Fun with Mathematics

Creative thoughts

From   R ^2   to   R ^2 – relation-II

          2  1           2  2

By taking different Pythagorean triples, one can construct

more such general relations for  R^2- relations  

                                               2  2

(5,12,13); (5m+13n)^2 +(12m+13n)^2

                                                  = (13m+17n)^2+(7n)^2

(8,15,17; (8m+17n)^2+(15m+17n)^2

                                                   =(17m+23n)^2+(7n)^2

(12,35,37); (12m+37n)^2+(35m+37n)^2

                                                 =(37m+47n)^2+(23n)^2

In an another method, for one smaller number, the biggest

number of the triple is multiplied with the other smaller

number and added with it, for the other smaller number ,

the biggest number is multiplied with the other smaller

number and subtracted with it. The sum of the squares

of them is equal to the sum of the squares of the biggest

number of the triple and its square. The general form of

this relation can be written as,

If a^2+b^2=c^2 than

(a+bc)^2+(b-ac)^2 = c^2 + (c^2)^2= c^2(c^2+1)

That is the sum, in this case, is equal to a product of

square of the biggest number of the triple and its next

higher number in the natural series.

It is interesting to know that a  R^2 is a self generator

of more                                    2  2                                                 

                                                      

such relations. If a^2+b^2=c^2+d^2, then

 (a+b)^2 +(a-b)^2= (c+d)^2+(c-d)^2

For example, 13^2+ 14^2 = 2^2+19^2

It gives, 27^2+1^2 = 21^2+ 17^2

Which in turn,

26^2+28^2 = 4^2+ 38^2, which is twice of the first new

relation generated.

                                       

To make  R^2 –relations with bigger and biggest numbers,one       

              2  2

can make use of two different Pythagorean triples instead of one.

With (a,b,c) and (x,y,z), we can generate,

(cz)^2= c^2(x^2+y^2) = (a^2+b^2)z^2

or, (cx)^2+(cy)^2 = (az)^2+(bz)^2

e.g., with (3,4,5) and (8,15,17) we get,

(3x17)^2+(4x17)^2 = (8x5)^2 +(15x5)^2

51^2+68^2=40^2+75^2

A general proof for Fermat’s last theorem

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

College of Engineering and Technology,

Amaravathipudur-630301)

Equations of the form a^n + b^n = c^n  have no solutions,

when n is a whole number greater than 2 and when a,b and c

are positive whole numbers.

When n = 1, a+b will always be equal to c, but when n ≥ 2,

a+b will always be greater than c. Let

                         a+b = c+k

where k is a small quantity, added with c to $balance  a+b.

Greater the value of n ,smaller will be c  and greater will be k.

By manipulating this logical relation, one can argue the

non-realistic nature of the equal power relation

a^3 +b^3 = c^3 .

From the above logical relation

                       a+b = c[1 + k/c]

 or                   (a+b)/c = [1+ k/c]

cubing both sides, we have

  [ a^3+b^3 +3ab(a+b)]/c^3 = [1+k/c]^3

If a^3+b^3 = c^3 is true, then the relation

1 + 3ab/c^2 [1+k/c]  =  [1+k/c]^3,

will be true. If the later is not true, the former will also

be not true.By rearranging the final relation, we get,

        1 = [1+k/c][(1+k/c)^2 – 3ab/c^2],

where the product of two factors is equal to 1.When one

of the factors [1+k/c] is greater than unity, the other

factors must necessarily be lesser then unity say (1-y),

where y is positive.

(1+x) (1-y) = 1 +x-y-xy = 1

or,  x= y(x+1)

When y is positive, it is an impossible relation.

Taking the final relation,

1 = [1+k/c][ 1 – (3ab/c^2- k^2/c^2 – 2k/c)]

 Comparing with the general relation,

x=k/c  and y = 3ab/c^2 – k^2/c^2-2k/c

k/c = [(3ab/c^2-k^2/c^2-2k/c)][(k/c)+1]

Since 3ab/c^2 –k^2/c^2-2k/c is taken as a positive quantity,

the above relation cannot be existing. In such relations

3ab/ck > [(k/c)+2]

A similar argument can be given for odd power relations.

For example, when n = 5,

a^5 + b^5 = c^5

a+b > c or a+b = c+ k

(a+b)^5 = a^5+b^5 + 5ab(a^3+b^3)+ 10a^2b^2(a+b)

                                           = c^5(1+k/c)^5

Where

a^3+b^3 = c^3(1+k/c)[(1+k/c)^2- 3ab/c^2]

and

a+b = c(1+k/c)

substituting these values and dividing by c^5, we have,

1       + (5ab/c^2)(1+k/c)[(1+k/c)^2-3ab/c^2] +

                       (10a^2b^2/c^4)(1+k/c)

                        = (1+k/c)^5

or, 1 = (1+k/c) {(1+k/c)^4-(5ab/c^2)[(1+k/c)^2-3ab/c^2]

                 - 10a^2b^2/c^2]}

1 = (1+k/c) { 1- [(5ab/c^2)(1+k/c)^2+ 5 a^2b^2/c^4-

                         k^4/c^4+4k^3/c^3+6k^2/c^2+ 4k/c]}

In the product of two factors, since one is greater than 1,

the other one must be less than one.

1 = (1+x)(1-y)= 1 +x – y - xy

                     x = y(x+1)

More details can be obtained from

meydhanam@gmail.com