No, this is the basic theory of Fermat's law.
Above cubic, there can be no integer solution.
Incremental solutions of integer solutions of Fermat equation x n+y n = z n
Zhuang Hongfei
(Liaoyang Railway Equipment Factory 1 1 1000)
The proof of the integer solution relationship of Fermat equation x n+y n = z n has been controversial in mathematics for many years. In this paper, the existence conditions of integer solutions of the side length of right triangle A 2+B 2 = C 2 are comprehensively analyzed by using the plane geometry method, and the evaluation of multivariate algebra plus elements is put forward. The "fixed a calculation rule" of integer solution of right triangle with side length A 2+B 2 = C 2 is given "Calculation rules of increasing ratio"; "Rules of fixed difference formula"; "A-value parity sequence rule"; It is the algebraic condition and practical method of square integer solution; This paper puts forward the concepts of absolute power and absolute non-power of unary algebra. In this paper, the problem of judging the integer solution of Fermat equation x n+y n = z n, the original ternary indefinite equation, is skillfully transformed into a univariate definite solution equation by using the property of increasing ratio of the same power and the property of difference formula of increasing term of integer power.
Keywords: incremental difference formula for solving absolute power absolute non-power adjacent integer power
Introduction: 162 1 year, the French mathematician Fermat proposed that the equation x n+y n = z n has infinite integer solutions when n=2, and when n >. And claimed that he made a wonderful proof at that time. This is what later generations called Fermat's Last Theorem. Today, the answer to this question is still complicated and lengthy, and the controversy is constant, which makes people unable to agree.
In this paper, a new intuitive and concise theory and practical method of square integer solution of Fermat equation is established by using the relationship between the side length and area of right triangle and square. In this paper, the algebraic method is used to analyze and prove the integer solution relationship of Fermat equation x n+y n = z n when exponent n > 2.
Definition 1. Fermat equation
People used to call the relationship of x n+y n = z n Fermat equation. Its deep meaning is that after the index n is set, its x, y and z are all integers.
In the side length of a right triangle, it is often obtained that A, B and C are all integer relationships, such as right triangles 3, 4 and 5. At this time, we can get 3 2+4 2 = 5 2 from Pythagorean theorem, so when the power is 2, Fermat equation and Pythagorean theorem are of the same order. When the exponent is greater than 2, the study of integer solutions of Fermat equation, from Euler to Dirichlet, has become a major branch of mathematics.
Definition 2. Incremental solution method
In the evaluation and calculation of multivariate algebra, unknown terms other than the original terms are introduced to form an equality relationship and participate in the evaluation and calculation. We call the method of finding unknown elements by multivariate algebraic expressions the method of adding elements.
Finding the value of multivariate algebraic expressions by adding elements can sometimes make very complicated problems extremely simple.
Next, we use the method of element addition to evaluate the integer solution relationship of three sides of a right triangle A 2+B 2 = C 2.
First, the integer solution of the side length of right triangle A 2+B 2 = C 2 "Fixed a calculation rule"
Theorem 1. If a, b and c are three sides of a right triangle, q is an incremental term, and Q≥ 1, then the condition is satisfied:
a≥3
b=(a^2-Q^2)÷2Q
c= Q+b
Then at this time, A 2+B 2 = C 2 is the integer solution;
It is proved that in the square area relation, the area is A 2 from the side length of A. If (A 2-Q 2) ÷ 2q = b (where Q is an incremental term and B and Q are integers), the area A 2 can be decomposed into A 2 = Q 2+QB+QB, and the decomposition relation can be as follows.
Q2 Qb
Its notch is just a square with side length b, and a side length can be obtained by supplementing the notch area b 2
quarterback
For a square with Q+b, let's take Q+B = C, and according to the pythagorean theorem of the relationship between the side lengths of a right triangle, we can know that a, b and c at this time are three integer side lengths of a right triangle.
So the theorem 1 is proved.
