2.3 Case n = 4.
When n = 4, (2.3) becomes
|
x4 - 4( a - b )x3 - 6( a2 - b2 )x2 - 4( a3 - b3 )x-( a4 - b4 ) = 0 |
| (2.20) |
The roots of this equation will be of the form
|
{ x - ( p + k1 + k2 ) }{ x - ( p - k1 + k2 ) }{ x - ( p - k2 ± jk3 ) } = 0 |
| (2.21) |
When expanded (2.21) becomes
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x4 - A3 x3 + A2 x2 - A1 x + A0 = 0 |
| (2.22) |
Where
|
| |
|
|
A2 = [ { ( p + k2
)2 - k12 } + 4( p +
k2 )( p - k2 ) + { ( p - k2 )2 + k32 } ]
|
|
|
|
A1 = 2[ { ( p + k2
)2 - k12 }( p - k2 ) + { ( p - k2 )2 + k32 }(
p + k2 ) ] |
|
|
|
A0 = { ( p + k2
)2 - k12 }{ ( p - k2 )2 + k32 }
|
| |
(2.23) |
Consider the co-efficient of x2 in (2.22). From (2.23) after multiplying
out, this becomes
|
A2 = 6p2 - 2k22 - k12 + k32 |
| (2.24) |
Comparing (2.24) with the co-efficient of x2 in (2.20) gives
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-6a2 + 6b2 = 6( a - b )2 - 2k22 - k12 + k32 |
| (2.25) |
Which reduces to
|
2k22 + k12 - k32 = 12a2 - 12ab = 12ap |
| (2.26) |
Now consider the co-efficient of x in (2.22). From (2.23), after multiplying out this becomes
|
A1 = - 4p3 + 2(2k22 + k12 - k32 )p - 2k2 (k12 + k32 ) |
| (2.27) |
Substituting (2.26) for the co-efficient of p and for k32 in the final term in (2.27) gives
|
A1 = 20a3 - 36a2b + 12ab2 + 4b3 + k2{24a( a - b ) - 4k12 - 4k22 } |
| (2.28) |
Comparing (2.28) with the co-efficient of x in (2.20) gives
|
- 4a3 + 4b3 = 20a3 - 36a2b + 12ab2 + 4b3 + k2{24a( a - b ) - 4k12 - 4k22 } |
| (2.29) |
and this reduces to
|
24a3 - 36a2b + 12ab2 + k2{24a( a - b ) - 4k12 - 4k22 } = 0 |
| (2.30) |
Dividing (2.30) throughout by 24a(a-b) then gives
|
a - |
b
2
|
+ k2 |
ì
í
î
|
1 - |
k12 + k22
6a(a - b )
|
ü
ý
þ
|
= 0 |
| (2.31) |
In the quotient in (2.31), the term ( k12 + k22 ) is of
the same order in a and b as the denominator and therefore this quotient must
be a pure number, q4. Eq.(2.31) therefore becomes
|
a - |
b
2
|
+ k2(1 - q4 ) = 0 |
| (2.32) |
So that
Substitution of (2.33) into the positive root of (2.21) then gives
|
x1 = a - b - |
a - b/2
1 - q4
|
+ k1 |
| (2.34) |
Substitution of (2.33) into the negative real root of (2.21) gives
|
x2 = a - b - |
a - b/2
1 - q4
|
- k1 |
| (2.35) |
Adding (2.34) and (2.35) then yields
|
|
x1 + x2
2
|
= x¢ = - |
æ
è
|
a - |
b
2
|
ö
ø
|
|
æ
è
|
q4
1 - q4
|
ö
ø
|
- |
b
2
|
|
| (2.36) |
and re-arranging for a yields
|
a = - |
æ
è
|
x¢ + |
b
2
|
ö
ø
|
|
æ
è
|
1 - q4
q4
|
ö
ø
|
+ |
b
2
|
|
| (2.37) |
The parameter a must be positive so that q4 must be greater than unity and
therefore (2.37) is rewritten as
|
a = |
æ
è
|
x¢ + |
b
2
|
ö
ø
|
|
æ
è
|
1 - |
1
q4
|
ö
ø
|
+ |
b
2
|
|
| (2.38) |
x/ is the average of the real roots and must therefore be either
integer or half integer. If x/ is integer, then (x/ + b/2) must be
half integer and a cannot then be integer because (1 - 1/q4) cannot be an
odd integer.
If x/ is half integer then (x/ + b/2) is a full integer and for a to be
integer, not only would (x/ + b/2) have to be an odd integer, but (1 - 1/q4) would also have to be half integer.
This would require q4 to be
exactly equal to 2. In this case from (2.31)
|
|
k12 + k22
6a( a - b )
|
= 2 |
| (2.39) |
so that
Then from (2.33) with q4 = 2, (2.40) becomes
|
k12 = 12a( a - b ) - |
æ
è
|
a - |
b
2
|
ö
ø
|
2
|
|
| (2.41) |
Which reduces to
But under this condition k1 would have to be an integer for x1 and
x2 to be integers and this is not possible from (2.42). Therefore
q4 cannot be equal to 2, so that a in (2.20) and therefore z in (2.1) cannot
be integers. Thus, subject to q4 exhibiting satisfactory
characteristics, this proves Fermat's Last Theorem for n = 4.
M2 Version 1.0.0
Ó
P.G.Bass, April 2009
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