Fun with Mathematics

One more general proof for Fermat's Last theorem
The equal power relation in the form a^n + b^n = c^n cannot be existing with integral values for a,b and c and for any value of n except 1 and 2.
This relation ,if it exists ,can then be expressed in the Pythagorean form as
(a^n/2)^2 +(b^n/2)^2 = (c^n/2)^2
where the Pythagorean triple is a power of numbers.
All Pythagorean triples (x,y,z) must satisfy a condition xy = 2n(z+n)
Applying this condition to our equal power relation in the converted form ,we have,
ab(ab)^1/2 = 2n [c(c)^1/2 +n]
When a and b are odd ab(ab)^1/2 will be odd, but cannot be equated with 2n[c(c)^1/2+n] which is even for all values of c and n.

If a is even and b is also even, then c must be even. It is reducible without violating the equality of the
power relation . The reduction is stopped only when one of the members of the triple becomes odd.
If a is even, b is odd, c should necessarily be odd, then n must be either even or odd.
In (x,y,z) if x is even and y and z are odd it can be written as
z^2 -y^2= x^2
then z-y = x-2n
squaring and simplification gives,
y(z-y) = 2n(x-n)
Applying this condition to our equal power relation in the converted form,
b(b)^1/2 [c(c)^1/2 - b(b)^1/2] = 2n [a(a)^1/2- n]
when c and b are odd  LHS will be odd but RHS is even for all odd and even values of n and a.
As it contradicts , we can conclude that the equal power relation cannot be existng

Fun with Mathematics

More about the Pythagorean triples

1.In all Pythagorean triples (a,b,c) with c>b>a ,both a and b together cannot be odd.
Proof:
If a^2+b^2 = c^2, then a+b = c+2n where n = 1,2,3,4....
Squaring both sides and after simplification
ab = 2n(c+n)
That is the product of a and b is even which means that both a and b cannot be odd.

2.           2n^2 + 2cn -ab = 0
solving for n, we get,
n = (-1/2) + [1/2(c^2+2ab)^1/2]
It implies that c^2 +2ab must be a square number and if c is even c^2+2ab will be an even square and
if c is odd, it will be an odd square.

3. Again 2n^2= [an-2cn]
c+2n = a+b
 cubing both sides and after simplification
c^3-a^3-b^3 = 3ab(a+b)-8n^3-6cn(c+2n)
3(a+b)(ab-2cn)- 8n^3
(a+b)6n^2-8n^3
It shows that c^3-a^3-b^3 is divisible by 2 and also by a square number n^2 where n = (1/2)(a+b-c) 

Fun with Mathematics

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.

Fun with Mathematics

Digital ending method of getting conditional Pythagorean triples

A Pythagorean triple (a,b,c) must satisfy the relation a^2 + b^2 = c^2
where c >a and c > b but a+b > c.
The values of a,b and c must be such that the digital endings of both sides must be same..
Table given below shows the digital ending of c for all possible digital endings of a and b.
Table.: Digital ending of c in a^2 + b^2 = c^2
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      x      5     0       x    4,6     x      0      5      x
2                       2      5      x     x       0    3,7     0      x      x      5
3                       3      0      x     x       5    2,8     5      x      x      0
4                       4      x      0     5       x    1,9     x      5      0      x
5                       5    4,6   3,7  2,8    1,9    0     1,9   2,8   3,7   4,6
6                       6     x       0     5       x    1,9     x      5      0      x
7                       7     0       x     x       5    2,8    5       x      x      0
8                       8     5       x     x       0    3,7    0       x      x      5
9                       9     x       5     0      x     4,6    x       0      5      x

It is noted that certain digital ending combinations for a and b are forbidden. For example both a,and b
cannot be ending with identical digits except 5
It is found that a^2 - a = a(a-1) is always equal to some multiples of 2,hence
a^2 = a+ an(a)
b^2 = b + 2n(b)
c^2 = c+2n(c)
It gives a condition
a +b = c + 2n
According to the requirement of digital endings of the equal power relation ,one can assume the following
typical relation as an example,
(10x+7)^2 + (10y+4)^2 = (10z+5)^2
Since a+b = c+2n, 10(x+y)+11 = 10z+5 + 2n
It demands that n = (3+5k) with k = 0,1,2,3......and x+y = z+k
With k = 0 , x+y = z
(10x+7)^2 + (10y+4)^2 = [10(x+y)+5]^2
it gives y = 2(x+1)/10x+1
When x = 0; y=2,z=2 ; 7^2+24^2=25^2
when x=1 ,y=4/11,z=15/11; 187^2+84^2 = 205^2
when x=2 , y = 6/21,z=48,21; 567^2 +144^2= 585^2
when x = 3, y=8/31,z=101/31 ; 1147^2+204^2 = 1165^2
When k = 1, x+y= z+1
(10x+7)^2 + (10y+4)^2 = [10(x+y)-5]^2
It gives y = 12x+2/10x-9
when x=1 ,y=14,z=14 ;17^2+144^2=145^2
when x=2,y=26/11,z=37/11;  297^2+304^2 = 425^2