Application example:
Example 1. Find the square integer solution of the side length of a right triangle when the side length is 15 by using the calculation rule of determining a?
Solution: Take an application example: A is 15, and the optional incremental term Q is 1. According to the calculation rules of a, we get:
a= 15
{ b=(a^- q^2)÷2q=( 15^2- 1^2)÷2 = 1 12
c = Q+b = 1+ 1 12 = 1 13
So the square integer solution is152+1122 =1132.
Then take a as 15, and choose the incremental term q as 3. According to the calculation rules of a, we can get:
a= 15
b=(a^2-q^2)÷2q=( 15^2-3^2)÷6=36
c=Q+b=3+36=39
So the square integer solution is15 2+36 2 = 39 2.
When a=3, 4, 5, 6, 7, etc. Function coverage will be solved by different values of q.
Second, the integer solution of the side length of right triangle A 2+B 2 = C 2 "the law of increasing ratio calculation"
Theorem 2. If A 2+B 2 = C 2 is the set of integer solutions of the side length of a right triangle, then (an) 2+(BN) 2 = (CN) 2 (where n= 1, 2, 3 ...) are all integer solutions.
It is proved that according to Pythagorean theorem, if A 2+B 2 = C 2 is an integer solution, a right triangle A C with integer sides will be obtained. According to the principle of equal ratio amplification of plane line segments, the triangle is enlarged to get 2a2c;
b 2b
3a, 3c, 4a, 4c ... The condition that a, b and c are integers, 2a, 2b, 2c;
3b 4b
3a、3b、3c； 4a, 4b, 4c… na, nb and nc are all integers.
So Theorem 2 was proved.
Application example:
Example 2. Prove that 303 2+404 2 = 505 2 is an integer solution?
Solutions; 3 2+4 2 = 5 2 obtained from right triangle 3 5 is an integer solution. According to the ratio,
four
For the algorithm, when the relationship of right triangle 3×1015×101is the side length, there must be.
4× 10 1
303 2+404 2 = 505 2 is an integer solution.
Third, the "rule of definite difference formula" for the integer solution of the side length of right triangle A 2+B 2 = C 2
3a + 2c + n = a 1
(where n = the difference between B and A, n= 1, 2,3 ...)
Theorem 3. If the right triangle a2+B2 = C 2 is an integer solution satisfying the relationship of b-a=n, then the square matrix AI 2+Bi 2 = CI 2 ... AI formed by the above formulas 3a+2c+ n = a 1, a2 and A3 are all obtained.
It is proved that if n is 1, 3 2+4 2 = 5 2 can be obtained from three sides 3, 4 and 5 of a right triangle, where n=b-a=4-3= 1. According to 3a+2c+ 1= a 1, the difference formula has the following rules.
At this time, A 1=3×3+2×5+ 1=20 is obtained.
20 2+212 = 29 2 Continue to calculate with the formula:
The a2 obtained at this time is 3× 20+2× 29+1=119.
1192+1202 =1692 continues to be calculated by the formula.
a3 = 3× 1 19+2× 169+ 1 = 696。
696^2+697^2=985^2
…
So the fixed difference is 1.
Now let's take n as 7, and we have a right triangle of 212+28 2 = 35 2, where n=28-2 1=7. According to 3a+2c+7 = a 1, the fixed difference formula is as follows:
At this time, a1= 3× 21+2× 35+7 =140 is obtained.
140 2+ 147 2 = 203 2 continue to use the formula to calculate:
A2=3× 140+2×203+7=833。
833 2+840 2 = 1 183 2 Continue to use the formula to calculate:
A3 = 3× 833+2×1183+7 = 4872, which is obtained at this time.
4872^2+4879^2=6895^2
…
Therefore, the fixed difference is 7.
Let n be 129, and we have a right triangle 387 2+5 16 2 = 645 2, where n=5 16-387= 129, according to 3a+2c+129 =.
At this time, a1= 3× 387+2× 645+129 = 2580 is obtained.
