1852, 10 year123 October, 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, so he wrote to his good friend, Sir Hamilton, a famous mathematician, for advice. Hamilton demonstrated the four-color problem after receiving Morgan's letter. But until the death of 1865 Hamilton, this 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. Many first-class mathematicians in the world have participated in the great battle of four-color conjecture. 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. Everyone thought that the four-color conjecture was solved from now on.
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. Later, more and more mathematicians racked their brains for this, but found nothing. Therefore, people began to realize that this seemingly simple topic is actually a difficult problem comparable to Fermat's conjecture: the efforts of previous mathematicians paved the way for later mathematicians to uncover the mystery of the four-color 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.
-
Fermat's last theorem is one of the three major mathematical problems in the modern world.
The New York Times, a recognized world newspaper, published a headline on June 24th, 1993.
About the news that the math problem has been solved, the news headline is "In the ancient math dilemma, someone finally called"
I found it. " The opening article of the first edition of The Times also attached a picture of long hair and wearing a medieval European robe.
Pictures of men. This ancient man was the French mathematician Pierre de Fermat.
Please refer to the appendix for the biography). Fermat is one of the most outstanding mathematicians in17th century, and he has made great achievements in many fields of mathematics.
Great contribution, because he is a professional lawyer, in order to commend his mathematical attainments, the world called him "amateur prince"
"Reputation, one day more than 360 years ago, Fermat was reading a book by the ancient Greek mathematician Diofendus.
When I was writing a math book, I suddenly wrote a seemingly simple theorem in the margin of the page.
Capacity is a problem about the positive integer solution of equation x2+y2 =z2. When n=2, it is called Pythagorean rule.
Li (also called Pythagorean Theorem in ancient China): x2+y2 =z2, where z represents the hypotenuse of a right angle, and X and Y are it.
The square of the hypotenuse of two strands, that is, a right triangle, is equal to the sum of the squares of its two strands. Of course, this equation has
Integer solutions (in fact, there are many), such as: x=3, y=4, z = 5；; x=6、y=8、z = 10； x=5、y= 12、z= 13…
Wait a minute.
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.
Fermat didn't explain the reason at that time, he just left this narrative, saying that he found the proof of this theorem wonderful.
Method, but there is not enough space on the page to write it down. The founder Fermat therefore left an eternal question, 300
Over the years, countless mathematicians have tried in vain to solve this problem. This Fermat, known as the century problem, is the most
The post-theorem has become a big worry in the field of mathematics, and it is eager to solve it quickly.
In the19th century, the Francis Institute of Mathematics in France provided a gold medal and two prizes in 18 15 and 1860.
Whoever solves this difficult problem will be given 300 francs, but unfortunately no one will get a reward. German mathematician Wolff
Skell (p？ Wolfskehl) provides100000 mark in 1908 to those who can prove the correctness of Fermat's last theorem.
The validity period is 100 year. At the same time, due to the Great Depression, this award has depreciated to 7500 marks, although
This 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 astronomical, with about 25960 digits).
Nevertheless, mathematicians have not found a universal proof. However, this unsolved mathematical problem for more than 300 years has finally been solved.
Yes, this math problem was solved by British mathematician andrew wiles. In fact, Willis is
The development of abstract mathematics in the last 30 years of the 20th century proves this point.
In 1950s, Yutaka Taniyama, a Japanese mathematician, first put forward a conjecture about elliptic curvature, which was later recorded by another mathematician.
Muragoro carried it forward. At that time, no one thought that this conjecture had anything to do with Fermat's last theorem. In the 1980s, Germany
Frey, a mathematician in China, linked Yutaka Taniyama conjecture with Fermat's Last Theorem, and what Willis did was based on this connection.
Prove that one form of Yutaka Taniyama's conjecture is correct, so is Fermat's last theorem. This conclusion
It was officially published by Willis at the seminar of Newton Institute of Mathematics, Cambridge University on June 1993, 2 1.
The report immediately shocked the whole mathematics field, and even the public outside the mathematics door wall paid infinite attention. But Willis's
The certificate was immediately found to have some defects, so it took Willis and his students another 14 months to correct it.
