The four-color conjecture was put forward by Britain. 1852, when Francis guthrie, who graduated from London University, came to a scientific research institute to do map coloring, he found an interesting phenomenon: "It seems that every map can be colored with four colors, which makes countries with the same border painted with different colors." Can this conclusion be strictly proved by mathematical methods? There has been no progress in his research with his younger brother Grace, who is in college.
1852 10 His younger brother asked his teacher, the famous mathematician De Morgan, for proof of this problem. Morgan couldn't find a solution to this problem either, so he wrote to his good friend, Sir Hamilton, a famous mathematician, for advice. Until Hamilton died in 1865, the problem was not solved.
1872, Kelly, the most famous mathematician in Britain at that time, formally put forward this question to the London Mathematical Society, so the four-color conjecture became a concern of the world mathematical community. During the two years from 1878 to 1880, Kemp and Taylor, two famous lawyers and mathematicians, respectively submitted papers to prove the four-color conjecture and announced that they had proved the four-color theorem.
1 1 years later, that is, 1890, the mathematician Hurwood pointed out that Kemp's proof and his accurate calculation were wrong. Soon, Taylor's proof was also denied. As a result, people began to realize that this seemingly simple topic is actually a difficult problem comparable to Fermat's conjecture.
Since the 20th century, scientists have basically proved the four-color conjecture according to Kemp's idea. 19 13 years, boekhoff introduced some new skills on the basis of Kemp, and American mathematician Franklin proved in 1939 that maps in 22 countries can be colored in four colors. 1950 someone has been promoted from 22 countries to 35 countries. 1960 proves that maps below 39 countries can be colored with only four colors; And then push it to 50 countries. It seems that this progress is still very slow. After the emergence of electronic computers, the process of proving the four-color conjecture has been greatly accelerated due to the rapid improvement of calculation speed and the emergence of man-machine dialogue. 1976, American mathematicians Appel and Harken spent 1200 hours on two different computers at the University of Illinois in the United States, made 1000 billion judgments, and finally completed the proof of the four-color theorem. The computer proof of the four-color conjecture has caused a sensation in the world. It not only solved a problem that lasted for more than 100 years, but also may become the starting point of a series of new ideas in the history of mathematics. However, many mathematicians are not satisfied with the achievements made by computers, and they are still looking for a simple and clear written proof method.
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Fermat's last theorem is one of the three major mathematical problems in the modern world.
Fermat is one of the most outstanding mathematicians in17th century. He has made great contributions in many fields of mathematics. Our bank is a professional lawyer. In recognition of his mathematical attainments, the world called him "amateur prince". One day more than 360 years ago, Fermat was reading a math book by Diofendos, an ancient Greek mathematician, when he suddenly had a whim and wrote a seemingly simple theorem. The content of this theorem is about the positive integer solution of an equation x2+y2 =z2. When n=2, it is the well-known Pythagorean Theorem (also called Pythagorean Theorem in ancient China): x2+y2 =z2, where z represents the hypotenuse of a right triangle, and x and y are its two branches, that is, the square of the hypotenuse of a right triangle is equal to the sum of the squares of its two branches. Of course, this equation has an integer solution.
Fermat claims that when n>2, there is no integer solution satisfying xn +yn = zn, such as the equation x3 +y3=z3.
Find an integer solution. At that time, Fermat did not explain why. He just left this narrative, saying that he found a wonderful way to prove this theorem, but there was not enough space on the page to write it down. Fermat, the initiator, thus left an eternal problem. For more than 300 years, countless mathematicians have tried to solve this problem in vain. This Fermat's last theorem, known as the century's difficult problem, has become a big worry in mathematics and is extremely eager to solve it.