2580 2+2709 2 = 374 1 2 Continue to calculate with the formula:
A2 = 3× 2580+2× 3741+129 =15351,which is obtained at this time.
153512+15480 2 = 218012 continue to use the formula to calculate:
a3 = 3× 1535 1+2×2 180 1+ 129 = 89784。
89784^2+899 13^2= 127065^2
…
Therefore, the fixed difference is 129.
Thus, the calculation rules of fixed difference n are established.
So Theorem 3 was proved.
Fourth, the square integer solution A 2+B2 = C 2 A value parity sequence rule:
Theorem 4. If A 2+B2 = C 2 is three integer sides of a right triangle, then the odd-even series relationship of one of the following values must be established;
(a) odd column a:
If Table A is an odd number of 2n+ 1 (n= 1, 2, 3 ...), then the relationship between square integer solutions of odd columns is:
a=2n+ 1
c=n^2+(n+ 1)^2
b=c- 1
Syndrome: When the conditions of this formula are n= 1, 2, 3 ... respectively:
3^2+4^2=5^2
5^2+ 12^2= 13^2
7^2+24^2=25^2
9^2+40^2=4 1^2
1 1^2+60^2=6 1^2
13^2+84^2=85^2
…
Therefore, the odd column A relation holds.
(2) Even series A:
If Table A is an even number of 2n+2 type (n= 1, 2, 3 ...), then the relationship between square integer solutions of a series of even numbers is:
a=2n+2
c= 1+(n+ 1)^2
b=c-2
Syndrome: When the conditions of this formula are n= 1, 2, 3 ... respectively:
4^2+3^2=5^2
6^2+8^2= 10^2
8^2+ 15^2= 17^2
10^2+24^2=26^2
12^2+35^2=37^2
14^2+48^2=50^2
…
Therefore, the relation A of even sequence holds.
Therefore, the relation of Theorem 4 holds.
Therefore, among the three sides of right triangle A, B and C:
The difference between b and a can be 1, 2, 3…
The difference between a and b can be 1, 2, 3…
The difference between c and a can be 1, 2, 3…
The difference between c and b can be 1, 2, 3…
There are infinite fixed difference square integer solutions;
Every fixed difference square integer has an infinite number of solutions.
Above, we give the algebraic conditions and practical methods of square integer solutions. We can also prove that the Fermat equation x n+y n = z n has no integer solution when the exponent n > 2. Proved as follows:
First of all, we prove that the law of increasing ratio holds in any power.
Theorem 5: If A, B and C are all different integers greater than 0, and M is an integer greater than 1, if the same power relation of A M+B M = C M+D M+E M holds, the same power relation still holds after the ratio of A, B, C, D and E increases.
It is proved that in the original formula of the theorem, a m+b m = c m+d m+e m, and the increasing ratio is n, n > 1,
Get: (n a) m+(nb) m = (NC) m+(nd) m+(ne) m.
The original formula is: n m (a m+b m) = n m (c m+d m+e m)
The original formula is obtained by eliminating n m on both sides.
Therefore, there is a calculation rule of increasing ratio between the same power and the difference formula, and it is still the same power after increasing the ratio.
So Theorem 5 was proved.
Theorem 6, if A, B and C are different integers and the relationship A M+B = C M holds, where B > 1 and B is not the same power of A and C, when A, B and C increase year-on-year, B is still not the same power of A and C. ..
Proof: take the original formula of theorem A M+B = C M
Taking the increase rate as n, n > 1, we get: (na) m+n MB = (NC) m.
The original formula is: n m (a m+b) = n MC m.
The original formula is obtained by eliminating n m on both sides.
Since b cannot be transformed into the same power of a and c, n^mb cannot be transformed into the same power of a and C.
Therefore, after the same increase rate of * * * contains several terms that are not the same power between the same power and the difference formula, the equation relationship still holds. Among them, the same power term is still the same power after increasing the proportion, and the same power term is still the same power after increasing the proportion.
So Theorem 6 was proved.