Correct it. 1September 1994 19 They finally handed over a complete and flawless scheme, and the nightmare of mathematics finally ended. 1997 6
In May, Willis won the Wolfskeil Prize from the University of G? ttingen. At that time,100000 FAK was about $2 million.
However, when Willis received it, it was only worth about $50,000, but Willis has gone down in history and will be immortal.
Prove Fermat's last theorem is correct
(that is, xn+yn = zn has no positive integer solution to n33)
Just prove that x4+ y4 = z4, xp+ yp = zp (P is an odd prime number) has no integer solution.
-
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 On June 7th, Goldbach wrote to tell the great Italian mathematician Euler this problem 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.
A basic problem in mathematics.
1, what is the number?
2. What are the four operations?
3. Why do addition and multiplication conform to the laws of exchange, association and distribution?
4. What are geometric figures?
Two unsolved problems.
1, found (11) 3+(1/2) 3+(1/3) 3+(1/4) 3+(65438
More generally:
Find out when k is odd.
( 1/ 1)^k+( 1/2)^k+( 1/3)^k+( 1/4)^k+( 1/5)^k+……+( 1/n)^k=？
Background:
Euler found:
( 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
background
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.
Prove:
ζ(s)= 1+( 1/2)^s+( 1/3)^s+( 1/4)^s+( 1/5)^s+…
(s belongs to complex number field)
The zeros of the definition function zeta (s) all have the real part of 1/2 except negative integer real numbers.
Background:
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?
Background:
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?
Background:
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?
Background:
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).
BACKGROUND: 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.
Background: Godel proved the incompleteness of the arithmetic system, which dashed Hilbert's idea of proving that the arithmetic axiom system is not contradictory with meta-mathematics.
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.
Background: German and French mathematicians made great progress in 1960s.
7. The generalization of Kroneck's theorem on the problem 12 Abelian field on arbitrary algebraic rational field.
Background: This problem has only some scattered results and 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.
Background: 1957 Soviet mathematicians solved the case of continuous function. If you need to parse the function, this problem has not been completely solved.
9. Question 15 The strict foundation of Schubert counting calculus.
Background: 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 number
Scientists believe that in addition to solving the mystery of prime number distribution, it is of great significance to analytic number theory, function theory,
Elliptic function theory, group theory and prime number test will all have substantial influence.
2. Young-Mills Theory and Mass Gap
Hypothetically)
In 1954, Yang-Mills gauge theory was put forward by and Mills.
At the beginning of mathematics, a normative theoretical framework was put forward, and later it gradually developed into quantum.
The important theory of physics also makes him an important figure in the foundation of modern physics.
In the theory put forward by Yang Zhenning and Mills, particles that transfer force will be produced. They
The difficulty is the mass of this particle. The result they got through mathematical deduction.
Yes, this particle has a charge, but no mass. However, the difficulty is that if this is charged.
Our particles have no mass, so why is there no experimental evidence? If we assume
If the particle has mass, the gauge symmetry will be destroyed. General physicists believe in quality.
Quantity, 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. already
Given that the size is n, if it can be determined that the calculation time is below cnd (c and D are positive real numbers) time.
When we are sure, we call it "polynomial time determination method". You can use this.
The problem solved by the algorithm is the P problem. On the other hand, if other factors are involved, such as the sixth sense.
The algorithm is called "non-deterministic algorithm", and this kind of problem is "NP problem", NP is
Abbreviation for uncertain polynomial time.
By definition, P problem is a part of NP problem. But is there a NP problem?
What about things that don't belong to the P-level? 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, it is produced in the process of seeking correction.
New results. French engineer Naville and British mathematician Stoke experienced rigorous mathematics.
Considering the viscous term, the Cashierville-Stoke equation is derived.
Starting from 1943, French mathematician Le Rey proved Naville Stowe.
After the global weak solution of Dirk equation, people always want to know.
Is this solution unique? The result is that if we presuppose the Naville-Stoke party,
If the solution of a process is 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? Then there is the certificate.
The solution will blow up in a limited time.
