# 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 a^2+b^2=c^2+d^2? Using Euclid's formula we can generate pytogorian...

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 a^2+b^2=c^2+d^2?

Using Euclid's formula we can generate pytogorian triples. Like wise is there any formula to generate a^2+b^2=c^2+d^2?

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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.

I found a way to generate some, not all of the pairs of triples.

Let j > k > 0 where j and k are integers.

Use the following to generate (m, n) and (m',n') pairs for use in the formula for generation of Pythagorean Triples:

m=2jk+j+k+1, n=j-k and m′=2jk+j+k, n′=j+k+1

This will produce two pairs of numbers that m^2 + n^2 = m'^2 + n'^2

This can then be used to generate the numbers

(m^2-n^2), (2mn) and (m^2+n^2) for one pt and (m'^2-n^2), (2m'n') and the same (m'^2+n'^2).

These numbers will generate a^2 + b^2 = c^2 = d^2 + e^2 if Pythagorean Triples are what you needed.

An example j = 2, k = 1 we get

m = 8, n = 1 and m'=7, n'=4 and 8^2 + 1^2 = 7^2 + 4^2 which are not Pythagorean Triples but

(8^2-1^2)^2 + (2(8)(1))^2 = (7^2 - 4^2)^2 + (2(4)(8))^2 = (8^2+1^2)^2 which is a Pythagorean Triple.

Despite generating all primitive triples, Euclid's formula does not produce all triples. This can be remedied by inserting an additional parameter *k* to the formula. The following will generate all Pythagorean triples (although not uniquely):

a=k(m^2-n^2)

b=2mn

c=k(m^2+n^2)

where *m*, *n*, and *k* are positive integers with *m* > *n*.