19th century, Francis Institute of Mathematics in France provided a gold medal and 300 francs to anyone who solved this problem twice in 18 15 and 1860. Unfortunately, no one can receive the prize. German mathematician Wolfskeil (p? Wolfskehl) provides 100000 marks in 1908 to those who can prove the correctness of Fermat's last theorem, and the validity period is100 years. In the meantime, due to the Great Depression, the bonus amount has been devalued to 7500 marks, but it still attracts many "math idiots". After the development of computers in the 20th century, many mathematicians can prove that this theorem holds when n is large. 1983, the computer expert Sloansky ran the computer for 5782 seconds, which proved that Fermat's last theorem was correct when n was 286243- 1 (Note 286243- 1 is an astronomical figure with about 25960 digits).
Nevertheless, mathematicians have not found a universal proof. However, this 300-year-old math unsolved case has finally been solved. Andrew wiles, an English mathematician, solved this mathematical problem. In fact, Willis used the achievements of the development of abstract mathematics in the last 30 years of the twentieth century to prove it. In 1950s, Yutaka Taniyama, a Japanese mathematician, first put forward a conjecture about elliptic curvature, which was later developed by another mathematician, Goro Shimamura. At that time, no one thought that this conjecture had anything to do with Fermat's last theorem. In 1980s, German mathematician Frey linked Yutai Taniyama's conjecture with Fermat's last theorem. What Willis did was to prove that one form of Yutai Taniyama's conjecture was correct according to this connection, and then deduced Fermat's last theorem. This conclusion was officially published by Willis at the seminar of Newton Institute of Mathematics, Cambridge University, USA in June 1993. This report immediately shocked the whole mathematics field, and even the public outside mathematics gave infinite attention. However, Willis' proof was immediately found to have some defects, so it took Willis and his students 14 months to correct it. 1September, 994, they finally handed over a complete and flawless plan, and the nightmare of mathematics finally ended. 1In June, 1997, Willis won the Wolfskeil Prize at the University of G? ttingen. At that time,1100,000 grams was about $2 million, and when Willis received it, it was only worth about $50,000, but Willis has been recorded in the history books and will be immortal.
To prove that Fermat's last theorem is correct (that is, xn+yn = zn has no integer solution to n33), it is only necessary to prove that x4+ y4 = z4, xp+ yp = zp (P is an odd prime number) has no integer solution.
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Goldbach conjecture, one of the three major mathematical problems in the modern world.
Goldbach is a German middle school teacher and a famous mathematician. He was born in 1690, and was elected as an academician of Russian Academy of Sciences in 1725. 1742, Goldbach found in teaching that every even number not less than 6 is the sum of two prime numbers (numbers that can only be divisible by themselves). For example, 6 = 3+3, 12 = 5+7 and so on. 1742 In June, Goldbach wrote to tell the great Italian mathematician Euler this question and asked him to help prove it. In his reply to him on June 30th, Euler said that he thought this conjecture was correct, but he could not prove it. Describing such a simple problem, even a top mathematician like Euler can't prove it. This conjecture has attracted the attention of many mathematicians. They began to check even numbers until they reached 330 million, which showed that the guess was correct. But for a larger number, the guess should be correct, but it can't be proved. Euler died without proof. Since then, this famous mathematical problem has attracted the attention of thousands of mathematicians all over the world. 200 years have passed and no one has proved it. Goldbach conjecture has therefore become an unattainable "pearl" in the crown of mathematics. It was not until the 1920s that people began to approach it. 1920, the Norwegian mathematician Bujue proved by an ancient screening method, and reached the conclusion that every even number with larger ratio can be expressed as (99). This method of narrowing the encirclement is very effective, so scientists gradually reduced the number of prime factors in each number from (99) until each number is a prime number, thus proving "Goldbach". 1924, mathematician Rad mahar proved (7+7); 1932, mathematician eissmann proved (6+6); In 1938, mathematician Buchstaber proved (55), and in 1940, he proved (4+4). 1956, mathematician vinogradov proved (3+3); In 1958, China mathematician Wang Yuan proved (23). Subsequently, Chen Jingrun, a young mathematician in China, also devoted himself to the study of Goldbach's conjecture. After 10 years of hard research, we finally made a major breakthrough on the basis of previous studies and took the lead in proving it (L 12). At this point, Goldbach conjecture is only the last step (1+ 1). Chen Jingrun's paper was published in 1973 Science Bulletin of China Academy of SciencesNo. 17. This achievement has attracted the attention of the international mathematics community, which has made China's number theory research leap to the leading position in the world. Chen Jingrun's related theory is called "Chen Theorem". 1in late March, 996, when Chen Jingrun was about to take off the jewel in the crown of mathematics, "when he was only a few feet away from the brilliant peak of Goldbach's conjecture (1+ 1), he fell down exhausted ..." Behind him, more people would climb this peak.