Solving this problem is not only helpful to mathematics, but also to physics and aerospace engineering, especially chaos.
Flow (turbulence) will have a decisive influence, as well as Naville-Stoke equation and Austria.
The great physicist Pozmann's Pozmann equation is also closely related to Naville's research.
Er-Stoke equation and Boltzmann equation.
The knowledge of the relationship between people is called the limit of fluid mechanics, which is acceptable.
Weil-Stoke equation itself has very rich connotations.
5. Poincare conjecture (Poincare conjecture)
Poincare conjecture is a difficult problem in topology. In mathematical terms: simple connection
Three-dimensional closed manifold and three-dimensional spherical homeomorphism.
In the mathematical sense, this is a seemingly simple problem, but it is not.
This frequently encountered problem was put forward by Poincare in 1904.
Later, it attracted many excellent mathematicians to devote themselves to this research topic.
Soon after Poincare's conjecture (Figure 4) was put forward, mathematicians would naturally
We extend it to high-dimensional space (n4) and call it generalized Poincare conjecture: simply connected.
≥
N(n4)-dimensional closed manifold, if N is used.
A sphere with dimension ≥ has the same basic group, so it must be homeomorphic to an N-dimensional sphere.
After nearly 60 years, in 196 1 year, the American mathematician Smale took
Clever method, he ignored the difficulties of three-dimensional and four-dimensional, and directly proved that five-dimensional (n5) is above.
≥
Generalized Poincare conjecture, so he won the Fields Prize in 1966. After 20 years.
Later, another American mathematician Friedman proved the four-dimensional Poincare hypothesis.
1986, he won the Fields Medal for this achievement. But for us, really.
The three-dimensional space where people live (n3) was still an unsolved mystery at that time.
=
Until April 2003, Russian mathematician perelman was at
MIT made three speeches, in which he answered many mathematicians' questions.
There are many indications that Feldman may have cracked the Poincare conjecture. A few days later, the new york Times appeared for the first time.
The second time, the news was disclosed to the public with the title "Russians solved famous mathematical problems". the same
The headline article published by MathWorld, an influential mathematics website, is Poincare conjecture.
Facts have proved that this time it is true! 」[ 14]。
The examination for mathematicians will not be completed until 2005, and so far, it has not been found.
Feldman couldn't collect the million-dollar loophole in the Clay Institute of Mathematics.
6. White and Swinerton-Dale conjecture (Burch and Swinerton-Dale)
Guess)
General elliptic curve equation y 2 = x 3+ax+b, when calculating the arc length of an ellipse.
You will encounter this curve. Since 1950s, mathematicians have discovered elliptic curves and number theory,
Geometry and cryptography are closely related. For example, wiles proved Fermat.
In the last theorem, the key step is to use the relationship between elliptic curve and module-that is, the Taniyama-Zhicun conjecture, which is related to the white matter-swinton-Dell conjecture.
Elliptic curve.
In the 1960 s, Bai Zhi and swinton Dale of the University of Cambridge, England, calculated some by computer.
Understanding of polynomial equation. There are usually infinite solutions, but how to calculate infinity?
And then what? Its solution is to classify first, and the typical mathematical method is the concept of congruence.
From this, we can get the congruence class, that is, the remainder after dividing by a number, which is infinite.
Multiple numbers cannot be all. Mathematicians naturally choose prime numbers, so this problem has nothing to do with
Zeta function of riemann conjecture. After a long period of calculation and data collection, he
Scientists have observed some laws and patterns, so they put forward this conjecture. They worked out the knot by computer.
Decisively speaking, an elliptic curve will have infinite rational points if and only if it is attached to the curve.
When zeta function zeta (s) =, the value is 0, that is zeta (1).
; When s 1= 0
7. Hodge conjecture
Any harmonic differential form on a nonsingular projective algebraic curve is an algebraic circle.
Rational combination of cohomology classes. 」
The last question, although not the most difficult of the seven major problems in the Millennium, may be
Energy is the most difficult thing for ordinary people to understand. Because there are too many profound, professional and abstract.
References:
100 basic problems of mathematics, mathematics and culture, review of 23 mathematical problems of Hilbert.