Fermat's Last Theorem- Applications

If x is given a whole integer. Then

x3 = z3 - y3 = (z - y) (z2 + zy + y2)

x(odd) = P01P02 and x3 = P013P023

Since x+y > z , x > z - y

The distribution of the prime factors of x3 among its two product factors should satisfy this condition.

z - y = P01 then z2 + zy + y2 =P012P023

z= P01 + y substituting this value of z in the other equation

z2 + zy + y2 = 3y2 + 3y P01 + P012 = P012P023

3y2 + 3y P01 + P012 - P012P023 = 0

This is a quadratic equation in terms of y, solving for y we get,

y = [ -3P01 + √ 9 P012 -12 (P012 - P012P023)] / 6

y = [ -3P01 + √ 12 P012P023 - 3P012 ] /6

If x = P01P02 = 3x7 , then

y = [-3 + √ 4113] / 2 and c = [3 + √ 4113] / 2

The conservation of odd-evenness of the cubical relations demands that z - y = 23n where n is a number.

when n=1 z - y= 8 and y is singly even,when n = 2 x - y = 64 and y is doubly even and so on.

y 3 = 23n P013 ........... (any number of odd prime factors)

y3 = z3 - x3 = (z - x) (z2 +zx + x2)

where z - x is even where as z2 +zx + x2 is odd, hence z - x = 23n or one or more prime factors may be attached with it.. All the remaining prime factors multiplied together must be equal to z2 +zx + x2 . For example

y = 2P01 then y3 = 23P013 = z3 - x3 = (z - x) (z2 +zx + x2)

z - x = 8 or z = x + 8. Substituting this value of z in z2 +zx + x2 = P013

3a2 +24a +64 - P013 = 0

or a =[ -12 + √ 3P013 - 48] / 3

If P01 = 7 , x = - 4 + √109 and z = 4 + √ 109

If P01 = 11, x = [ -12 + √3945]/3 and z = x+ 8 = [ 12 + √ 3945]/3

If x is given a whole integer. Then

x3 = z3 - y3 = (z - y) (z2 + zy + y2)

x(odd) = P01P02 and x3 = P013P023

Since x+y > z , x > z - y

The distribution of the prime factors of x3 among its two product factors should satisfy this condition.

z - y = P01 then z2 + zy + y2 =P012P023

z= P01 + y substituting this value of z in the other equation

z2 + zy + y2 = 3y2 + 3y P01 + P012 = P012P023

3y2 + 3y P01 + P012 - P012P023 = 0

This is a quadratic equation in terms of y, solving for y we get,

y = [ -3P01 + √ 9 P012 -12 (P012 - P012P023)] / 6

y = [ -3P01 + √ 12 P012P023 - 3P012 ] /6

If x = P01P02 = 3x7 , then

y = [-3 + √ 4113] / 2 and c = [3 + √ 4113] / 2

The conservation of odd-evenness of the cubical relations demands that z - y = 23n where n is a number.

when n=1 z - y= 8 and y is singly even,when n = 2 x - y = 64 and y is doubly even and so on.

y 3 = 23n P013 ........... (any number of odd prime factors)

y3 = z3 - x3 = (z - x) (z2 +zx + x2)

where z - x is even where as z2 +zx + x2 is odd, hence z - x = 23n or one or more prime factors may be attached with it.. All the remaining prime factors multiplied together must be equal to z2 +zx + x2 . For example

y = 2P01 then y3 = 23P013 = z3 - x3 = (z - x) (z2 +zx + x2)

z - x = 8 or z = x + 8. Substituting this value of z in z2 +zx + x2 = P013

3a2 +24a +64 - P013 = 0

or a =[ -12 + √ 3P013 - 48] / 3

If P01 = 7 , x = - 4 + √109 and z = 4 + √ 109

If P01 = 11, x = [ -12 + √3945]/3 and z = x+ 8 = [ 12 + √ 3945]/3

## No comments:

## Post a Comment