From Pythagoras Theorem to Fermat’s Last Theorem and the Relationship between the Equation of Degree n with One Unknown

The most interesting and famous problem that puzzled the mathematicians all around the world is much likely to be the Fermat’s Last Theorem. Howev-er, since the Theorem was proposed, people can’t find a way to solve the problem until Andrew Wiles proved the Fermat’s Last Theorem through a very difficult method called Modular elliptic curves in 1995. In this paper, I firstly constructed a geometric method to prove Fermat’s Last Theorem, and in this way we can easily get the conclusion below: If a and b are integer and a = b, n ∈  and 1 n > , the value of c satisfies the function n n n a b c + = that can never be integer; if a , b and c are integer and a b ≠ , n is integer and 2 n > , the function n n n a b c + = cannot be


Introduction
The Fermat's Last Theorem was proposed by French famous mathematician Pierre de Fermat in 1637, it was called the last theorem because it was the theorem of Fermat that can be proved at last, which means to prove the theorem is very difficult. The Fermat's Last Theorem states: there is no positive integer a, b and c to satisfy the function n n n a b c + = (1) when n is integer and 2 n > [1]. Many mathematicians paid attention to this theorem, and they found it not as easy as it looks like. In 1753, the famous Swiss mathematician Euler said in a letter to Goldbach that he proved the Fermat conjecture at n = 3, and his proof was published in the book Algebra Guide in 1770 [2]. Fermat himself proved the Fermat conjecture at n = 4 [3]. In 1825, the German mathematician Dirichlet and the French mathematician Legendre independently proved that Fermat's theorem was established at n = 5, using the extension of the method used by Eulerasin [2]. In 1844, Kummer proposed the concept of "ideal number", he proved that for all prime indices n less than 100, Fermat's theorem was established, and this study came to a stage [4]. But the mathematicians still struggled with Fermat's theorem in the first two hundred years of the conjecture with little progress. What's more, many theorem were proposed in order to prove the Fermat's Last Theorem, such as Modell conjecture, Taniyama-Shimura theorem. After proving the Taniyama-Shimura theorem, Andrew Wiles finally got a way to prove the Fermat's Last Theorem in 1995 [5].
At first, people wanted to prove that the Fermat's Last Theorem was established at different indices n, but the indices n is infinite, this method is meant to be failed. Then, people tried to propose another theorem to indirectly prove the Fermat's Last Theorem, but the relationship between two theorems is not very clear, thus the proof is hard to be verified.
To prove the Fermat's Last Theorem, I got inspiration from the Pythagoras Theorem. As we all know the Pythagoras Theorem: the sum of the squares of the two right-angled sides of a right-angled triangle is equal to the square of the hypotenuse, let the length of two right-angled sides be a and b, and the length of hypotenuse is c, then , what is the relationship between n and θ? This paper discusses the relationship between n and θ, and in this geometric method, we can easily prove the Fermat's Last Theorem.

Geometric Construction
A triangle has three sides, a, b and c, respectively. Firstly, let us discuss an easy condition: a = b.

2) When
1 p q = , p and q are positive integer.
t, s, p, q are positive integer, so 2 q is an even number, then t is an even number, let t = 2k (k is positive integer), is also an even number, then s must be an even number. Above all, we can prove that t and s are even number, which is contradictory with ( ) , 1 t s = , therefore, (1), only if n = 1, c can be integer, the relationship between n and θ is:

The Proof of Fermat's Last Theorem
The Fermat's Last Theorem is: When n is integer and 2 n > , the function (1) has no positive integer solution, which means a , b and c can't be positive integer at the same time or when a, b, and c is positive integer, n is integer and ( ) n n n n a b c a b + = < + , so c a b < + Therefore, a, b, and c can certainly form a triangle ∆OAB [7], and the triangle ∆OAB is shown in Figure 3.
In the section of 1.1, we have proved that the value of c in the function (1) is irrational even if n is rational ( 1 n > ) when a b r = = and r is positive integer. To prove the Fermat's Last Theorem, we have to prove another condition: if a, b and c is positive integer, and a b ≠ , the function n n n a b c + = (2) can not be established when n is integer and 2 n > .
Similarly, we can also construct the geometric method as same as the section of 2.1. The geometric graph is shown in Figure 3 a , b and t is integer, so ( )( ) is integer, then cosθ must be rational. Let cos R θ = , and R is rational.  integer, n is integer and n>2, if a , b is positive integer, the relationship between a , b, c and n must be n n n a b c + < , no matter c is integer or not. ing to Euclid's Elements, big angle to big side) [7], so if arcsin ,