Several unsolved problems.
1, find (11) 3+(1/2) 3+(1/3) 3+(1/4) 3+(65438 is more general.
When k is odd, find (11) k+(1/2) k+(1/3) k+(1/4) k+.
Euler worked out:
( 1/ 1)^2+( 1/2)^2+( 1/3)^2+( 1/4)^2+( 1/5)^2+…+( 1/n)^2=(π^2)/6
When k is an even number.
2. Transcendence of E+π
This is a special case of Hilbert's seventh question.
The transcendence of e π has been proved, but so far no one has proved the transcendence of e+π.
3. Prime number problem.
Proof: ζ (s) =1+(1/2) s+(1/3) s+(1/4) s+(1/5) s+
The zeros of the definition function zeta (s) all have the real part of 1/2 except negative integer real numbers. This is the Riemann conjecture. This is Hilbert's eighth question. American mathematicians have calculated the first 3 million zeros of ζ(s) function by computer, which really accords with the conjecture. Hilbert thinks that the solution of Riemann conjecture can make us strictly solve Goldbach conjecture (any even number can be decomposed into the sum of two prime numbers) and twin prime number conjecture (there are infinitely many prime numbers with a difference of 2).
The extended question is: the expression formula of prime number? What is the essence of prime numbers?
4. Is there an odd perfect number?
A perfect number is equal to the sum of its factors.
The first three perfect numbers are:
6= 1+2+3
28= 1+2+4+7+ 14
496= 1+2+4+8+ 16+3 1+62+ 124+248
The known 32 perfect numbers are all even numbers.
1973 concludes that if n is an odd perfect number, then:
n & gt 10^50
5. Except 8 = 2 3 and 9 = 3 2, are there no two consecutive integers that can be expressed as powers of other positive integers?
This is Catalonia's guess (1842). 1962, China mathematician Ke Zhao independently proved that there is no continuous integer that can be expressed as the power of other positive integers. 1976, Dutch mathematicians proved that any two positive integer powers greater than a certain number are discontinuous. So just check whether any positive integer power less than this number is continuous. But because this number is too large, there are more than 500 bits, which is beyond the calculation range of the computer. So this conjecture is almost correct, but no one can confirm it so far.
6. Given a positive integer n, if n is even, change it to n/2, and if it becomes odd after division, multiply it by 3 plus 1 (that is, 3n+ 1). Repeat this operation, and after a limited number of steps, will you definitely get 1?
This old conjecture (1930). People have never found a counterexample through a lot of checking computations, but no one can prove it.
Three unsolved problems in Hilbert's 23 problem.
1, problem 1 continuum hypothesis. There is no other cardinality between the cardinality of all positive integers (called countable set) and the cardinality of real number set (called continuous set).
1938 Austrian mathematician Godel proved that this hypothesis is not falsifiable in the axiomatic system of set theory, that is, the axiomatic system of Zeromolo-Fo Runkle. 1963, American mathematician Cohen proved that this assumption cannot be proved to be correct in this axiomatic system. So far, no one knows whether this assumption is right or wrong.
