A general proof for Fermat’s last theorem

(Dr.M.Meyyappan,Professor of Physics,Sri Raaja Raajan

College of Engineering and Technology,

Amaravathipudur-630301)

Equations of the form a^n + b^n = c^n  have no solutions,

when n is a whole number greater than 2 and when a,b and c

are positive whole numbers.

When n = 1, a+b will always be equal to c, but when n ≥ 2,

a+b will always be greater than c. Let

                         a+b = c+k

where k is a small quantity, added with c to $balance  a+b.

Greater the value of n ,smaller will be c  and greater will be k.

By manipulating this logical relation, one can argue the

non-realistic nature of the equal power relation

a^3 +b^3 = c^3 .

From the above logical relation

                       a+b = c[1 + k/c]

 or                   (a+b)/c = [1+ k/c]

cubing both sides, we have

  [ a^3+b^3 +3ab(a+b)]/c^3 = [1+k/c]^3

If a^3+b^3 = c^3 is true, then the relation

1 + 3ab/c^2 [1+k/c]  =  [1+k/c]^3,

will be true. If the later is not true, the former will also

be not true.By rearranging the final relation, we get,

        1 = [1+k/c][(1+k/c)^2 – 3ab/c^2],

where the product of two factors is equal to 1.When one

of the factors [1+k/c] is greater than unity, the other

factors must necessarily be lesser then unity say (1-y),

where y is positive.

(1+x) (1-y) = 1 +x-y-xy = 1

or,  x= y(x+1)

When y is positive, it is an impossible relation.

Taking the final relation,

1 = [1+k/c][ 1 – (3ab/c^2- k^2/c^2 – 2k/c)]

 Comparing with the general relation,

x=k/c  and y = 3ab/c^2 – k^2/c^2-2k/c

k/c = [(3ab/c^2-k^2/c^2-2k/c)][(k/c)+1]

Since 3ab/c^2 –k^2/c^2-2k/c is taken as a positive quantity,

the above relation cannot be existing. In such relations

3ab/ck > [(k/c)+2]

A similar argument can be given for odd power relations.

For example, when n = 5,

a^5 + b^5 = c^5

a+b > c or a+b = c+ k

(a+b)^5 = a^5+b^5 + 5ab(a^3+b^3)+ 10a^2b^2(a+b)

                                           = c^5(1+k/c)^5

Where

a^3+b^3 = c^3(1+k/c)[(1+k/c)^2- 3ab/c^2]

and

a+b = c(1+k/c)

substituting these values and dividing by c^5, we have,

1       + (5ab/c^2)(1+k/c)[(1+k/c)^2-3ab/c^2] +

                       (10a^2b^2/c^4)(1+k/c)

                        = (1+k/c)^5

or, 1 = (1+k/c) {(1+k/c)^4-(5ab/c^2)[(1+k/c)^2-3ab/c^2]

                 - 10a^2b^2/c^2]}

1 = (1+k/c) { 1- [(5ab/c^2)(1+k/c)^2+ 5 a^2b^2/c^4-

                         k^4/c^4+4k^3/c^3+6k^2/c^2+ 4k/c]}

In the product of two factors, since one is greater than 1,

the other one must be less than one.

1 = (1+x)(1-y)= 1 +x – y - xy

                     x = y(x+1)

More details can be obtained from

meydhanam@gmail.com