An Equation Mathematicians Have Stared At for 113 Years

Added: 2026-05-10

[00:00:00]Three integers squared on the left, three times their product on the right.

[00:00:06]That's the entire equation. The trivial guess works. 1 1 1, both sides equal three. Try 1 1 2. The left gives 1 + 1 + 4. The right gives 3 * 1 * 1 * 2. Both sides give six. Try 1 2 5. The left gives 1 + 4 + 25, 30. The right gives 3 * 1 * 2 * 5, 30. 1 5 13 also works. 2 5 29 also works. 5 13 194 also works. The numbers are exploding. The solutions don't stop coming, and nobody in 146 years has proven the basic question about how they're organized. Andrei Markov wrote this equation down in 1879.

[00:00:58]He was studying continued fractions, the way irrational numbers get approximated by fractions like 22 over 7 or 355 over 113. Buried in the analysis was a single integer equation. Three variables, all positive whole numbers. When Markov enumerated solutions, the structure caught him off guard. The triples weren't scattered, they sat on a tree.

[00:01:23]Every triple traced back to a parent.

[00:01:25]Every triple had children. The whole infinite family branched out from one root. From that tree, a question emerged, a simple yes or no question.

[00:01:35]Markov could not answer it. Frobenius could not answer it. Nobody for the next 113 years has been able to answer it.

[00:01:44]Today, three things: where the tree comes from, why every solution sits on it, and why this equation refuses to close.

[00:01:53]The trick is older than the equation.

[00:01:55]It's called Vieta jumping. Take the equation. Now fix two of the variables.

[00:02:00]Hold X and Y still. Treat Z as the only unknown. What's left is a quadratic in Z. Move everything to one side. Z squared minus 3XY times Z plus X squared plus Y squared equals 0. That's a standard quadratic. The coefficient of Z is minus 3XY. The constant term is X squared plus Y squared. Vieta's formulas tell us about the roots without solving for them. Sum of the two roots equals 3XY. Product of the two roots equals X squared plus Y squared. Suppose Z is one of those roots. Call the other root Z prime. The sum gives us Z plus Z prime equals 3XY. Solve for Z prime. Z prime equals 3XY minus Z. Stop and look at this. We started with one solution. We just constructed another one. Take any triple XYZ that satisfies the equation.

[00:02:57]Replace Z with 3XY minus Z. The new triple satisfies the equation, too. Same X, same Y, new Z. That's Vieta jumping.

[00:03:07]One algebraic step that turns any solution into a different solution. Test it on 111. Replace the third coordinate.

[00:03:16]3 * 1 * 1 - 1 is 2. New triple, 112.

[00:03:21]Verified earlier. Test it on 112.

[00:03:24]Replace the second coordinate. 3 * 1 * 2 - 1 is 5. New triple, 152. Sort it, 125.

[00:03:34]The same move applies in any of the three variable slots. Every Markov triple has three Vieta jumps available.

[00:03:41]One per coordinate.

[00:03:44]Start at 111.

[00:03:46]Vieta jump produces 112.

[00:03:49]The same triple, no matter which coordinate you replace because all three coordinates are equal. From 1 1 2, jump the third coordinate and you go back to 1 1 1. That's the parent. Jump the first coordinate. 3 * 1 * 2 - 1 is 5. Triple becomes 1 2 5. Jump the second coordinate. Same answer by symmetry.

[00:04:14]From 1 2 5, jump the third coordinate to get back to the parent. Jump the first coordinate. 3 * 2 * 5 - 1 is 29. New triple, 2 5 29. Jump the second coordinate of 1 2 5. 3 * 1 * 5 - 2 is 13. New triple, 1 5 13. Two genuinely new children at every node except at the trunk. Run a few levels out. The Markov numbers fill in, 1 2 5 13 29 34 89 169 194 233. Each Markov number lives at some specific spot on the tree.

[00:04:58]We've shown the tree exists. We haven't shown it contains every solution. Take any positive integer solution to the equation. Sort the coordinates so X is at most Y is at most Z. Now via to jump the largest coordinate, replace Z with the other root. That other root has a clean form. The product of the two roots is X squared plus Y squared. The product also equals Z times the new value. So the new value is X squared plus Y squared all divided by Z. For the triple one 1 1, this gives 1 + 1 over 1, 2. The new value is bigger than Z. The descent fails. That's how we know 1 1 1 is the floor. For every other triple, X squared plus Y squared is strictly less than z squared. We'll see why in a moment. Plug that into the formula. The new value is less than z. Why is x squared plus y squared less than z squared away from 1 1 1? Look at the original equation. 3 x y z equals x squared plus y squared plus z squared. Subtract z squared from both sides. 3 x y z minus z squared equals x squared plus y squared. Factor the left.

