2.2  Case n = 3.

When n = 3, (2.3) reduces to


x3 - 3( a - b )x2 - 3( a2 - b2 )x - ( a3 - b3 ) = 0
(2.4)

From the above discussion, the roots of (2.4) are of the form


{ x - ( p + k1 ) } ì
í
î 		x - æ
è 		p - k1
2
   ±  jk2 ö
ø 		ü
ý
þ 		= 0
(2.5)

where p = (a - b), and the k parameters are secondary variables of the a's and b's that contribute to generating the correct co-efficients of (2.4).

Expanding (2.5) gives


x3 - 3px2 + é
ë 		ì
í
î 		æ
è 		p - k1
2
 ö
ø 		2

 
 + k22 ü
ý
þ 		+ ( 2p - k1 )( p + k1 ) ù
û 		x - ì
í
î 		æ
è 		p - k1
2
 ö
ø 		2

 
 + k22 ü
ý
þ 		( p + k1 ) = 0
(2.6)

In (2.6) consider the co-efficient of x. Multiplying out this reduces to


A1 = 3p2 - 3
4
 k12 + k22
(2.7)

Comparing (2.7) with the co-efficient of x in ( 2.4)


- 3a2 + 3b2 = 3( a - b )2 - 3
4
 k12 + k22
(2.8)

which reduces to


k22 - 3
4
 k12 = - 6a( a - b ) = - 6ap
(2.9)

Now consider the final term in ( 2.6). Multiplying out this reduces to


A0 = - p3 - æ
è 		k22 - 3
4
 k12 ö
ø 		p + k1 æ
è 		k22 + k1
4
 ö
ø
(2.10)

Substituting ( 2.9) for the co-efficient of p in ( 2.10) and for k22 in the final term of ( 2.10) gives


A0 = - p3 + 6ap2 - k1( - 6ap + k12 )
(2.11)

Now compare ( 2.11) with the final term in ( 2.4) thus


- a3 + b3 = - ( a - b )3 + 6a( a - b )2 - k1( - 6a( a - b ) + k12 )
(2.12)

and this reduces to


6a3 - 9a2b + 3ab2 + 6k1 a( a - b ) - k13 = 0
(2.13)

Dividing through by the co-efficient of k1 now gives


a - b
2
 + k1 æ
è 		1 - k12
6a( a - b )
 ö
ø 		= 0
(2.14)

Now, k1 is of unity order in a and b so that k12 in the numerator of the quotient in ( 2.14) is of the same order in a and b as the denominator. Also because of the

independent presence of unity, the term  ì
í
î 		1 -  k12
6a( a - b )
 ü
ý
þ 		  must be a pure number. Consequently, the quotient in this term must also be a pure number,



i.e. q3, where the subscript denotes the order of the equation being analysed, i.e. ( 2.4). The parameter q3 is not a constant, but a "non-dimensional" function of the a's and b's. Thus ( 2.14) becomes


a - b
2
 + k1( 1-q3 ) = 0
(2.15)

So that


k1 =
-  æ
è 		a -   b
2
 ö
ø
( 1 - q3 )
(2.16)

Now, substitution of ( 2.16) into the positive root of ( 2.5) gives


x = a - b -
æ
è 		a -   b
2
 ö
ø
( 1 - q3 )
(2.17)

This must be a positive root so that q3 must be greater than unity, and thus ( 2.17) is re-written as


x = a - b +
æ
è 		a -   b
2
 ö
ø
( q3 - 1 )
(2.18)

Re-arranging for a gives


a = æ
è 		x + b
2
 ö
ø 		æ
è 		1 - 1
q3
ö
ø 		+ b
2
(2.19)

Now, because x is an integer, (x+b/2) must be half integer. Therefore, for a to be an integer, the term (1 - 1/q3) would have to be an odd integer. This is clearly impossible because with q3 > 1, the term (1 - 1/q3) must be fractional. Thus a in ( 2.2) and therefore z in ( 2.1) cannot be integers. Thus, subject to q3 exhibiting satisfactory characteristics, this proves Fermat's Last Theorem for n = 3.



M2 Version 1.0.0
Ó P.G.Bass, April 2009

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