Absolute power and absolute non-power properties of unary algebra
Definition 3, an absolute power.
In an unknown algebraic expression, if the unknown value is an integer greater than 0, the value of the algebraic expression is a complete power, and we call it an absolute power. For example: n 2+n+1,n 2+4n+4,
N 2+6n+9, ... is an absolute quadratic power; And n 3+3n 2+3n+ 1, n 3+6n 2+ 12n+8, ... are absolute cubic powers.
The expansion term in the general form of (n+b) m (m > 1, where b is a constant term).
Definition 4 is definitely not a power.
In an algebraic expression with an unknown number, if all the values of the unknown number are integers greater than 0, then the value of the algebraic expression is not a complete power, and we call the algebraic expression absolutely non-power. For example: n 2+ 1, n 2+2, n 2+2n, ... is definitely a non-quadratic power; And n 3+1,n 3+3n 2+1,n 3+3n+1,n 2+3n+1,n 3+6n 2+8 ...
When there are few terms in a unary algebra, it is easy to judge whether the algebra is absolutely unpowered. For example, n 2+n is absolutely non-power and n 7+n is absolutely non-power. However, when there are many terms in contemporary numbers, the conditions for obtaining absolute idempotence will become more and more harsh.
The general form of the unary absolutely unpowered formula is: subtract a term from the expansion term of (n+b) m (m > 2, b is a constant term).
Reasoning: A power algebraic formula that is not an absolute M-power formula and an absolute non-M-power formula will inevitably get a complete M-power number when the unknown quantity takes a certain value. For example, 3n 2+4n+ 1 is not absolute non-cubic. When n= 1, 3n 2+4n+1= 8 = 2 3, 3n 2+3n+1is not an absolute non-quadratic power. When n=7,
Reasoning: unary algebra without power term is not unique to any power. 2n+ 1=9=3^2,2n+ 1=49=7^2……4n+4=64=8^2,4n+4=256= 16^2 ……2n+ 1=27=3^3,2n+ 1= 125=5^3……
It is proved that the unary algebraic expression has m absolute non-square powers;
In the unary algebraic expression, the algebraic expression takes different values for the unknown and will get different calculation results. The corresponding relationship between the unknown number and the algebraic calculation result is unique, and the equation is reversible, which is a purely definite solution relationship. This is the algebraic axiom of unary algebra. That is to say, algebraic expression can be solved by substituting unknown values, and given algebraic values, unknown values can be solved in turn. Using these properties of unary algebra, odd-even classification, remainder classification and power classification of integers can be realized.
When the constant term is 1, the fixed form of the four-term algebraic expression of a complete cubic number is (n+1) 3 = n 3+3n 2+3n+1,where * * consists of four monomial elements with two power terms. For three of these algebraic expressions, if we lock any three of them, we can get three different unary algebras which must contain power terms, n 3+3n 2+1,n 3+3n+1,3n 2+3n+/kloc-0. For these three kinds of algebra, because these three kinds of algebra and the original cubic algebra form a fixed algebraic relationship with single definite difference, the existence of this algebraic relationship has nothing to do with unknown values. This relationship is:
(n+ 1)^3-3n= n^3+3n^2+ 1
(n+ 1)^3-3n^2= n^3+3n+ 1
(n+ 1)^3-n^3=3n^2+3n+ 1
So we get: when n= 1, 2, 3, 4, 5 …
n^3+3n^2+ 1≠(n+ 1)^3
n^3+3n+ 1≠(n+ 1)^3
3n2+3n+ 1≠(n+ 1)^^3
That is to say, the values of these three algebraic expressions cannot be equal to the complete cubic number in the form of (n+ 1)3-.
