Yet another proof for Fermat's last theorem
Dr.M.Meyyappan,Dean of Science
Sri Raja Raajan College of Engineering and Technology
Amaravathipudur-630301, Tamilnadu,India
The numeral relation in the form a^n +b^n = c^n is found to exist only when a,b,c and n are all whole
integers provided n = 1 or 2.
When n=1; a+b=c ,then a^n+b^n < c^n for n > or = 2
When n =2 ; a^2+b^2 = c^2 then a^n+b^n> c^n for n <2
a^n+b^n < c^n for n >2
The problem in the conditional relation a^n+b^n=c^n begins from n = 3 and it prevails for all higher
values n >3 .
a^n+b^n =/=c^n when n > or = 3
Let us assume that a^3 +b^3 = c^3 exists with a,band c are all whole numbers. Since sum of two
cubes is equal to a cube, c>a,c>b, but c <a+b .
The cube of a number x can be expressed as x^3 = (x-1)x(x+1) +x
For any product of three successive numbers ,6 will invariably be a factor, and hence,
x^3 = 6n(x) + x
Applying this fact to the cubical relation assume,we have
a+b-c = 6[n(c)-n(a)-n(b)] = 6n
or a+b = c+6n
For a given a and b, that is for given digital endings of a and b, the digital ending of c cannot
have all possible value, but is fixed ,because the digital ending of c^3 must be equal to the
sum of the digital ending of a^3 and b^3.
Table given below shows that digital ending of c for all possible digital endings of a and b
Digital ending of c in a^3 +b^3 = c^3
a / 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
According to the requirement of digital endings of the equal power relation, one can assume the
following typical relation as an example.
(10x+7)^3 + (10y+4)^3 = (10z+3)^3
Fermat's last theorem demands that x,y,z all cannot be whole numbers simultaneously to satisfy the
relation. Since c>b>a (or c>a>b) ,one can assume that c =b+m, where m = 1,2,3,......
The acceptable and permitted values of m according to digital ending requirement are
9,19,29..... or in general (10k +9) with k =0,1,2,3,.......
It demands that z = y +1
Since a+b = c+6n
10(x+y)+11 = 10z +3 + 6n
The digital ending requirement allows only certain quantized value n = (3+5m)
where m = 0,1,2,3,......It demands that x+y= z+1 or x = 2
By introducing these derived conditions, we can express the typical cubical relation in terms of one
variable y.
27^3 + (10y+4)^3 = (10y+13)^3
90y^2 + 153y - 585 = 0
Solving this quadratic equation for y,
y = [-153 + (2340090)^1/2]/180
= 1.8374708..... or -3.5374708.....
It gives
27^3 + (22.374708...)^3 = (31.374708...)^3
It clearly shows that when a is taken as whole number, then both b and c cannot be whole
integers in the relation a^3+b^3 = c^3. Hence Fermat's last theorem is true for n = 3.
Dr.M.Meyyappan,Dean of Science
Sri Raja Raajan College of Engineering and Technology
Amaravathipudur-630301, Tamilnadu,India
The numeral relation in the form a^n +b^n = c^n is found to exist only when a,b,c and n are all whole
integers provided n = 1 or 2.
When n=1; a+b=c ,then a^n+b^n < c^n for n > or = 2
When n =2 ; a^2+b^2 = c^2 then a^n+b^n> c^n for n <2
a^n+b^n < c^n for n >2
The problem in the conditional relation a^n+b^n=c^n begins from n = 3 and it prevails for all higher
values n >3 .
a^n+b^n =/=c^n when n > or = 3
Let us assume that a^3 +b^3 = c^3 exists with a,band c are all whole numbers. Since sum of two
cubes is equal to a cube, c>a,c>b, but c <a+b .
The cube of a number x can be expressed as x^3 = (x-1)x(x+1) +x
For any product of three successive numbers ,6 will invariably be a factor, and hence,
x^3 = 6n(x) + x
Applying this fact to the cubical relation assume,we have
a+b-c = 6[n(c)-n(a)-n(b)] = 6n
or a+b = c+6n
For a given a and b, that is for given digital endings of a and b, the digital ending of c cannot
have all possible value, but is fixed ,because the digital ending of c^3 must be equal to the
sum of the digital ending of a^3 and b^3.
Table given below shows that digital ending of c for all possible digital endings of a and b
Digital ending of c in a^3 +b^3 = c^3
a / 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
According to the requirement of digital endings of the equal power relation, one can assume the
following typical relation as an example.
(10x+7)^3 + (10y+4)^3 = (10z+3)^3
Fermat's last theorem demands that x,y,z all cannot be whole numbers simultaneously to satisfy the
relation. Since c>b>a (or c>a>b) ,one can assume that c =b+m, where m = 1,2,3,......
The acceptable and permitted values of m according to digital ending requirement are
9,19,29..... or in general (10k +9) with k =0,1,2,3,.......
It demands that z = y +1
Since a+b = c+6n
10(x+y)+11 = 10z +3 + 6n
The digital ending requirement allows only certain quantized value n = (3+5m)
where m = 0,1,2,3,......It demands that x+y= z+1 or x = 2
By introducing these derived conditions, we can express the typical cubical relation in terms of one
variable y.
27^3 + (10y+4)^3 = (10y+13)^3
90y^2 + 153y - 585 = 0
Solving this quadratic equation for y,
y = [-153 + (2340090)^1/2]/180
= 1.8374708..... or -3.5374708.....
It gives
27^3 + (22.374708...)^3 = (31.374708...)^3
It clearly shows that when a is taken as whole number, then both b and c cannot be whole
integers in the relation a^3+b^3 = c^3. Hence Fermat's last theorem is true for n = 3.
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