10 Hard Math Problems That Continue to Stump Even the Brightest Minds

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For now, you can take a crack at the hardest math problems known to man, woman, and machine. ✅ More from Popular Mechanics :

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The Collatz Conjecture

hardest math problems

In September 2019, news broke regarding progress on this 82-year-old question, thanks to prolific mathematician Terence Tao. And while the story of Tao’s breakthrough is promising, the problem isn’t fully solved yet.

A refresher on the Collatz Conjecture : It’s all about that function f(n), shown above, which takes even numbers and cuts them in half, while odd numbers get tripled and then added to 1. Take any natural number, apply f, then apply f again and again. You eventually land on 1, for every number we’ve ever checked. The Conjecture is that this is true for all natural numbers (positive integers from 1 through infinity).

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Tao’s recent work is a near-solution to the Collatz Conjecture in some subtle ways. But he most likely can’t adapt his methods to yield a complete solution to the problem, as Tao subsequently explained. So, we might be working on it for decades longer.

The Conjecture lives in the math discipline known as Dynamical Systems , or the study of situations that change over time in semi-predictable ways. It looks like a simple, innocuous question, but that’s what makes it special. Why is such a basic question so hard to answer? It serves as a benchmark for our understanding; once we solve it, then we can proceed onto much more complicated matters.

The study of dynamical systems could become more robust than anyone today could imagine. But we’ll need to solve the Collatz Conjecture for the subject to flourish.

Goldbach’s Conjecture

hardest math problems

One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes.” You check this in your head for small numbers: 18 is 13+5, and 42 is 23+19. Computers have checked the Conjecture for numbers up to some magnitude. But we need proof for all natural numbers.

Goldbach’s Conjecture precipitated from letters in 1742 between German mathematician Christian Goldbach and legendary Swiss mathematician Leonhard Euler , considered one of the greatest in math history. As Euler put it, “I regard [it] as a completely certain theorem, although I cannot prove it.”

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Euler may have sensed what makes this problem counterintuitively hard to solve. When you look at larger numbers, they have more ways of being written as sums of primes, not less. Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23. So it feels like Goldbach’s Conjecture is an understatement for very large numbers.

Still, a proof of the conjecture for all numbers eludes mathematicians to this day. It stands as one of the oldest open questions in all of math.

The Twin Prime Conjecture

hardest math problems

Together with Goldbach’s, the Twin Prime Conjecture is the most famous in Number Theory—or the study of natural numbers and their properties, frequently involving prime numbers. Since you've known these numbers since grade school, stating the conjectures is easy.

When two primes have a difference of 2, they’re called twin primes. So 11 and 13 are twin primes, as are 599 and 601. Now, it's a Day 1 Number Theory fact that there are infinitely many prime numbers. So, are there infinitely many twin primes? The Twin Prime Conjecture says yes.

Let’s go a bit deeper. The first in a pair of twin primes is, with one exception, always 1 less than a multiple of 6. And so the second twin prime is always 1 more than a multiple of 6. You can understand why, if you’re ready to follow a bit of heady Number Theory.

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All primes after 2 are odd. Even numbers are always 0, 2, or 4 more than a multiple of 6, while odd numbers are always 1, 3, or 5 more than a multiple of 6. Well, one of those three possibilities for odd numbers causes an issue. If a number is 3 more than a multiple of 6, then it has a factor of 3. Having a factor of 3 means a number isn’t prime (with the sole exception of 3 itself). And that's why every third odd number can't be prime.

How’s your head after that paragraph? Now imagine the headaches of everyone who has tried to solve this problem in the last 170 years.

The good news is that we’ve made some promising progress in the last decade. Mathematicians have managed to tackle closer and closer versions of the Twin Prime Conjecture. This was their idea: Trouble proving there are infinitely many primes with a difference of 2? How about proving there are infinitely many primes with a difference of 70,000,000? That was cleverly proven in 2013 by Yitang Zhang at the University of New Hampshire.

For the last six years, mathematicians have been improving that number in Zhang’s proof, from millions down to hundreds. Taking it down all the way to 2 will be the solution to the Twin Prime Conjecture. The closest we’ve come —given some subtle technical assumptions—is 6. Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer.

