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The Clay Mathematics Institute officially designated the title Millennium Problem for the seven unsolved mathematical problems, the Birch and Swinnerton-Dyer conjecture, Hodge conjecture, Navier–Stokes existence and smoothness, P versus NP problem, Riemann hypothesis, Yang–Mills existence and mass gap, and the Poincaré conjecture at the ...
Many mathematical problems have been stated but not yet solved. These problems come from many areas of mathematics, such as theoretical physics, computer science, algebra, analysis, combinatorics, algebraic, differential, discrete and Euclidean geometries, graph theory, group theory, model theory, number theory, set theory, Ramsey theory, dynamical systems, and partial differential equations.
1983 Selberg trace formula. Hejhal's proof of a general form of the Selberg trace formula consisted of 2 volumes with a total length of 1322 pages. Arthur–Selberg trace formula. Arthur's proofs of the various versions of this cover several hundred pages spread over many papers. 2000 Almgren's regularity theorem. Almgren's proof was 955 pages ...
Goldbach’s Conjecture. 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 ...
Huge breakthroughs in math and science are usually the work of many people over many years. Seven math problems were given a $1 million bounty each in 2000, and just one has been solved so far.
A college student just solved a seemingly paradoxical math problem—and the answer came from an incredibly unlikely place. Skip to main content. 24/7 Help. For premium support please call: 800 ...
For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f 2 (4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting ...
Problem II.8 of the Arithmetica asks how a given square number is split into two other squares; in other words, for a given rational number k, find rational numbers u and v such that k 2 = u 2 + v 2. Diophantus shows how to solve this sum-of-squares problem for k = 4 (the solutions being u = 16/5 and v = 12/5 ).