, equal 1. These can be thought of as points on a circle, with multiplication rotating them around on the circle. These number systems live in the complex numbers, but one can consider their maximal real subfield, the part that lives in the real numbers.
In the mid 1800s, Ernst Kummer first conjectured in letters to Leopold Kronecker that for any odd prime number p, the prime p does not divide the class number of the maximal real subfield of the p-th cyclotomic field—in other words, the number system we get when we adjoin a p-th root of 1 to the rational numbers.
Kummer’s conjecture remains unproven. Harry Vandiver rediscovered and popularized it in the early 1900s. Today the still-open problem is known as the Kummer-Vandiver conjecture or sometimes just the Vandiver conjecture. Kummer verified his conjecture by hand for primes less than 200, and Vandiver later verified it for primes less than 600.
With modern computing, mathematicians have verified the conjecture for primes up to two billion. Of course, this doesn’t prove the conjecture—it simply means that if a counterexample exists, it will involve a prime larger than current computational limits can check. What I find most fascinating about this conjecture is its surprising connection to algebraic K-theory, a highly abstract field of mathematics developed by Daniel Quillen in the ’70s, which at first glance seems unrelated to such a concrete problem in number theory. —Mona Merling, University of Pennsylvania
How To Construct Interesting Algebraic Subvarieties
I work in algebraic geometry, particularly over the complex numbers. An algebraic variety is defined as the zero locus (the set of points where an equation takes on the value zero) of given polynomial equations in several variables. The most important question we are faced with is: How do we construct interesting algebraic subvarieties of a given algebraic variety? Of course, the important precision here is “interesting” because we can always construct subvarieties by simply adding extra equations, but they will be “uninteresting.”
The Hodge conjecture was formulated by W.V.D. Hodge in the mid-20th century and amended into the generalized Hodge conjecture by Alexander Grothendieck in the late 1960s. If true, these conjectures make the incredibly beautiful and well-structured theory of Hodge structures a perfect bridge between topology and algebraic geometry. Although their formulation requires some sophisticated knowledge, they have a strong predictive power and can be tested on very simple instances. For example, the generalized Hodge conjecture predicts the existence of a lot of interesting surfaces in hypersurfaces of degree d, defined by a single degree d equation in projective space of dimension at least 2d. This small special case, which can be stated without appealing to any notion of topology, is completely open except for very small values of d.
Concerning the Hodge conjecture itself, essentially only one case is known, namely the case of co-dimension 1 subvarieties. The proof is wonderfully simple but also very deep and not generalizable to higher co-dimensions. —Claire Voisin, French National Center for Scientific Research (CNRS)
Even after Thousands of Years of Study, Diophantine Equations are Extremely Hard
In school, we learn that the solutions to a quadratic equation, ax2 + bx + c = 0, are:
Our ability to solve or even understand algebraic equations is surprisingly limited when we move beyond this well-known case, even though thinking about them (obsessively) has historically been an extremely fertile ground for the development of deep and vast systems of ideas. For example, the difficulty of writing down the solutions to an innocent equation such as x5 – x + 1 = 0 led to the theory of groups, in the absence of which a substantial portion of modern theoretical physics, with its extensive reliance on a methodical understanding of symmetry, would not be possible.
I’ve spent most of my career thinking about algebraic equations with one more unknown, such as y2 = x5 – x + 1. Most of the time, if someone hands you an equation such as this, it’s hard enough to find all solutions in rational numbers to that one equation. (If you are in the habit of trying things out, you might have found the special solution x = 0, y = 0.)
The main problem in the area, however, does not concern any single equation or even a whole class of equations. It will sound very much like computer science:
Construct a computer algorithm that takes any such equation as input and writes down all the rational solutions:
f(x,y) = 0 → DE algorithm →{all rational solutions to f(x,y) = 0}
The “DE” here abbreviates Diophantine equation, named after the Egyptian mathematician whose book from the popularized the study of such rational solutions. The key challenge, then, is to construct such a DE algorithm, which sounds simple but turns out to be ridiculously difficult.
The DE algorithm problem encompasses a key portion of the conjecture of Bryan John
Birch and Peter Swinnerton-Dyer, the resolution of which will earn an award of $1 million. It’s also a fact (a theorem of Gerd Faltings) that for most equations in two unknowns, there are only finitely many rational solutions. It’s strange then that in most cases, we have no idea how to find this finite set.
The DE algorithm problem comprises some of humankind’s oldest known mathematical challenges, which people from all over the world have studied for millennia. Thus, it’s shocking to me how little is known. Perhaps human intelligence is not good enough for this. —Minhyong Kim, International Center for Mathematical Sciences
How Many Faces Can a Four-Dimensional Polyhedron Have?
I love studying polyhedra (3D shapes with flat sides). I am not so interested in “metrical” questions, questions about volume or area of sides. I am interested in the “combinatorics” of polyhedra—that is, how the vertices (corners), edges and sides fit together. You have probably heard about the Platonic solids, the three-dimensional polyhedra where all the sides are congruent polygons and the same number of them meet at each vertex. (Think cubes and dodecahedra.) But there are much more interesting shapes that qualify as 3D polyhedra, and they play an important role in applications, such as optimization and graphics.
These applications come from the fact that 3D polyhedra can be described as the solutions of a set of linear inequalities in three variables. But those applications often have more than three variables. What about four variables? We call such a thing a four-dimensional polyhedron. Maybe you have heard of the hypercube? It is sometimes pictured as a cube within a cube with corresponding vertices connected up. Four-dimensional polyhedra have vertices, edges and two-dimensional sides. They also have 3D sides; I’ll call these facets. How many of each can a 4D polyhedron have?