The Riemann Hypothesis

hardest math problems

Today’s mathematicians would probably agree that the Riemann Hypothesis is the most significant open problem in all of math. It’s one of the seven Millennium Prize Problems , with $1 million reward for its solution. It has implications deep into various branches of math, but it’s also simple enough that we can explain the basic idea right here.

There is a function, called the Riemann zeta function, written in the image above.

For each s, this function gives an infinite sum, which takes some basic calculus to approach for even the simplest values of s. For example, if s=2, then 𝜁(s) is the well-known series 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly 𝜋²/6. When s is a complex number—one that looks like a+b𝑖, using the imaginary number 𝑖—finding 𝜁(s) gets tricky.

So tricky, in fact, that it’s become the ultimate math question. Specifically, the Riemann Hypothesis is about when 𝜁(s)=0; the official statement is, “Every nontrivial zero of the Riemann zeta function has real part 1/2.” On the plane of complex numbers, this means the function has a certain behavior along a special vertical line. The hypothesis is that the behavior continues along that line infinitely.

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The Hypothesis and the zeta function come from German mathematician Bernhard Riemann, who described them in 1859. Riemann developed them while studying prime numbers and their distribution. Our understanding of prime numbers has flourished in the 160 years since, and Riemann would never have imagined the power of supercomputers. But lacking a solution to the Riemann Hypothesis is a major setback.

If the Riemann Hypothesis were solved tomorrow, it would unlock an avalanche of further progress. It would be huge news throughout the subjects of Number Theory and Analysis. Until then, the Riemann Hypothesis remains one of the largest dams to the river of math research.

The Birch and Swinnerton-Dyer Conjecture

hardest math problems

The Birch and Swinnerton-Dyer Conjecture is another of the six unsolved Millennium Prize Problems, and it’s the only other one we can remotely describe in plain English. This Conjecture involves the math topic known as Elliptic Curves.

When we recently wrote about the toughest math problems that have been solved , we mentioned one of the greatest achievements in 20th-century math: the solution to Fermat’s Last Theorem. Sir Andrew Wiles solved it using Elliptic Curves. So, you could call this a very powerful new branch of math.

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In a nutshell, an elliptic curve is a special kind of function. They take the unthreatening-looking form y²=x³+ax+b. It turns out functions like this have certain properties that cast insight into math topics like Algebra and Number Theory.

British mathematicians Bryan Birch and Peter Swinnerton-Dyer developed their conjecture in the 1960s. Its exact statement is very technical, and has evolved over the years. One of the main stewards of this evolution has been none other than Wiles. To see its current status and complexity, check out this famous update by Wells in 2006.

The Kissing Number Problem

hardest math problems

A broad category of problems in math are called the Sphere Packing Problems. They range from pure math to practical applications, generally putting math terminology to the idea of stacking many spheres in a given space, like fruit at the grocery store. Some questions in this study have full solutions, while some simple ones leave us stumped, like the Kissing Number Problem.

When a bunch of spheres are packed in some region, each sphere has a Kissing Number, which is the number of other spheres it’s touching; if you’re touching 6 neighboring spheres, then your kissing number is 6. Nothing tricky. A packed bunch of spheres will have an average kissing number, which helps mathematically describe the situation. But a basic question about the kissing number stands unanswered.

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First, a note on dimensions. Dimensions have a specific meaning in math: they’re independent coordinate axes. The x-axis and y-axis show the two dimensions of a coordinate plane. When a character in a sci-fi show says they’re going to a different dimension, that doesn’t make mathematical sense. You can’t go to the x-axis.

A 1-dimensional thing is a line, and 2-dimensional thing is a plane. For these low numbers, mathematicians have proven the maximum possible kissing number for spheres of that many dimensions. It’s 2 when you’re on a 1-D line—one sphere to your left and the other to your right. There’s proof of an exact number for 3 dimensions, although that took until the 1950s.

Beyond 3 dimensions, the Kissing Problem is mostly unsolved. Mathematicians have slowly whittled the possibilities to fairly narrow ranges for up to 24 dimensions, with a few exactly known, as you can see on this chart . For larger numbers, or a general form, the problem is wide open. There are several hurdles to a full solution, including computational limitations. So expect incremental progress on this problem for years to come.

The Unknotting Problem

hardest math problems

The simplest version of the Unknotting Problem has been solved, so there’s already some success with this story. Solving the full version of the problem will be an even bigger triumph.

