Fermat’s last theorem (continuation)
(M.Meyyappan., Professor of Physics,Sri Raaja Raajan College
Engineering and Technology,Amaravathipudur-630301,
Sivagangai dist,Tamilnadu)
There are few more proofs to arrive at the same conclusion.
In this issue two proofs are described
There is yet another proof for the Fermat’s last theorem.
If the two cube roots summed up end with say 3 and 5
Respectively, then the resultant cube root should end with 8,
since the sum of the cubes end with 2.
(10x+3)^3 +(10y+5)^3 = (10z+8)^3
since a+b = c+6m, we have the condition x+y= z+3p
Expanding them,
1000(x^3+y^3) +100(9x^2+15y^2)+ 10(27x +75y)+(27+125)
= 1000z^3 + 100(24z^2) +10 (192z) +512
where the last digit ends with 2 in both sides.
Now computing the digit of the 10-th decimal place in both sides and equating them,
7x+5y+2+2 +1 (carry of the last digit)= 2z +1
7x+5y +5 = 2z+1
substituting for z, we get,
5x+3y+4 = -6p
for positive values of x and y, this relation cannot be existing.
Proof.4
The difference between the cube of a number and its cube root
is always divisible by 6.
x^3-x = x(x^2-1) = (x-1)x(x+1)
The product of any three successive numbers will always be
divisible by 6 .
a^3 + b^3 = c^3
a^3 + b^3 = a+b + 6n(a) + 6n(b)
c^3 = c + 6n(c)
a^3+b^3-c^3 = (a+b-c) + 6 [(n(a)+n(b) –n(c)]
If a+b-c = 0 a+b = c,which cannot be since c^3 = a^3+b^3.
If n(a)+n(b)-n(c) =0 n(c) = n(a)+n(b)
a^3+b^3 = a+b+ 6[n(a)+n(b)]= a+b+6n(c)
since c ≠ a+b , c^3 ≠ a+b +6n(c)
In all cases, a+b-c ≠ 0 and n(a)+n(b)-n(c) ≠ 0
Which necessitates that a^3 +b^3 ≠ c^3
Probably this may be the missing details
in Fermat’s scribbles !
(meydhanam@gmail.com)
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