When n= 1, 2, 3, 4, 5 ..., then the value of (n+1) 3 = n 3+3n 2+3n+1 is the cube of all integers starting from 2, while integers less than 2 are only1.
n^3+3n^2+ 1=5≠ 1
n^3+3n+ 1=5≠ 1
3n^2+3n+ 1=7≠ 1
It is concluded that when n= 1, 2, 3, 4, 5 …, the algebraic expressions n 3+3n 2+ 1, n 3+3n+ 1 are not equal to all integers. These algebraic expressions are cubic absolutely unpowered expressions.
Through the above methods, we can prove the unary algebraic expressions: n 4+4n 3+6n 2+ 1, n 4+4n 3+4n+ 1, n 4+6n 2+4n+ 1, 4n 3+6n. These algebraic expressions are quartic absolutely unpowered expressions.
It can be proved that the expansion term of unary algebraic expression (n+ 1) m whose power is greater than 5 can get m different unary algebraic expressions, and when n= 1, 2, 3, 4, 5 … these algebraic expressions are absolute non-square powers of m degree.
Now we give the difference formula of the square power increase term of two adjacent integers n and n+ 1 by algebraic method;
When the power is 2, there is: (n+1) 2-n 2.
=n^2+2n+ 1-n^2
=2n+ 1
Therefore, the difference formula of the square number of quadratic adjacent integers is 2n+ 1.
Since 2n+ 1 contains no power relation, all odd powers can be expressed as 2n+ 1, so when 2n+ 1 is a complete square number, there must be n 2+(2 √ 2n+1) 2 = (n+1. However, the square integer solution of xyz coprime of z-x > 1 cannot be obtained by the law of increasing ratio. The method of obtaining these square integer solutions is as follows:
When (n+2) 2-n 2 = 4n+4 is a complete square number, the ratio of all square integers with z-x=2 increases after solving;
When (n+3) 2-n 2 = 6n+9 is a complete square number, the ratio of all square integers with z-x=3 is obtained.
When (n+4) 2-n 2 = 8n+ 16 is a complete square number, the increase ratio of all square integers with z-x=4 is obtained.
……
This increasing relationship of constant terms applies to all integers. When n= 1, 2, 3 …, all square integer solutions of integers can be obtained.
Therefore, when the exponent is 2, Fermat's equation x n+y n = z n holds.
At the same time, because all odd powers can be expressed as 2n+ 1, and some even powers can be expressed as 4n+4, 6n+9, 8n+ 16 ..., therefore, there must be an integer solution relationship of x 2+y n = z 2.
To the third power, there are: (n+1) 3-n 3.
=n^3+3n^2+3n+ 1-n^3
=3n^2+3n+ 1
So the difference formula of cubic number of cubic adjacent integers is 3n 2+3n+ 1.
Because 3n 2+3n+ 1 is the missing formula of (n+ 1) 3, it still contains a power relation and is a cubic absolute non-power formula. Therefore, when n is an arbitrary integer, the value of 3n 2+3n+ 1 is not a complete cubic number, so there is no N 3+(3 √ 3N2+3N+1) 3 = (n+1) 3 between integers, that is, z-x = 65438+. The Fermat equations of xyz coprime with z-x > 1 cannot be expressed by the law of increasing ratio. These cubic Fermat equations are expressed as follows:
From (n+2) 3-n 3 = 6n2+12n+8 Therefore, n is an arbitrary integer whose value is not completely cubic;
(n+3) 3-n 3 = 9N2+27N+27, so n is an arbitrary integer whose value is not a complete cubic number;
From (n+4) 3-n 3 =12n2+48n+64 Therefore, n is an arbitrary integer whose value is not a complete cubic number;
……
This increasing relationship of constant terms applies to all integers. When n= 1, 2, 3 …, the cubic power relation of Fermat equation will cover all the integers after increasing the proportion.
Therefore, the Fermat equation x n+y n = z n has no integer solution when the exponent is 3.
There is a fourth power; (n+ 1)^4-n^4
=n^4+4n^3+6n^2+4n+ 1-n^4
=4n^3+6n^2+4n+ 1
Therefore, the difference formula between the fourth power and the fourth power of adjacent integers is 4n 3+6n 2+4n+ 1.