You probably haven’t heard of the math subject Knot Theory . It ’s taught in virtually no high schools, and few colleges. The idea is to try and apply formal math ideas, like proofs, to knots, like … well, what you tie your shoes with.

For example, you might know how to tie a “square knot” and a “granny knot.” They have the same steps except that one twist is reversed from the square knot to the granny knot. But can you prove that those knots are different? Well, knot theorists can.

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Knot theorists’ holy grail problem was an algorithm to identify if some tangled mess is truly knotted, or if it can be disentangled to nothing. The cool news is that this has been accomplished! Several computer algorithms for this have been written in the last 20 years, and some of them even animate the process .

But the Unknotting Problem remains computational. In technical terms, it’s known that the Unknotting Problem is in NP, while we don ’ t know if it’s in P. That roughly means that we know our algorithms are capable of unknotting knots of any complexity, but that as they get more complicated, it starts to take an impossibly long time. For now.

If someone comes up with an algorithm that can unknot any knot in what’s called polynomial time, that will put the Unknotting Problem fully to rest. On the flip side, someone could prove that isn’t possible, and that the Unknotting Problem’s computational intensity is unavoidably profound. Eventually, we’ll find out.

The Large Cardinal Project

hardest math problems

If you’ve never heard of Large Cardinals , get ready to learn. In the late 19th century, a German mathematician named Georg Cantor figured out that infinity comes in different sizes. Some infinite sets truly have more elements than others in a deep mathematical way, and Cantor proved it.

There is the first infinite size, the smallest infinity , which gets denoted ℵ₀. That’s a Hebrew letter aleph; it reads as “aleph-zero.” It’s the size of the set of natural numbers, so that gets written |ℕ|=ℵ₀.

Next, some common sets are larger than size ℵ₀. The major example Cantor proved is that the set of real numbers is bigger, written |ℝ|>ℵ₀. But the reals aren’t that big; we’re just getting started on the infinite sizes.

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For the really big stuff, mathematicians keep discovering larger and larger sizes, or what we call Large Cardinals. It’s a process of pure math that goes like this: Someone says, “I thought of a definition for a cardinal, and I can prove this cardinal is bigger than all the known cardinals.” Then, if their proof is good, that’s the new largest known cardinal. Until someone else comes up with a larger one.

Throughout the 20th century, the frontier of known large cardinals was steadily pushed forward. There’s now even a beautiful wiki of known large cardinals , named in honor of Cantor. So, will this ever end? The answer is broadly yes, although it gets very complicated.

In some senses, the top of the large cardinal hierarchy is in sight. Some theorems have been proven, which impose a sort of ceiling on the possibilities for large cardinals. But many open questions remain, and new cardinals have been nailed down as recently as 2019. It’s very possible we will be discovering more for decades to come. Hopefully we’ll eventually have a comprehensive list of all large cardinals.

What’s the Deal with 𝜋+e?

hardest math problems

Given everything we know about two of math’s most famous constants, 𝜋 and e , it’s a bit surprising how lost we are when they’re added together.

This mystery is all about algebraic real numbers . The definition: A real number is algebraic if it’s the root of some polynomial with integer coefficients. For example, x²-6 is a polynomial with integer coefficients, since 1 and -6 are integers. The roots of x²-6=0 are x=√6 and x=-√6, so that means √6 and -√6 are algebraic numbers.

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All rational numbers, and roots of rational numbers, are algebraic. So it might feel like “most” real numbers are algebraic. Turns out, it’s actually the opposite. The antonym to algebraic is transcendental, and it turns out almost all real numbers are transcendental—for certain mathematical meanings of “almost all.” So who’s algebraic , and who’s transcendental?

The real number 𝜋 goes back to ancient math, while the number e has been around since the 17th century. You’ve probably heard of both, and you’d think we know the answer to every basic question to be asked about them, right?

Well, we do know that both 𝜋 and e are transcendental. But somehow it’s unknown whether 𝜋+e is algebraic or transcendental. Similarly, we don’t know about 𝜋e, 𝜋/e, and other simple combinations of them. So there are incredibly basic questions about numbers we’ve known for millennia that still remain mysterious.

Is 𝛾 Rational?

hardest math problems

Here’s another problem that’s very easy to write, but hard to solve. All you need to recall is the definition of rational numbers.

