2009/020) Pythagorean triplets.- Algebraic way

There is an algebraic way getting Pythagorean triplets

we know

x^2+y^2 = z^2 become a Pythagorean triplet if we have x y and z all to be integers.

If we can find x,y and z all to be rational then multiplying them by LCM of the denominator we are through.

To sort it we start with

n^2+(2n+1) = (n+1)^2 ..1

we know that n^2 and (n+1)^2 are squares and if we chose 2n+1 to be a square then we are through by proper transformation.

We cannot choose n to be an integer as we shall loose some values in the process so we shall put n = p/q but that is la

However we must have to chose 2n+1 to be a whole square say m^2 to get a Pythagorean triplet

So 2n = m^2 – 1

To avoid denominator in 1 we multiply (1) by 4 to get

4n^2 + 4(2n+1) = (2n+2)^2

Or (m^2-1)^2 + 4m^2 = (m^2+1)^2

so
2m, (m^2-1) and (m^2+1) satisfy the condition.(a^2+b^2 = c^2)

But m is not necessarily an integer and by choosing m = p/q and multiplying by q^2 we get

2pq, (q^2-p^2) and (q^2+p^2) satisfy the equation and if p and q are relatively prime then we get triplet with no common factor and hence it is a primitive triple..

Multiply this by any other integer and we shall get another triple(not primitive)