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"og:description": "Citing , here are the [seven millennium problems in mathematics](http://www.claymath.org/millennium-problems), for each there is a $ 1M prize, by Clay Mathematics Institute on May 24, 2000: " ## [Yang–Mills and Mass Gap](http://www.claymath.org/millennium-problems/yang%E2%80%93mills-and-mass-gap) Experiment and computer simulations suggest the existence of a "mass gap" in the solution to the quantum versions of the Yang-Mills equations. But no proof of this property is known. ## [Riemann Hypothesis](http://www.claymath.org/millennium-problems/riemann-hypothesis) The prime number theorem determines the average distribution of the primes. The Riemann hypothesis tells us about the deviation from the average. Formulated in Riemann's 1859 paper, it asserts that all the 'non-obvious' zeros of …"
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"item:title": ".:en:Solve The 7 Millenium Problems In Mathematics.:cn:解决数学中7个千年问题"
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.:en
Citing , here are the [seven millennium problems in mathematics](http://www.claymath.org/millennium-problems), for each there is a $ 1M prize, by Clay Mathematics Institute on May 24, 2000:
"
## [Yang–Mills and Mass Gap](http://www.claymath.org/millennium-problems/yang%E2%80%93mills-and-mass-gap)
Experiment and computer simulations suggest the existence of a "mass gap" in the solution to the quantum versions of the Yang-Mills equations. But no proof of this property is known.
## [Riemann Hypothesis](http://www.claymath.org/millennium-problems/riemann-hypothesis)
The prime number theorem determines the average distribution of the primes. The Riemann hypothesis tells us about the deviation from the average. Formulated in Riemann's 1859 paper, it asserts that all the 'non-obvious' zeros of the zeta function are complex numbers with real part 1/2.
## [P vs NP Problem](http://www.claymath.org/millennium-problems/p-vs-np-problem)
If it is easy to check that a solution to a problem is correct, is it also easy to solve the problem? This is the essence of the P vs NP question. Typical of the NP problems is that of the Hamiltonian Path Problem: given N cities to visit, how can one do this without visiting a city twice? If you give me a solution, I can easily check that it is correct. But I cannot so easily find a solution.
## [Navier–Stokes Equation](http://www.claymath.org/millennium-problems/navier%E2%80%93stokes-equation)
This is the equation which governs the flow of fluids such as water and air. However, there is no proof for the most basic questions one can ask: do solutions exist, and are they unique? Why ask for a proof? Because a proof gives not only certitude, but also understanding.
## [Hodge Conjecture](http://www.claymath.org/millennium-problems/hodge-conjecture)
The answer to this conjecture determines how much of the topology of the solution set of a system of algebraic equations can be defined in terms of further algebraic equations. The Hodge conjecture is known in certain special cases, e.g., when the solution set has dimension less than four. But in dimension four it is unknown.
## [Poincaré Conjecture](http://www.claymath.org/millennium-problems/poincar%C3%A9-conjecture) <- [solved](https://en.wikipedia.org/wiki/Poincar%C3%A9_conjecture)!
In 1904 the French mathematician Henri Poincaré asked if the three dimensional sphere is characterized as the unique simply connected three manifold. This question, the Poincaré conjecture, was a special case of Thurston's geometrization conjecture. Perelman's proof tells us that every three manifold is built from a set of standard pieces, each with one of eight well-understood geometries.
## [Birch and Swinnerton-Dyer Conjecture](http://www.claymath.org/millennium-problems/birch-and-swinnerton-dyer-conjecture)
Supported by much experimental evidence, this conjecture relates the number of points on an elliptic curve mod p to the rank of the group of rational points. Elliptic curves, defined by cubic equations in two variables, are fundamental mathematical objects that arise in many areas: Wiles' proof of the Fermat Conjecture, factorization of numbers into primes, and cryptography, to name three.
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