Rational numbers can be written in the form p/q, where p and q are integers. So, 42 and -11/3 are rational, while 𝜋 and √2 are not. It’s a very basic property, so you’d think we can easily tell when a number is rational or not, right?

Meet the Euler-Mascheroni constant 𝛾, which is a lowercase Greek gamma. It’s a real number, approximately 0.5772, with a closed form that’s not terribly ugly; it looks like the image above.

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The sleek way of putting words to those symbols is “gamma is the limit of the difference of the harmonic series and the natural log.” So, it’s a combination of two very well-understood mathematical objects. It has other neat closed forms, and appears in hundreds of formulas.

But somehow, we don’t even know if 𝛾 is rational. We’ve calculated it to half a trillion digits, yet nobody can prove if it’s rational or not. The popular prediction is that 𝛾 is irrational. Along with our previous example 𝜋+e, we have another question of a simple property for a well-known number, and we can’t even answer it.

Headshot of Dave Linkletter

Dave Linkletter is a Ph.D. candidate in Pure Mathematics at the University of Nevada, Las Vegas. His research is in Large Cardinal Set Theory. He also teaches undergrad classes, and enjoys breaking down popular math topics for wide audiences.

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The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously difficult to solve. So what is the Collatz Conjecture and what makes it so difficult? Veritasium investigates. 

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Unsolved Problems

There are many unsolved problems in mathematics. Some prominent outstanding unsolved problems (as well as some which are not necessarily so well known) include

1. The Goldbach conjecture .

2. The Riemann hypothesis .

3. The conjecture that there exists a Hadamard matrix for every positive multiple of 4.

4. The twin prime conjecture (i.e., the conjecture that there are an infinite number of twin primes ).

5. Determination of whether NP-problems are actually P-problems .

6. The Collatz problem .

7. Proof that the 196-algorithm does not terminate when applied to the number 196.

8. Proof that 10 is a solitary number .

11. Finding an Euler brick whose space diagonal is also an integer.

12. Proving which numbers can be represented as a sum of three or four (positive or negative) cubic numbers .

14. Determining if the Euler-Mascheroni constant is irrational .

15. Deriving an analytic form for the square site percolation threshold .

16. Determining if any odd perfect numbers exist.

The Clay Mathematics Institute ( http://www.claymath.org/millennium/ ) of Cambridge, Massachusetts (CMI) has named seven "Millennium Prize Problems," selected by focusing on important classic questions in mathematics that have resisted solution over the years. A $7 million prize fund has been established for the solution to these problems, with $1 million allocated to each. The problems consist of the Riemann hypothesis , Poincaré conjecture , Hodge conjecture , Swinnerton-Dyer Conjecture , solution of the Navier-Stokes equations, formulation of Yang-Mills theory, and determination of whether NP-problems are actually P-problems .

In 1900, David Hilbert proposed a list of 23 outstanding problems in mathematics ( Hilbert's problems ), a number of which have now been solved, but some of which remain open. In 1912, Landau proposed four simply stated problems, now known as Landau's problems , which continue to defy attack even today. One hundred years after Hilbert, Smale (2000) proposed a list of 18 outstanding problems.

K. S. Brown, D. Eppstein, S. Finch, and C. Kimberling maintain webpages of unsolved problems in mathematics. Classic texts on unsolved problems in various areas of mathematics are Croft et al. (1991), in geometry , and Guy (2004), in number theory .

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The Simplest Math Problem No One Can Solve (Collatz Conjecture)

The Simplest Math Problem No One Can Solve (Collatz Conjecture)

Concerning the above math problem, Paul Erdös said that “Mathematics is not yet ripe for such questions.” He also offered $500 USD for its solution. Jeffrey Lagarias said in 2010 that this math problem: “Is an extraordinarily difficult problem, completely out of reach of present day mathematics.” Richard Guy has this on his list of “Don’t try to solve these problems.”

What is the simplest equation no one can solve?

3x+1 popularly called the Collatz conjecture is the simplest math problem no one can solve. Even though it’s easy for almost anyone to understand it’s also at the same time impossible to solve. It’s was named after Lothar Collatz in 1973. This problem has many origin stories and many names. It’s also known as the Ulam conjecture, Kakutani’s problem, Thwaites conjecture, Hasse’s algorithm, the Syracuse problem, and simply 3N+1.

