Great Fermat theorem Print E-mail
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Written by Âàëåðèé Ïåòðîâ   
Âîñêðåñåíüå, 08 Èþíü 2008
The text of the original in Russian


Theorem: Suppose there are three numbers that satisfy the equation: 
                                                           zn = xn + yn (1). 
Required to prove that when n> 2, this equation has no integer solutions 
Proof. 
We know that three numbers that satisfy the equation (1) shall meet the following conditions: 
one of the numbers, for example, z, must be even, the other two - odd; 
number must be mutually prime, ie pairs should not have common factors; 
No two numbers can not be equal to each other. 
Suppose for definiteness that z> x> y. 
Obviously, the number of z less than the sum of two numbers, ie 
                                                                z
Suppose there are three segments of length z, x, y, satisfying the condition (2). Then, because of the well-known theorems in these sections can be constructed as a triangle on the sides. We know that the triangle between the parties which is the ratio of (1), where n> 2 acute. Then for the sides of the triangle is the ratio arising from a theorem of cosines: 
                                                      z2 = x2 + y2 - 2xycosα: 
where α - the angle between the sides x and y. 
Building an acute-angled triangle ABC with sides AB = x, BC = y, AC = z. Put the point of acute triangle ABC A perpendicular to the opposite side of BC, as shown in the figure.
 teor.jpg
Fig. 1. Acute triangle 

From the triangle BC1C find cosα = m1 / BC = m1 / y. Substituting the value of cosα (2), we get: 
                                                       z2 = x2 + y2 - 2xym1 / y 
                                                       z2 = x2 + y2 - 2xm1 (3) 
Thus, for the same triangle at the same time we have two different relationships between the parties: (1) and (3). Then the essence of the theorem can be expressed differently: it is required to prove that no integer solutions of equation (3) are not for the equation (1). 
Multiply equation (3) zn-2. Obtain: 
                                               zn-2z2 = zn-2x2 + zn-2y2 - 2xzn-2m1 (4) 
Let zn-2 = xn-2 + a = yn-2 + b, where a and b - some whole numbers, for these equity. Then, substituting the value zn-2 (4), we get: 
                                              zn = (xn-2 + a) x2 + (yn-2 + b) y2 - 2x (xn-2 + a) m1 
                                              zn = xn + ax2 + yn + by2 - 2x (xn-2 + a) m1 (5) 
Subtracting (1) of (5), we get: 
                                                   0 = ax2 + by2 - 2x (xn-2 + a) m1 (6) 
Thus if any integer numbers x and y equation (6) will be equal to zero, the solution of this equation (ie the value of numbers x and y) will be at the same time the solution of the equation (1), because in this case possible to equation (5) to transform the equation (1). 

Solving this equation, we get: 
                                                        by2 = 2x (xn-2 + a) m1 - ax2 
                                                        by2 = x [2 (xn-2 + a) m1 - ax] 
Writing for ease of computation 2 (xn-2 + a) m1 - ax = k. Obtain: 
                                                                              by2 = kx, 
where should:
yr.jpg
ò.å. korx.jpg is one of the factors of y
Thus, the integer solutions of equation (6) are possible only under the condition that korx.jpg  is one of the factors of y, which contradicts the initial condition problem. Therefore, under any values of the numbers x and y, satisfying the initial conditions problem, equation (6) is not equal to zero. Therefore, under any values of the numbers x and y, satisfying the initial conditions problem, equation (5) can not be transformed into equation (1). Therefore, under any values of the numbers x and y, satisfying the initial conditions problem, equation (1) can not have any integral solutions. What was required to prove. 

Petrov, VV 
Prospekt Lenina 30, Apt. 9 
Nikolaev 54029 
Ukraine  
 

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