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
B. Now, suppose can
be true with same p.
We have ,
make the function f(x)=
the above two expressions are , p>3.
D. Because (p>3)
E. Let us see the above ( ),
if ( )
p>3, surely ( )
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]
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)/3mod 19.
This is a contradiction.
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
s1, . 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
ALL RIGHTS RESERVED