Because 4n 3+6n 2+4n+ 1 is the missing formula of (n+ 1) 4, it still contains a power relation, and it is an absolute quartic non-square power formula. Therefore, when n is an arbitrary integer, the value of 4n 3+6n 2+4n+ 1 is not a perfect quartic number, so there is no N 4+(4 √ 4N3+6N2+4N+1) 4 = (n+1) 4 between integers, that is, Z-. The Fermat equations of xyz coprime with z-x > 1 cannot be expressed by the law of increasing ratio. The expression of these fourth-order Fermat equations is as follows:
By (n+1) 4-n 4 = 8n3+24N2+32n+16 Therefore, n is an arbitrary integer, and its value is not completely quartic;
From (n+1) 4-n 4 =12n3+54N2+108n+81Therefore, n is an arbitrary integer, and its value is not completely quartic;
By (n+1) 4-n 4 =16n3+96n2+256n+256 Therefore, n is an arbitrary integer, and its value is not completely quartic;
……
This increasing relationship of constant terms applies to all integers. When n= 1, 2, 3 …, the fourth power relation of Fermat equation will cover all integers after increasing the proportion.
Therefore, the Fermat equation x n+y n = z n has no integer solution when the exponent is 4.
When the power is m, the difference formula of the powers of adjacent integers is:
(n+ 1)^m-n^m
=n^m+mn^m- 1+…+…+mn+ 1-n^m
=mn^m- 1+…+…+mn+ 1
Therefore, the difference formula of the addition term of the m-power adjacent integer is Mn m- 1+…+…+Mn+ 1.
Because Mn m- 1+…+Mn+ 1 is the missing term formula of (n+ 1) m, it still contains a power relation, which is an absolute non-square power formula of m degree. Therefore, when n is an arbitrary integer, the value of Mn m- 1+…+Mn+ 1 is not a complete power of m, so there is no N m+(m √ Mn m-1+…+Mn+1) between integers. The Fermat equations of xyz coprime with z-x > 1 cannot be expressed by the law of increasing ratio. The method of expressing these Fermat equations by the power of m is as follows:
From (n+2) m-n m = 2mn m-1+…+2m-1Mn+2m Therefore, n is an arbitrary integer whose value is not a complete power of m;
From (n+3) m-n m = 3mnmm-1+…+3m-1Mn+3m Therefore, n is an arbitrary integer, and its value is not a complete power of m;
From (n+4) m-n m = 4mnmm-1+…+4m-1mn+4m Therefore, n is an arbitrary integer, and its value is not a complete power of m;
……
This increasing relationship of constant terms applies to all integers. When n= 1, 2, 3 …, the m-th power relation of Fermat equation will cover all the integers after increasing the proportion.
Therefore, the Fermat equation x n+y n = z n has no integer solution when the exponent is m.
So when the exponent n > 2, the Fermat equation x n+y n = z n will never have an integer solution.
Therefore, Fermat's last theorem, which lasted for more than 300 years, is an elementary number, just like Goldbach's conjecture.
Understand the problem.
Responder: danchide friendship-trainee magician level 2 8-5 13:57
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Other answers *** 2
It is only necessary to prove that x 4+y 4 = z 4, x p+y p = z p (p is an odd prime number) has no integer solution.
Responder: lshhy- senior magician level 6 8-4 17: 14.
In history, many people have not made achievements in their main jobs, but they have made great achievements in their leisure time. Fermat is a typical example. Pierre de fermat (160 1 ~ 1665) is mentioned today mainly because he is a politician or a judge, but because he is an excellent amateur mathematician. Fermat has made great achievements in many fields of mathematics, but what really made him famous in the world was the conjecture called Fermat's last theorem by later generations.
The expression of Fermat's last theorem is simple: for positive integers, it is impossible to write a power higher than 2 times as the sum of two powers of the same power. In other words, the equation xn+yn = Zn has no positive integer solution when n > 2. In the margin of a book, Fermat wrote: I have a very beautiful proof of this proposition. The margin here is too small to write.
