Is the pytogorian triple a^2+b^2=c^2 unique? and Is there is any formula to generate the numbers which satisfy the equation...

No they are not unique.  The smallest
is

(63)^2 + (16)^2 =  65^2 = (33)^2 +
(56)^2

I have not been able to find a formula for
them.

If you do not have to have the sum a
square,

(8)^2 + (1)^2 = (7)^2 + (4)^2  (64 + 1 = 49 + 16)
and this gives the seeds for the above equation by using the formula for generating
pythagorean squares,
if n = 8, m = 1 you get 63^2 + 16^2 = 65^2,
and
if n = 7, m = 4 you get (33)^2 + (56)^2 =
65^2

Here are some others:

(2,
11), (5, 10)
(3, 14), (6, 13)
(2, 9), (6, 7)
(4, 17), (7,
16)
(1, 18), (6, 17), (10, 15)
(5, 20), (8, 19), (13,
16)
(3, 11), (7, 9)
(6, 23), (9, 22)
(4, 13), (8,
11)
(7, 26), (10, 25), (14, 23)
(3, 28), (8, 27)
(5, 15),
(9, 13)
(8, 29), (11, 28)
(2, 16), (8,
14)

These are pairs of m and n, so use the formula for
generating Pythagorean triples to get the actual
numbers.

(16^2-2^2)^2 + (2(2)(16))^2 =
(2^2+16^2)^2
(14^2-8^2)^2 + (2(8)(14))^2 = (14^2+8^2)^2
and
(2^2+16^2) = (14^2+8^2)

No they are not
unique.