 Welcome To Numbertheory Web Site     Site Owner: Li Ke Xiong
Email: likexiong@126.com The " Primary Proof " For Fermat Last Theorem

A. At the beginning, we think it is necessary to repeat our Definitions in the paper of " The Development of the Concept of Congruence "
Definition I (A). mod . integer a, b, n and s. (B). Never can "p" be as the denominator divisor of (and in) any finite expression,
otherwise meaningless.  (In any finite expressions, when "p" is a denominator divisor,
1/p is 1/0, hence the finite expression is meaningless, mod .
We should say, if A is a basic fact, B should be true, too. A and B can not be
separated.

B. Now, suppose can be true with same p. We have   , make the function f(x)= the above two expressions are   , p>3.

C. D. Because  (p>3) E. Let us see the above (      ) , if (      )   p>3, surely (       )  mod p.
if suppose (       ) not mod p, (        ) not  without any doubt.
a. So let us suppose (      )  mod p, we have (         )     ,
please note when x is an integer, mod p.
thus (      )    = -2/3(27-8)=-2/3X19 mod p (p>3)
The only possibility for (        )   (p>3) is p=19.
b.Let us suppose (         )   (p>3)
(        ) = due to having supposed   so (         )  3+9-3+(p-2)/(p+2)X27-[2+8/3-2+(p-2)/(p+2)X8]
=19[1/3+(p-2)/(p+2)]  (p>3)
1. If p=19, we get 1/3+(19-2)/(19+2) mod 19, we get 24 , or 1/3+(19-2)/(19+2) 1/3+(-2)/2=1/3-1=(-2)/3 mod 19.
2. If , we get 1/3+(p-2)/(p+2)  , p>3 surely 1/3+(p-2)/(p+2) ,
that is 1/3-1 mod p [Please note: (p-2)/(p+2) -1 ,pod p], so -2/3 mod p, p>3
-2 mod p, obviously this is also a contradiction.

F. Let us see ,
a. Let us suppose  mod p k,m and s are integers 1
s 1, . We know should be a rational number, otherwise can not match
f(3)-f(2) is rational.  So we can write b. Needless to say    is also a meaningless one.

G. From the above E and F, we can see f(3)-f(2) never  . The conclusion is: there is
no such "p" to make   , p>3 to be true together at the same time. Site Owner: Li Ke Xiong
Email: likexiong@126.com