2. Question 2 Compatibility of arithmetic axioms.
Godel proved that the arithmetic system is incomplete, which shattered Hilbert's idea of using meta-mathematics to prove that the arithmetic axiom system is not contradictory.
3. Question 7: Unreasonability and transcendence of some figures. See 2 above.
5, problem 8 prime number problem. See article 2 ter above.
6. The coefficient of the problem 1 1 is the quadratic form of any algebraic number.
Mathematicians in Germany and France made great progress in the 1960s.
7. The generalization of Kroneck's theorem on the problem 12 Abelian field on arbitrary algebraic rational field.
This problem has only some scattered results, and it is far from being completely solved.
8. Question 13: It is impossible to solve the general algebraic equation of degree 7 only with binary functions.
1957 Soviet mathematicians solved the case of continuous functions. If you need to parse the function, this problem has not been completely solved.
9. Question 15 The strict foundation of Schubert counting calculus.
The number of intersections in algebra. Related to algebraic geometry.
10, Problem 16 Topology of algebraic curves and surfaces.
Algebraic curves need to contain the maximum number of closed branch curves. And the maximum number and relative position of limit cycles of differential equations.
1 1, problem 18 constructing space with congruent polyhedron.
The problem of the closest arrangement of infinite polyhedron in a given form has not been solved yet.
12, the 20th general boundary value problem.
Boundary value problems of partial differential equations are developing vigorously.
13, further development of the variational method in question 23.
Four thousand and seven difficult problems
In 2000, the American Clay Association for the Advancement of Mathematics proposed. In memory of 23 questions raised by Hilbert a hundred years ago. The reward for each question is millions of dollars.
1, Riemann conjecture. See page 3 of 2.
Through this conjecture, mathematicians believe that the mystery of prime number distribution can be solved. This problem is one of Hilbert's 23 unsolved problems. By studying Riemann conjecture, mathematicians believe that it will not only solve the mystery of prime number distribution, but also have a substantial impact on analytic number theory, function theory, elliptic function theory, group theory, prime number test and so on.
2. Young-Mills theory and mass gap hypothesis.
In 1954, Yang-Mills gauge theory was put forward with Mills. Yang Zhenning started with mathematics and put forward a normative theoretical framework, which gradually developed into an important theory of quantum physics, making him an important figure in laying the foundation for modern physics. In the theory put forward by Yang Zhenning and Mills, there will be particles that transmit force, and the difficulty they encounter is the quality of this particle. As a result of their mathematical deduction, this particle has a charge and no mass. However, the difficulty is that if this charged particle has no mass, why is there no experimental evidence? If the particle is assumed to have mass, the gauge symmetry will be destroyed. Most physicists believe in quality, so how to fill this loophole is a very challenging mathematical problem.
3.P versus NP problem.
With the increase of calculation scale, the kind of problem that the calculation time will increase in a polynomial way is called "P problem". P of the p problem is the first letter of polynomial time. Given a size of n, if it can be determined that the calculation time is below cnd (C and D are positive real numbers), we call it "polynomial time determination method". The problem that this algorithm can solve is the P problem. On the other hand, if there are other factors, such as the sixth sense, the algorithm is called "non-deterministic algorithm" and this kind of problem is "NP problem". NP is the abbreviation of non-deterministic polynomial time. By definition, P problem is a part of NP problem. But is there anything in NP that doesn't belong to P? Or will the NP problem eventually become a P problem? This is the quite famous PNP problem.