A basic question about three-dimensional polyhedra is how many vertices, edges and sides can they have? There is a complete answer to that question, discovered over a hundred years ago by Steinitz. If v, e and s represent the number of vertices, edges and sides of a three-dimensional polyhedron, then v – e + s = 2, v and s are each at least 4, 2e ≥ 3v, and 2e ≥ 3s. The remarkable thing is that if you give me any three integers that satisfy these conditions, I can build a polyhedron with v vertices, e edges and s sides.
Let’s say a 4D polyhedron has v vertices, e edges, s 2D sides and f facets. There are conditions, such as: v – e + s – f = 0, v ≥ 5, f ≥ 5, 2e ≥ 4v, 2s ≥ 4f and some others. But we don’t know the complete set. I can give a set of integers that satisfy all the conditions we know but for which there is no 4D polyhedron. And we don’t even have a guess for what the complete set of conditions is. We do know that a complete set must have some nonlinear inequalities.
One other nice thing about these numbers is a certain symmetry. Given the list of numbers v, e, s or v, e, s, f, we can reverse them and get the numbers for another polyhedron. For example, while the dodecahedron has v = 20, e = 30, s = 12, the icosahedron has v = 12, e = 30, s = 20. The hypercube has v = 16, e = 32, s = 24, f = 8. Another 4D polyhedron, the cross-polytope, has v = 8, e = 24, s = 32, f = 16.
This question of how many faces a 4D polyhedron can have has plagued me for decades. —Margaret Bayer, University of Kansas
The HRT Conjecture
In 1996 Christopher Heil, Jayakumar Ramanathan and Pankaj Topiwala posed what is now known as the Heil-Ramanathan-Topiwala (HRT) conjecture. They stated that any finite set of time-frequency shifts of a nonzero, square-integrable function on the real line is linearly independent.
The HRT conjecture is deceptively simple because it uses the linear algebra notion of linear independence. As such, the HRT conjecture is very easy to state but has proved extremely difficult to solve. In simple terms, a finite set of vectors is linearly independent if the only linear combination that results in the zero vector is the trivial one, where all coefficients are zero. For the HRT conjecture, these vectors are functions generated by a basic operation: the time-frequency shift of a fixed function. Specifically, the time-frequency shift of a nonzero function g by a point (p, q) in the plane is the function obtained by translating g by the first coordinate p and multiplying the result by a complex exponential function with frequency q.
So far, progress on the HRT conjecture has been limited and doesn’t yet offer a definitive answer to its validity. In addition, the known solutions to special cases of the conjecture employ tools from different areas of mathematics and typically fall into two categories. In the first, restrictions are placed on the points in the plane used as time-frequency parameters, while the function is chosen arbitrarily. For instance, the conjecture is true when the points are selected from a lattice, a condition that always holds when dealing with any three distinct points. In the second category, restrictions are imposed on the function, while the set of points remains arbitrary. There are also other cases where restrictions are imposed on both the function and the set of points. The conjecture remains unproven, however, even when dealing with an arbitrary nonzero square-integrable function and any arbitrary set of four distinct points in the plane. —Kasso Akochayé Okoudjou, Tufts University
The Schoenflies Problem
It’s amazing that we don’t know the answer to the following question: Does a smooth sphere in space bound a ball? This is known as the Schoenflies problem. The trick is that the sphere can be any dimension—e.g., a one-dimensional sphere is a circle, and a 2D sphere is the usual surface of a 3D ball—and sits smoothly inside a space that’s one dimension larger.
We actually do know that the answer to this problem is yes when the ambient space has any dimension other than four. But the question remains open for smooth 3D spheres in 4D space.
I find it unsettling to have a gap in the middle of solved values. One might think, “Surely the answer is yes. Why should dimension four be different from all the other dimensions?”
On the other hand, 4D topology often is different from the study of spaces of other dimensions. For example, there are infinitely many different smooth 4D objects that are continuously, but not smoothly, equivalent to standard 4D space. That doesn’t happen in any other dimension.
I would hazard a guess that the answer to the Schoenflies problem is “no” in dimension four—which would be very exciting and would also mean I have no idea where to start because the situation would be different from all previous cases. —Maggie Miller, University of Texas at Austin
Distance Between Knots in Three-Manifolds
I study three-manifolds, or spaces that look like 3D Euclidean space when you zoom in but that have more structure when you zoom out. An analogy would be how a sphere looks like a plane when you zoom in, but when you zoom out, you discover it has more structure to it. In knot theory one studies knotted loops in 3D space (so no loose ends), which includes examples such as the figure eight knot and the trefoil knot. It can be hard to directly visualize three-manifolds because they don’t fit inside 3D space. Because they look like 3D space from the inside, however, you can still talk about things such as knots and ask: How different can knots be in these new settings?
One question that really interests me is how far away knots in other three-manifolds can be from knots in 3D space. In my area, one often measures distance between knots by how complicated a surface between them has to be. It amazes me that despite the variety of knots in 3D space, we have already shown that there are knots in three-manifolds that require very complicated surfaces to get to any knot in 3D space. At the same time, a lot of those knots were specifically constructed for our techniques to work, and we have a lot to learn to answer that question for arbitrary knots. —Seppo Niemi-Colvin, Indiana University Bloomington