This is the most dangerous problem in mathematics, one that young mathematicians are warned not to waste their time on. It’s a simple conjecture that not even the world’s best mathematicians have been able to solve. This math problem is both famous and infamous as no respectable mathematician is still trying to find a solution for it. It’s thought to be unsolvable.

The Collatz conjecture is the most famous unsolved problem in all of mathematics. This conjecture asks whether repeating two simple arithmetic operations will at some point transform every positive integer into 1. It uses sequences of integers in which each number (or term) is obtained from the previous number as follows: if the previous number is even, the next number is one half of the previous number. If the previous number is odd, the next number is 3 times the previous number plus 1. The conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence.

Consider the following operation on an arbitrary positive integer:

If the number is even, divide it by two. If the number is odd, triple it and add one.

Now form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.

The Collatz conjecture is that this process will eventually reach the number 1, regardless of which positive integer is chosen to start with.

We apply two rules. If the number is odd, we multiply by three and add one. So three times seven is 21, plus one is 22. If the number is even, we divide by two. So 22 divided by two is 11. Now, we keep applying these two rules. 11 is odd, so we multiply by 3, 33, and add 1, 34. Even, divide by two, 17, odd. Multiply by 3, 51, add 1, 52, even. Divide by two, 26, still even. Divide by two, 13, odd. So we multiply by 3, 39, add one, and that’s 40, which is even, so we divide by two, 20, divide by two, 10, divide by two, five, odd. Multiply by three, 15, add one, 16, divide by two that’s eight, and then four, two, and one. Now, one is odd, so we multiply by three and add one, which equals four. But four goes to two, goes to one, so we’re in a loop, and the lowest number is one.

Now, the conjecture is this: every positive integer, if you apply these rules, will eventually end up in the four, two, one loop. [1]

The two things that are so fascinating about this math problem is that you can start with any number and take it through the iterations and you will get wildly different pathways for each number as some can go very high before getting trapped in the 4-2-1 loop and reverting to 1. The amount of steps that it takes for a number to rise to highs and then revert to 1 also varies widely even with numbers starting next to each other.

When you place the trajectories of the different numbers on charts as they rise and revert back to 1 as they go through the math formula the patterns they create are random. The patterns the numbers make look similar to stocks that revert back to the mean after parabolic run ups in price.

Both are examples of geometric Brownian motion. That means if you take the log and remove the linear trend, the fluctuations are random. It’s like flipping a coin each step. If the coin is heads, the line goes up, tails, it goes down. 3x+1 is just like the random volatility of the stock market. Over long-enough periods, the stock market indexes tend to trend upwards, while 3x+1 eventually trends down to 1.

The below chart shows a histogram of the total stopping times (TST) of the integers from 1 to 100 million where the x-axis represents the stopping time and the y-axis represents the frequency. We can see that the largest stopping time does not exceed 500. The most frequent TST is around 175. This provides strong evidence that the Collatz conjecture is true. Since stopping times seem to always be small, computer programs can be written to determine whether the Collatz sequence of a given integer reaches 1 or not. This method has been used to verify the conjecture for all positive integers up to 5 * 260.

The Simplest Math Problem No One Can Solve (Collatz Conjecture)

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

The smallest  i  such that  a i  <  a 0   is called the stopping time of  n . Similarly, the smallest  k  such that  a k  = 1  is called the total stopping time of  n .  If one of the indexes  i  or  k  doesn’t exist, we say that the stopping time or the total stopping time, respectively, is infinite. [2]

Why is 3x+1 impossible?

What is the longest collatz sequence.

The longest Collatz chain below five million contains 597 elements and starts with 3732423. [3]

Can the Collatz conjecture be solved?

The Collatz conjecture is a brain teasing piece of trivia. The mathematical methods used to determine an answer are very interesting. Mathematicians still don’t really know how to decide the true value of the Collatz conjecture with the methods we know to calculate a solution. Every possible answer will involve either a novel application of our current processes or new methods that might also apply to other math problems. That’s why this problem is so interesting.

The Simplest Math Problem No One Can Solve (Collatz Conjecture)

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Some Math Problems Seem Impossible. That Can Be a Good Thing.