2) When
(which is contradictory with function (16)) Therefore, if b > 1 and 1 n b ≥ + , the relationship between a , b, c and n must be n n n a b c + < , the function (16) can't be satisfied.
2) 2 1 n b < < + , then If there exists a positive integer c to satisfy the function (17), then t must be a integer, and the value of t is the solution for the equation of degree n with one unknown in function (18), and the solution of t in function (18) is also the solution of t in function (17).
Obviously, one of the solution of t in function (17) must be: In this paper, we only discuss the condition of 0 t b < < , so the value of t satisfies the requirement is: Therefore, we have found a solution for the equation of degree n with one unknown in function (18), and the solution of t is the function (19) and (20).
For example: let a = 2, b = 1, and n = 5, so the function (18) is equal to the function shows below: According to function (20), we can easily get the solution of t in function (22) is  (24): Let's discuss the question in the next Section 2.3. According to function (18), we can transform it to the function (25): t is a positive integer, so the left side of the equal sign in function (25) must be a positive integer, therefore, the value of n b t must be a positive integer.
Let the value of n b m t = , and m is a positive integer, then Therefore, we have to find the value of t in the function (26), and determine if the value of t can be an integer. If t can't be a positive integer when 3 n ≥ , then the Fermat's last Theorem is right. If n = 1, the function (26) is equal to 1 0 m − = , then If n = 2, the function (26) is equal to Therefore, if n = 1 and n = 2, the function n n n a b c + = can have integer solution.
In conclusion, if n = 3, and the solution of t in function (26) is positive integer, but the value of b can't be a positive integer, thus, the Fermat's Last Theorem is established when n = 3.

The Solution for the Equation of Degree n with One Unknown
How to find the solution for the arbitrarily equation of degree n with one unknown as function (24): This paper only discusses the real solution of function (24). As we already known, the solution of t in the function (18) is So, the function (24) can be deformed as function (27): The coefficients in function (27) are shown below: 1 1 n n n n n n n n n n n n n n n n n n The solution of t in the function (27) If we want the function (27) to be established, then all the value of t(k) must be equal, and we can find the solution for the equation of degree n with one unknown. We can set the value of a and 0 a ≠ , so here are n equations and we have to find the solution of 1 2 , , , n b b b  , and the solution can be found. 1) Part one: let's find the solution for the random equation of degree 3 with one unknown, the equation is shown below: The function (29) can be deformed as: In order to make the function (29) to be established, the value of ( ) , then:  3  3  3  1  3  2  2  2   2  2   3  3  3  3  3   3   3  3 3 2   3  1  2  2  1 2 3  3  3  3  3  3  3   2  2  3  3  2   27  9  2  3  3 3 27 According to the Cardano Formula of the General Solution of Cubic Algebraic Equations [8], the real solution of function (31) is: Therefore, the solution of t satisfy the function in (29) is The value of u in the function (35) must be a integer, therefore,  In conclusion, if n = 4, the solution of t in function (26) is irrational, thus, the Fermat's Last Theorem is established when n = 4.
2) Part two: let's discuss the solution for the random equation of degree 4 with one unknown, the equation is shown below: The function (32) can be deformed as:

A t at a t a t b A A a t A A a t B A b
Similarly, in order to make the function (33) to be established, the value of ( ) 3 8 16    We can set a value for k and 0 k ≠ , p, q, and r are the known number, therefore, we can find the value of u and v are:

A A A p A A A A A A A A A q A A B A A A A A A A A A A A
Now, the problem is transformed to find the solution x for function (37), according to the results of Part one, we can get the solution x is:   1.18412432 or 1.5591424 2

A B A A A A A A A A A A A A r a m A
The value of u in the function (41) must be a integer, therefore, must be a square number, then Clearly, the function (42) can't be established, therefore, the value of w is irrational, and the solutions of t satisfy the function in (40) is In conclusion, if n = 5, the solution of t in function (26) is irrational, thus, the Fermat's Last Theorem is established when n = 5.

A t at a t a t a t b A A a t ct c t b A A a A A a c t B Ab A A a b
In order to make the function (43) to be established, the value of ( ) The function (45) can be arranged as:  4  2  2  3  5  1  5  3  5   5   3  2  4  2  2  5  2  3  3  5  1  5  3  5  5  3  5   5   10   10  3   10  3  5  10  3   10  5  10  3 10 The function (46) can be deformed as: Let 2 w x = , then the function (47) can be deformed as: Therefore, if p, q, r, s satisfy the function (49), the solution of the function (48) are easy to be found.
But, if those conditions can't be satisfied, then the function (46) can't be deformed as function (47), so we can deform the function (46) to function (50): The unknown number d, e, f, g satisfy the relationship (51) below: The value of u in the function (58) must be a integer, therefore, The value of u in the function (61) must be a integer, therefore,   The value of N must be the factor of r, thus the value of R, K could be integer.  If N = 1, 2, 7, 14, 3 3 14 N N   +     is an odd number; If N is integer and 0 N ≠ , 1, 2, 7, and 14, 3 3 14 is a fraction number, therefore, s is a positive integer, thus b is irrational, which is contradictory with b is a integer. Therefore, the Fermat's Last Theorem is established when n = 10.
The values of b satisfy the condition under different n is concluded in the Table 1. Table 1 indicated that if 2 < n < 21, in order to have a positive integer solution  In order to prove the value of c satisfies the function (1) can't be positive integer when a, b, and n are positive integer and n > 2, in this section, we find the solution for the random equation of degree n with one unknown (when n = 3 and 4), and proved that the Fermat's theorem was established at n = 3, 4 and 5,