A simple proof for Fermat’s last theorem
Proof.1
The relation a^n + b^n = c^n where a,b,c and n are all integral
values is true when n takes value either 1 or 2 ,but not existing
for all values n ≥ 3.
Let us suppose that a^3 + b^3 = c^3 is existing. Since cube of
a number x can be expressed as
x^3 = 6k + x
where k is a number ,all cubical relations irrespective of their
reality must satisfy a condition, the sum of the cube roots in
one side must be greater than by 6 or multiple of 6 with the
sum of the cube roots in the other side of the numeral relation.
Applying this unique condition to our cubical relation we have,
a+b = c +6m
Rearranging the terms after squaring both sides,
a^2+b^2 –c^2-36m^2= 2 [6cm-ab]
(a+c)(a-c) +(b-6m)(b+6m)= (a+c)[-(b-6m)]+(b-6m)(b+6m)
= (b-6m)[b+6m-a-c]
(a-6m)(b-6m)= (ab-6cm)
cubing both sides of the unique conditional relation,
a^3+b^3+3ab(a+b)= c^3 +216m^3 + 18cn(c+6m)
If a^3+b^3 = c^3 is true, then
72m^3 – (a+b)[ab-6cm]=0
72m^3 = (a+b)(b-6m)(a-6m) must also be true.
When m = 1
72 = (a+b) (a-6)(b-6)
If a and b are odd then c must be even.If a is odd and b is even,
then c must be odd.If the ending digit of a and b is fixed, then c
cannot have all possible ending digit.a^3+b^3=c^3 demands
that the ending digit of the sum of the cubes of a and b must
be equal to the ending digit of the cube of c.
The following table shows the ending digit of c for the given
ending digit of a and b
ending ending digit of a
digit of_______________________________________________
b 0 1 2 3 4 5 6 7 8 9
______________________________________________________
0 0 1 2 3 4 5 6 7 8 9
1 1 8 9 2 5 6 3 4 7 0
2 2 9 6 5 8 7 4 1 0 3
3 3 2 5 4 1 8 7 0 9 6
4 4 5 8 1 2 9 0 3 6 7
5 5 6 7 8 9 0 1 2 3 4
6 6 3 4 7 0 1 8 9 2 5
7 7 4 1 0 3 2 9 6 5 8
8 8 7 0 9 6 3 2 5 4 1
9 9 0 3 6 7 4 5 8 1 2
For example, if a = 10x+3,b= 10y +5, then c= 10z+8
Substituting this conditional values,we have,
72 = [10(x+y)+8][(10x-3)(10y-1)]
= [10(x+y)+8][10(x-1)+7][10(y-1)+9]
Since the product of 3 numbers ending with 8,7 and 9 does not
end with 2 ,this auxiliary condition stands as a proof for the
non-existence of the initially assumed cubical relation.
This is true for all values of m. For example when m = 2
576 = [10(x+y)+8][10(x-1)+1][10(y-1)+3]
where LHS ends with 6 while RHS ends with 4,which is
quite absurd for numeral equations.
Proof.2
If a ends with 3 and b ends with 5, then c ends with 8.
Let a = 10x+3, b= 10y+5 and c = 10z +8
Substituting these values in the unique condition of the
cubical relation a+b= c+6m we have,
10(x+y)+8 = 10z + 8 + 6m
10(x+y) = 10z + 6m
m cannot take all possible values. The acceptable values
of m are
m = 5p where p = 1,2,3,4,……
or x+y = z +3p
when p = 1 x+y = z +3
Since a<c and b<c and a+b>c
X= z then y =3 or in general x = z-q then y = q+3
[10(z-q)+3]^3 +[10(q+3)+5]^3 = (10z+8)^3
When q=0
(10z+3)^3 +35^3 = (10z+8)^3
1000z^3 +900z^2 +270z +27 + 42875
= 1000z^3 + 2400z^2 +1920z +512
or 1500z^2+1650z – 42390 = 0
50z^2 +55z – 1413 = 0
The quadratic equation in z does not have any integral solutions .
The possible solution is
Z = [-55 ±√ 3025 +282600]/100 = 4.7943895
= [-11+ 5√457]/20
This is true for all values of p and q, which concludes that
a^3 + b^3 = c^3 is not true.
There two more proofs,which we will see in the next issue.
I hope my argument is sensible. (meydhanam@gmail.com)
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