2011/021) Pythagorean triplet with Gaussian Integer

Here we discuss the Pyhagorean triplet using Gaussian integers

As I have discussed in the previous blog entry about Gaussian integers that Gaussian integers are complex number of the form α = u + vi, where u and v are ordinary integers d i is the square root of minus one.

A primitive Pythagorean triple is one in which a and b are coprimes

That is integers (a,b,c) form a primitive Pythagorean triple where a^2+b^2= c^2 and

a and b are coprime

as a and b are co primes both cannot be even and only one can be odd. Both cannot be odd as

a mod 2 = 1 and b mod 2 = 1 shall give a^2 + b^2 mod 4 = 2 which is not a perfect square.

Now c^2 = (a^2+ b^2) = (a+ib)(a-ib)

Now both factors a + ib and a-ib cannot share any common factor that is they most be coprime to each other

So both of them must be perfect squares
Let
(a+ib) = (u+iv)^2 (and so (a-ib) = (u-iv)^2)

squaring RHS we get

a = u^2-v^2
b= 2uv

and c = (u+iv)(u-iv) = (u^2+v^2)

so we get

(u^2-v^2, 2uv, u^2+v^2) as Pythagorean triplet