Since then, countless wise men, including great mathematicians Euler and Cauchy, have done their best for this. Although they can take a small step forward every time, they can't finally prove Fermat's last theorem. For more than 300 years, many people claimed to have found a solution to this difficult problem, but it was overturned every time. As far as Fermat's last theorem itself is concerned, proof is of little significance to the development of mathematics. But on the one hand, this is a challenge to wisdom; On the other hand, mathematicians have gained many unexpected gains in the process of proving Fermat's last theorem, and some new branches and methods of mathematics have emerged in the study of it. So the proof of Fermat's last theorem has always been proved by people.
Worries.
There are many episodes about Fermat's Last Theorem, among which German Paul Wolfskeil set up a special fund for Fermat's Last Theorem. According to people's popular saying, Wolfskeil tried to end his life because he was lovelorn. Some time before he thought everything was ready to shoot himself at midnight, he found a paper on Fermat's Last Theorem. As it happens, Wolfskeil himself is a math enthusiast, so he unconsciously got lost in the paper and missed the scheduled suicide time. Later, Wolfskeil gave up the idea of suicide and left a will before he died, giving a large sum of wealth as a prize to the first person who proved Fermat's Last Theorem, valid until 2007.
Andrew wiles, a professor at Princeton University, published his proof of Fermat's Last Theorem in 1993 after seven years of painstaking research. His certificate was confirmed on 1995, and he finally won the prize left by Wolfskell.
Wiles's proof is more than 100 pages, involving a lot of the latest mathematical knowledge, and only a handful of people in the world can understand it at present. So there is such a controversy: some people think that this can't be the proof that Fermat thought of in the past, and there should be a simpler proof that has not been discovered; But there are also many people who tend to think that Fermat didn't actually find anything, or just thought of a wrong method.
1637, when Fermat was reading the Latin translation of Diophantine arithmetic, he wrote beside the eighth proposition in Volume 1 1: "It is impossible to divide a cubic number by the sum of two cubic numbers, it is impossible to divide a quartic power by the sum of two quartic powers, and it is even more impossible to divide a power higher than quadratic by the sum of two powers of the same order. In this regard, I am sure that I have found a wonderful proof, but unfortunately the space here is too small to write down. " After all, Fermat didn't write a proof, and his other conjectures made great contributions to mathematics, which inspired many mathematicians' interest in this conjecture. The related work of mathematicians enriches the content of number theory and promotes its development.
For many different N's, Fermat's Last Theorem has already been proved. But mathematicians are still confused about the general situation of the first 200 years.
1908, Germany Vlfsk announced that it would give 65438+ million marks as a prize to the first person who proved the theorem within 100 years after his death.
1983, gerd faltings proved the model conjunction, and concluded that when N >: 2 (n is an integer), there is no coprime A, B, C, which makes an+bn = cn.
1986, Gerhard Frey put forward the "ε conjecture": if A, B and C make an+bn = cn wrong, that is, Fermat's last theorem, then the elliptic curve.
y2 = x(x-an)(x + bn)
Will be a counterexample of Taniyama intellectual village's conjecture. Frey's guess was immediately confirmed by Kenneth Rebett. This conjecture shows the close relationship between Fermat's Last Theorem and elliptic curve and module form.
1995 wiles and Taylor proved the Taniyama conjecture in a special case, and the Frey elliptic curve is just within this special case, thus proving Fermat's last theorem.
Wiles's process of proving Fermat's Last Theorem is also very dramatic. It took him seven years to obtain most of the evidence without being known. Then in June of 1993, he published his certificate at an academic conference, which immediately became the headlines of the world. But in the process of examining and approving the certificate, the experts found a very serious mistake. Wiles and Taylor then spent nearly a year trying to remedy it, and finally succeeded in a method abandoned by wiles in September 1994. Their proof was published in the Mathematical Yearbook of 1995.
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