Fun with mathematics-14

 Fermat’ last theorem-5

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

of Engineering & Technology, Amaravathipudur-630301

Tamilnadu.India)

Proof.7

Like cube of a number n is  in the  form of 6x + n, the fifth

power of a number is in the form of 30x +n. The

representation of any power number in this form is useful

 to prove the Fermat’ last theorem.

a^5 +b^5 = c^5 is supposed to be true.

Or a^5+b^5-c^5=0

a^5 = 30[x(a)] + a

b^5 = 30[x(b)]+b

a^5 +b^5 = 30[x(a)+x(b)]+(a+b)

c^5 = 30[x(c)] +c

 

substituting these values in the supposed relation,

a^5+b^5-c^5= (a+b-c)+30[x(a)+x(b)-x(c)]

If a+b-c=0, a+b=c then a^5+b^5≠ c^5

as (a+b)^5 contains additional terms besides

a^5+b^5.

If x(a)+x(b)-x(c) = 0, x(a)+x(b) = x(c)

c^5= 30 x(c)  +c

       = 30[x(a)+x(b)]+c

as c ≠ a+b ,it is not equal to a^5 +b^5.

This method can even be generalized.

a^n+b^n=c^n

is supposed to be existing.

a^n = kx(a) + a

b^n = kx(b) + b

 c^n = kx(c) + c

a^n + b^n- c^n  = k[x(a)+x(b)- x(c)] + (a+b-c)

 If a+b= c then a^n +b^n ≠ c^n

If x(a)+x(b)= x(c), then a^n+b^n≠ c^n.