4. Naville-Stokes equation.
Because Euler's equation is too simplified, we seek to modify it, and new results are produced in the process of modification. Naville, a French engineer, and Stoke, a British mathematician, obtained the Naville-Stoke equation after strict mathematical deduction and considering the viscous term. Ever since the French mathematician Leray proved the global weak solution of the Naville-Stoke equation in 1943, people have been wondering whether this solution is unique. The result is that if the solution of Naville-Stoke equation is assumed to be a strong solution, then this solution is unique. So the question becomes: how big is the gap between the weak solution and the strong solution, and is it possible for the weak solution to be equal to the strong solution? In other words, can we get the full-time smooth solution of Naville-Stoke equation? It is further proved that the solution will blow up in a finite time. Solving this problem not only contributes to mathematics, but also contributes to physics and aerospace engineering, especially turbulence. In addition, the Naville-Stoke equation is closely related to the Pozmann equation of the great Austrian physicist Pozmann. The knowledge of studying the relationship between Naville-Stokes equation and Pozmann equation is called the limit of fluid mechanics, which explains the Naville-Stokes equation itself.
5. Poincare conjecture (Poincare conjecture)
Poincare conjecture is a difficult problem in topology. In mathematical jargon, a simply connected three-dimensional closed manifold and a three-dimensional spherical homeomorphism. Mathematically, this is a seemingly simple but very difficult problem. Since Poincare proposed it in 1904, this research topic has attracted many excellent mathematicians. Shortly after the Poincare conjecture (Figure 4) was put forward, mathematicians naturally extended it to high-dimensional space (n4), which we call generalized Poincare conjecture: a simply connected closed manifold with dimension ≥n(n4) must be homeomorphic to the N-dimensional sphere if it has the same basic group as the N-dimensional sphere. After nearly 60 years, in 196 1 year, American mathematician Sumaru directly proved the generalized Poincare conjecture (n5) with more than five dimensions through ingenious methods, so he won the Fields Prize of 1966. Twenty years later, another American mathematician Friedman proved the four-dimensional Poincare conjecture and won the Fields Prize for this achievement with 1986. But the three-dimensional space (n3) where we really live was still an unsolved mystery at that time. Until April 2003, Russian mathematician perelman gave three lectures at MIT and answered many mathematicians' questions. There are many signs that perelman may have cracked the Poincare conjecture. A few days later, The New York Times announced the news to the public for the first time on the topic of "Russians solve famous math problems". On the same day, the headline article published by the influential mathematics website MathWorld was "Poincare conjecture proved, this time it is true! 」[ 14]。 The review of mathematicians will not be completed until 2005. So far, the loophole that Filleman can't receive millions of dollars from Clay Institute of Mathematics has not been found.
6. White matter and swinton-Dale conjecture will encounter the general elliptic curve equation y 2 = x 3+ax+b when calculating the arc length of an ellipse. Since 1950s, mathematicians have found that elliptic curves are closely related to number theory, geometry and cryptography. For example, in wiles's proof of Fermat's Last Theorem, one of the key steps is to make use of the relationship between elliptic curves and modules, that is, the Taniyama-Zhicun conjecture, and both Shirakawa's and swinton-Dale's conjectures are related to elliptic curves.
In 1960s, Bai Zhi and Swinerton Dale of Cambridge University in England calculated some polynomial equations with computers. There are usually infinite solutions, but how to calculate infinity? The solution is to classify first, and the typical mathematical method is the concept of congruence, from which it is concluded that the congruence class, that is, the remainder after dividing a number, cannot exist in every infinite number. Mathematicians naturally choose prime numbers, so this problem is related to the Zeta function of Riemann conjecture. After a long period of calculation and data collection, they observed some laws and patterns, and thus put forward this conjecture. They assert from the results of computer calculation that the elliptic curve will have infinite rational points if and only if the zeta function zeta (s) = attached to the curve takes the value of 0, that is zeta (1); When s 1= 0
7. Hodge conjecture
"Any harmonic differential form on a nonsingular projective algebraic curve is a rational combination of homology classes on an algebraic circle. This last question, though not the most difficult of the seven questions in the Millennium, may be the most difficult for ordinary people to understand. Because there are too many profound professional and abstract reference materials:
100 basic problems of mathematics, mathematics and culture, review of 23 mathematical problems of Hilbert.