November 18, 2020

math problems you can't solve

BIG MOUTH for Quanta Magazine


Construct a convex octagon with four right angles.

It probably says a lot about me as a teacher that I assign problems like this. I watch as students try to arrange the right angles consecutively. When that doesn’t work, some try alternating the right angles. Failing again, they insert them randomly into the polygon. They scribble, erase and argue. The sound of productive struggle is music to a teacher’s ears.

Then they get suspicious and start asking questions. “You said four right angles. Did you really mean three?” “Are you sure you meant to say convex?” “Four right angles would basically make a rectangle. How can we get four more sides in our octagon?” I listen attentively, nodding along, acknowledging their insights.

Finally someone asks the question they’ve been tiptoeing around, the question I’ve been waiting for: “Wait, is this even possible?”

This question has the power to shift mindsets in math. Those thinking narrowly about specific conditions must now think broadly about how those conditions fit together. Those working inside the system must now take a step back and examine the system itself. It’s a question that’s been asked over and over in the history of math, by those working on problems ranging from squaring the circle to circumambulating the city of Königsberg . And it’s a question that helps us shape what mathematics is and how we understand it.

For example, finding an octagon with certain properties is a very different mathematical task than showing that no such octagon could possibly exist. In playing around with different octagons, we might just stumble upon one that has four right angles.

But luck doesn’t play a role in proving that such an octagon can’t exist. It takes deep knowledge, not just of polygons, but of mathematics itself. To consider impossibility, we need to understand that just asserting that a thing exists doesn’t make it so. Mathematical definitions, properties and theorems all live in a tension born of interconnectedness. In trying to imagine our octagon with four right angles, we work inside those interconnected rules.

But to realize our octagon is impossible, we need to step back and look at the big picture. What mathematical and geometric principles might be violated by an octagon with four right angles? Here, the polygon angle sum theorem is a good place to start.

The sum of the interior angles of an n -sided polygon is given by the formula:

S = ( n – 2) × 180º

This is because every n -sided polygon can be cut into ( n − 2) triangles, each with total internal angles of 180º.

For an octagon, this means the interior angles add up to (8 – 2) × 180º = 6 × 180º = 1080º. Now, if four of those angles are right, each with a measure of 90º, that accounts for 4 × 90º = 360º of the angle sum, which leaves 1080º – 360º = 720º to divide up among the octagon’s remaining four angles.

That means the average measure of those four remaining angles must be:

$latex \frac{720º}{4}$ = 180º

But the interior angles of a convex polygon must each measure less than 180º, so this is impossible. A convex octagon with four right angles cannot exist.

Proving impossibility in this way requires stepping back and seeing how different mathematical rules — like the polygon angle sum formula and the definition of a convex polygon — exist in tension. And since proofs of impossibility rely on thinking broadly across rules, there’s often more than one way to construct the proof.

If an octagon had four right angles, walking around just those angles would bring us full circle: It would be as though we had walked completely around a rectangle. This insight leads us to a rule that gives us a different proof of impossibility. It is known that the sum of the exterior angles of a convex polygon is always 360º. Since an exterior angle of a right angle is also a right angle, our four right angles would take up the entire 360º of the octagon’s exterior-angle measure. This leaves nothing for the remaining four angles, again establishing that our octagon is impossible.

Proving that something is impossible is a powerful act of mathematics. It shifts our perspective from that of rule follower to that of rule enforcer. And to enforce the rules, you must first understand them. You need to know not just how to apply them, but when they don’t apply. And you also need to be on the lookout for situations where rules might conflict with one another. Our octagon exploration exposes the interplay between polygons, convexity, right angles and angle sums. And it highlights how  S = ( n – 2) × 180º isn’t just a formula: It’s one condition in a world of competing conditions.

Proofs of impossibility can help us better understand all areas of math. In school, probability lessons often begin with flipping lots of imaginary coins. I invite students to create an unfair coin — one that is biased toward coming up heads or tails — that has the following property: When the coin is flipped twice, the results of the two flips are more likely to be different than the same. In other words, you’re more likely to get heads and tails than to get heads and heads or tails and tails.

After some tinkering and a little productive frustration, the students arrive at an interesting hypothesis: Different outcomes are never more likely than the same outcome. Some algebra illuminates this and hints at an underlying symmetry.