[00:06:17]z times 3 x y minus z equals x squared plus y squared. For x squared plus y squared to be less than z squared, the left has to be less than z squared. That means 3 x y minus z is less than z or 3 x y is less than 2 z. That inequality holds for every Markov triple except 1 1 1. The proof is direct. If 3 x y were at least 2 z, the equation would force x equals y equals z equals 1. Otherwise, 2 z dominates. The point away from the root Vieta jumping the largest coordinate produces a strictly smaller largest coordinate. The smaller triple is still a Markov triple. Sort it. Jump the new largest. Get something even smaller. The largest coordinate is a positive integer that strictly decreases at every step. It can't decrease forever. Where does the descent stop? At the unique triple where Vieta jumping the largest produces something the same size or larger. That's 1 1 1. Every Markov triple flows down to 1 1 1 in infinitely many Vieta jumps. Reverse the flow. Start at 1 1 1. Build up the tree by Vieta jumping. The tree contains every solution. There are no orphans.

[00:07:40]Markov numbers, the values that appear as coordinates of some triple, form a specific sequence: 1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610.

[00:08:00]Each Markov number M is the largest coordinate of some triple. 13 is the largest of 1, 5, 13, 29 is the largest of 2, 5, 29. Frobenius asked in 1913, is M the largest coordinate of exactly one triple? Look at the small numbers. They each appear at the top of a unique triple. The pattern holds for every example anyone has ever written down.

[00:08:29]Computer searches have pushed the verified bound far. Every Markov number up to 10 to the 105th has been checked.

[00:08:36]No counterexample. Specific infinite families have been proven. Markov numbers that are prime powers, Markov numbers of the form twice a prime, Markov numbers of the form four times a prime. The conjecture holds for these, but the general statement is open. 113 years of mathematicians staring at it.

[00:08:58]Whatever the answer is, the proof must use a tool we haven't found yet. Every standard technique, induction, descent, generating functions, modular arithmetic, has been tried. None reaches the bottom.

[00:09:12]The reason this matters reaches outside number theory. Hurwitz proved a theorem in 1891. For every irrational alpha, there are infinitely many fractions P over Q such that the difference between alpha and P over Q is less than 1 over root 5 * Q squared. The constant root 5 is sharp. Replace it with anything bigger and the theorem fails. Which irrational forces the constant to be exactly root 5, the golden ratio, phi.

[00:09:44]Now, exclude phi and everything algebraically equivalent to phi, meaning every number of the form a phi plus b all over c phi plus d with ad minus bc equal to plus or minus 1. Run Hurwitz again on what's left. The constant improves from root 5 to root 8. Exclude the next worst class. Root 8 becomes the root of 221 over 25, about 3.04.

[00:10:12]Keep going. The constants form a sequence. The square root of 9 minus 4 over m squared, where m runs through the Markov numbers. m equals 1 minus golden ratio class, constant root 5. m equals 2 minus constant root 8. m equals 5 minus constant root of 221 over 25. m equals 13, the next constant in the sequence.

[00:10:39]Markov triples and badly approximable irrationals are governed by the same tree. An equation about three integers controls how badly the real line approximates itself with rationals.

[00:10:51]Two specific paths through the tree have famous names. Take the branch that always Vieta jumps the smallest coordinate. The triples come out as 1 1 2 1 2 5 1 5 13 1 13 34 1 34 89 1 89 233.

[00:11:15]The largest coordinates are 1 2 5 13 34 89 233.

[00:11:23]Every other Fibonacci number starting from the second. Every odd index Fibonacci number is a Markov number.

[00:11:30]Vieta jumping is just the Fibonacci recurrence in disguise. Take a different path. Always Vieta jump the middle coordinate. The triples become 1 1 2 1 2 5 2 5 29 5 29 433. The largest coordinates 1 2 5 29 169 985. Pell related numbers. With recurrence, each one is six times the previous minus the one before. Two completely different recurrences. Same equation, same tree, different branches.

[00:12:06]Now leave the real line entirely. Take a torus. Remove one point. Put a hyperbolic metric on what's left. The result is a once punctured torus. A surface of constant negative curvature with one hole. On this surface, there are simple closed geodesics. Loops that start and end at the same place. Never cross themselves. Follow the shortest path between every pair of their points.

[00:12:33]Each one has a hyperbolic length.

[00:12:35]Theorem. The cosh of half the length of any simple closed geodesic equals 3/2 times a Markov number. Every simple closed geodesic on the once punctured torus corresponds to a Markov triple.

[00:12:49]And the Vieta jumping that grows the tree of triples corresponds to a basic operation on the surface. A den twist that slides one curve past another.

[00:12:59]Three integers in an equation. Lengths of curves on a hyperbolic donut. The same combinatorial tree controls both.

[00:13:08]An equation Markov wrote down to study continued fractions. 1879.

[00:13:14]Three positive integers. X² + Y² + Z² = 3 times their product. Solutions live on a binary tree. Every triple descends to 1 1 1. Vieta jumping is the engine of the Markov numbers 1 2 5 13 29 34 89 169 and onward govern more than three places in mathematics rational approximations to irrationals closed geodesics on a hyperbolic surface freeze patterns of width one every odd index Fibonacci number sitting on the same branch and the basic question for banious asked in 1913 is each Markov number uniquely the maximum of one triple is still open an equation with a one-line definition a complete description of where its solutions live and a single yes or no question that has stayed open for 113 years.