open problems with monetary rewards 2014 NAU High School Math Day Dana C. Ernst Northern Arizona University October 28, 2014

open problems with monetary rewards There is a history of individuals and organizations offering monetary rewards for solutions—affirmative or negative—to difficult open problems. Their reasons could be to draw other mathematicians’ attention, to express their belief in the magnitude of the difficulty of the problem, to challenge others, etc. In the words of the Clay Mathematics Institute: “…to elevate in the consciousness of the general public the fact that in mathematics, the frontier is still open and abounds in important unsolved problems; to emphasize the importance of working towards a solution of the deepest, most difficult problems; and to recognize achievement in mathematics of historical magnitude.” 1

millennium prize problems

millennium prize problems The Millennium Prize Problems are seven problems in mathematics that were stated by the Clay Mathematics Institute in 2000. 1. Poincaré Conjecture 2. P versus NP 3. Hodge Conjecture 4. Riemann Hypothesis 5. Yang–Mills Existence and Mass Gap 6. Navier–Stokes Existence and Smoothness 7. Birch and Swinnerton–Dyer Conjecture A correct solution to any of the problems results in a $1,000,000 prize. The Poincaré Conjecture was solved by Grigori Perelman, but he declined the award in 2010. The other problems remain unsolved. 3

wolfskehl and fermat’s last theorem

wolfskehl and fermat’s last theorem Theorem (Fermat’s Last Theorem) No three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. “I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain.” – Fermat (but in Latin) Paul Friedrich Wolfskehl (1856–1906), was a physician and mathematician. In 1905, he bequeathed 100,000 marks (about $1,700,000 in 1997) to the first person to prove Fermat’s Last Theorem. On June 27, 1997, the prize was finally won by Andrew Wiles, but as a result of currency reform in Germany in 1948, the award was only worth 75,000 marks. 5

the beal conjecture

beal’s conjecture Conjecture If ax + by = cz, where a, b, c, x, y, and z are positive integers and x, y and z are all greater than 2, then a, b, and c must have a common prime factor. Texas billionaire D. Andrew Beal stated this conjecture in 1993. Beal has been trying to prove his theorem ever since, offering cash rewards in steadily increasing amounts: ∙ 1997: $5,000 ∙ 2000: $100,000 ∙ 2013: $1,000,000 7

conway’s problems

john h. conway John H. Conway (born 26 December 1937) is a British mathematician active in the theory of finite groups, knot theory, number theory, combinatorial game theory, and coding theory. Conway is currently Professor of Mathematics at Princeton University. Conway maintains a list of open problems and for each problem on the list, he is offering $1,000 to the first person that provides a correct solution. If you solve one of his problems, you can reach him by sending snail mail (only) in care of the Department of Mathematics at Princeton University. 9

sylver coinage game The Sylver Coinage Game is a game in which 2 players alternately name positive integers that are not the sum of nonnegative multiples of previously named integers. The person who names 1 is the loser! 10

sylver coinage game Sample game between A and B ∙ A opens with 5. Can’t play: 5, 10, 15, . . . ∙ B names 4. Can’t play: 4, 5, 8, 9, 10, 12→. Remaining: 1, 2, 3, 6, 7, 11. ∙ A names 11. Remaining: 1, 2, 3, 6, 7. ∙ B names 6. Remaining: 1, 2, 3, 7. ∙ A names 7. Remaining: 1, 2, 3. ∙ B names 2. Remaining: 1, 3. ∙ A names 3. Remaining: 1. ∙ B is forced to name 1 and loses. Problem If player 1 names 16, and both players play optimally thereafter, then who wins? 11

climb to a prime Let n be a positive integer. Write the prime factorization in the usual way, e.g., 60 = 22 · 3 · 5, in which the primes are written in increasing order, and exponents of 1 are omitted. Then drop exponents straight down and omit all multiplication signs, obtaining a number f(n). Now repeat. So, for example, f(60) = f(22 · 3 · 5) = 2235. Next, because 2235 = 3 · 5 · 149, it maps, under f, to 35149, and since 35149 is prime, it maps to itself. Thus 60 → 2235 → 35149 → 35149, so we have climbed to a prime, and we stop there forever. Problem Is it true that every number eventually climbs to a prime? The number 20 has not been verified to do so. Observe that 20 → 225 → 3252 → 223271 → · · · , eventually getting to more than 100 digits without yet reaching a prime! 12

the thrackle problem A doodle on a piece of paper is called a thrackle if it consists of certain distinguished points, called spots, and some differentiable (i.e., smooth) curves, called paths, ending at distinct spots and so that every pair of paths hit exactly once, where hit means having a common point at which they have distinct tangents and which is either an endpoint of both or an interior point of both. The thrackle to the left has 6 spots and 6 paths. Problem Can a thrackle have more paths than spots? 13

sources

sources http://mathoverflow.net/questions/66084/ open-problems-with-monetary-rewards http://www.claymath.org http://en.wikipedia.org/wiki/Millennium_Prize_Problems http://en.wikipedia.org/wiki/Paul_Wolfskehl http://www.ams.org/notices/199710/barner.pdf http://www.bealconjecture.com http://en.wikipedia.org/wiki/John_Horton_Conway http://www.cheswick.com/ches/conway1000.pdf http://en.wikipedia.org/wiki/Sylver_coinage http://www.thrackle.org 15