A simple and natural proof for Fermat's Last Theorem - 2
2.Conservation of odd-evenness in equal odd power relations
Before to understand the simple,natural and marvellous proof (may be the one what Fermat missed to add in the margin), let us be conversant with the inherent mathematical properties associated with equal power relations xn + yn = zn. When n is odd the conservation of odd-evenness does not permit the integer solutions.
This can be exemplified with the general form of cubical relation itself.
x3 + y3 = z3
(x+y) (x2 - xy + y2) = 8nz3
If x and y are both odd, then x+y will be even, while x2 - xy + y2 will be odd. This relation shows that the evenness of (x+y) is exactly same as theat of the evenness of z3.If z is singly even its value will be 8nz3, if it is doubly even, it will be 64nz3 and so on so fourth. If z is signly even z3 = 8 nz3 where nz is odd.Hence x+ y = 8 or 8nz [ x + y > z (=2nz) ] and (x2 - xy + y2) =nz3 or nz2 [(x2 - xy + y2) < z2(= 4nz2) ]
If x+y = 8, then y = 8- x ,on substitution in the other relation we get
3x2 - 24 x + 64 - nz3 = 0
It gives x = [12 - √3 nz3 -48 ] / 3 and y = [12 + √3nz3 -48] / 3 , both x and y will be irrational for all real and integer values of nz.
If x+y = 8nz, then y = 8nz - x
3x2 -24 nz x + 64 nz 2 - nz 2 = 0
3x2 -24 nz x + 63 nz 2 = 0
It gives x =[ 8nz +i nz √ 20 ] / 2 and y = [8nz - i nz √ 20 ] / 2
Both x and y become complex.
For any even integer values for z, both x and y are either irrational or complex. This is true even for odd integer values for z.
The concept of conservation of odd-evenness proves the FLT for n = 3.
If z is doubly even then z3 = 64 nz3 so that x+y= 64 or 64 nz and (x2 - xy + y2) = nz3 or nz2
It gives the value of x and y in terms of nz
x = [96 + i √ 3nz3 - 3072]/ 3 and y = 96 + i √3nz3 - 3072]/ 3
2.Conservation of odd-evenness in equal odd power relations
Before to understand the simple,natural and marvellous proof (may be the one what Fermat missed to add in the margin), let us be conversant with the inherent mathematical properties associated with equal power relations xn + yn = zn. When n is odd the conservation of odd-evenness does not permit the integer solutions.
This can be exemplified with the general form of cubical relation itself.
x3 + y3 = z3
(x+y) (x2 - xy + y2) = 8nz3
If x and y are both odd, then x+y will be even, while x2 - xy + y2 will be odd. This relation shows that the evenness of (x+y) is exactly same as theat of the evenness of z3.If z is singly even its value will be 8nz3, if it is doubly even, it will be 64nz3 and so on so fourth. If z is signly even z3 = 8 nz3 where nz is odd.Hence x+ y = 8 or 8nz [ x + y > z (=2nz) ] and (x2 - xy + y2) =nz3 or nz2 [(x2 - xy + y2) < z2(= 4nz2) ]
If x+y = 8, then y = 8- x ,on substitution in the other relation we get
3x2 - 24 x + 64 - nz3 = 0
It gives x = [12 - √3 nz3 -48 ] / 3 and y = [12 + √3nz3 -48] / 3 , both x and y will be irrational for all real and integer values of nz.
If x+y = 8nz, then y = 8nz - x
3x2 -24 nz x + 64 nz 2 - nz 2 = 0
3x2 -24 nz x + 63 nz 2 = 0
It gives x =[ 8nz +i nz √ 20 ] / 2 and y = [8nz - i nz √ 20 ] / 2
Both x and y become complex.
For any even integer values for z, both x and y are either irrational or complex. This is true even for odd integer values for z.
The concept of conservation of odd-evenness proves the FLT for n = 3.
If z is doubly even then z3 = 64 nz3 so that x+y= 64 or 64 nz and (x2 - xy + y2) = nz3 or nz2
It gives the value of x and y in terms of nz
x = [96 + i √ 3nz3 - 3072]/ 3 and y = 96 + i √3nz3 - 3072]/ 3
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