Let’s say that the coin is biased toward heads. We’ll call the probability of flipping heads $latex \frac{1}{2}$ + k , where 0 < k ≤ $latex \frac{1}{2}$. The fact that k > 0 guarantees that heads is more likely than tails, which has probability  $latex \frac{1}{2}$ – k , since the two probabilities must add up to 1.

If we flip the coin twice, the probability of getting two heads or two tails will be

$latex  \left(\frac{1}{2}+k\right)^{2}+\left(\frac{1}{2}-k\right)^{2}$.

Here we’re adding the probability of getting two heads (on the left) with the probability of getting two tails (on the right). Using algebra we can simplify the probability of getting the same result on both flips:

$latex  \left(\frac{1}{2}+k\right)^{2}+\left(\frac{1}{2}-k\right)^{2}$ = $latex \frac{1}{4}$ + k + k ² + $latex \frac{1}{4}$ –   k  +  k ² = $latex \frac{1}{2}$ + 2 k².

Since k > 0 , we know that $latex \frac{1}{2}$ + 2 k ² > $latex \frac{1}{2}$, which means it’s more likely than not that the outcomes of the flips will be the same. In fact, we see that even if  k = 0 (when the coin is fair), the probability of the same outcomes is exactly $latex \frac{1}{2}$, making the probability of different flips also $latex \frac{1}{2}$. The same outcome will never be less likely than different outcomes.

As with the octagon problem, we see competing mathematical tensions at work: Altering the likelihood of getting one side of the coin changes the likelihood of getting the other, and this interconnectedness governs what’s possible in terms of the two-flip outcomes. We expose those tensions by trying to do the impossible.

We can expose these tensions in every area of math. Try to find six consecutive integers that sum to 342, and some perseverance will lead to a better understanding of parity. (The fact that consecutive integers alternate between even and odd affects what their sums can be.) The search for a cubic polynomial with integer coefficients that has three non-real roots will teach you about the importance of complex conjugates — pairs of complex numbers whose product and sum are always real. And if you try to inscribe a non-square rhombus in a circle, you’ll walk away having discovered an important property of cyclic quadrilaterals — that the opposite angles in a quadrilateral whose vertices lie on a circle must sum to 180 degrees.

Confronting the impossible invites us to explore the boundaries of our mathematical worlds. The impossible itself is already a kind of generalization, so it’s only natural to keep generalizing: An octagon can’t have four right angles, but what about a decagon? What about a convex polygon with n > 4 sides? These kinds of questions push against the boundaries of our mathematical worlds and deepen our understanding of them.

If we push further, the impossible can even inspire the creation of new mathematical worlds. To prove the impossibility of squaring the circle — a problem that’s at least 2,000 years old — we needed the modern theory of transcendental numbers that cannot be roots of integer polynomials. To solve the bridges of Königsberg problem, Euler turned islands and bridges into vertices and edges, bringing to life the rich fields of graph theory and network theory, with their many applications. Taking the square root of −1 led to an entirely new system of arithmetic. And the logician Kurt Gödel changed the landscape of math forever when he proved that it’s impossible to prove that everything that is true is true.

So the next time you’re stuck on a math problem, ask yourself: “Is this possible?” Struggling with impossibility could give you a better understanding of what actually is possible. You might even create some new math along the way.

1. Find the area of the triangle whose side lengths are 46, 85 and 38.

2. Let f ( x ) = 2 x ³ + bx ² + cx + d . Find integers b , c and d such that  f  $latex  \left(\frac{1}{4}\right)$ = 0.

3. Find a perfect square, all of whose digits are in the set {2, 3, 7, 8}.

Click for Answer 1:

It’s fun to use something like Heron’s formula to compute the area of this non-triangle. Interesting questions will follow from this!

Click for Answer 2: There are different ways to establish the impossibility of this polynomial. For example, these conditions violate the rational root theorem, which says that any rational roots of a polynomial must be a ratio of a factor of the constant term divided by a factor of the leading coefficient.

Click for Answer 3: A curious fact about perfect squares shows us that this task is impossible. The units digit of a perfect square can only be 0, 1, 4, 5, 6 or 9. This can be shown by simply squaring every possible units digit and observing the possible results. Since no perfect square can end in 2, 3, 7 or 8, no perfect square exists with only those digits.

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