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
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
So that
Now, substitution of ( 2.16) into the positive root of ( 2.5) gives
This must be a positive root so that q3 must be greater than unity, and
thus ( 2.17) is re-written as
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
|