Example of Complex Pythagorean Triples

Question

I am looking for example of a Pythagorean Triple with Gaussian Integers.

I followed the links and looked at followings :

Relation to Gaussian integers in https://en.m.wikipedia.org/wiki/Pythagorean_triple

Links and Google searches mentioned in: Generating all the Pythagorean triples by factorizing using complex numbers

And

References for generating Pythagorean triple using complex number

Just looking for an example of a Pythagorean Triple as Gaussian integers, must be missing something obvious in all the above and their mentioned links to not come up with Just one example.

Answer 1

Examples :
$$(-4+1i)^2+(4+8i)^2=(4+7i)^2$$ $$(-4-1i)^2+(4-8i)^2=(4-7i)^2$$

Multiples :
$$(-8+2i)^2+(8+16i)^2=(8+14i)^2$$ $$(-12-3i)^2+(12-24i)^2=(12-21i)^2$$

By the way , we can see that all "Ordinary" Pythagorean triples are also "Complex" Pythagorean triples : (1) by setting Imaginary Part to $0$ or (2) by multiplying by $i$ or (3) by multiplying by arbitrary gaussian integer $z$ :
$$(3+0i)^2+(4+0i)^2=(5+0i)^2$$ $$(0+3i)^2+(0+4i)^2=(0+5i)^2$$ $$(3z)^2+(4z)^2=(5z)^2$$

ADDENDUM :

Some more Examples with Positive Integers :
$$( 1 + 5 i)^2+( 9 + 3 i)^2=( 8 + 4 i)^2$$ $$( 2 + 3 i)^2+( 6 + 2 i)^2=( 6 + 3 i)^2$$ $$( 2 + 9 i)^2+( 7 + 6 i)^2=( 6 + 10 i)^2$$ $$( 2 + 9 i)^2+( 9 + 2 i)^2=( 6 + 6 i)^2$$ $$( 3 + 6 i)^2+( 6 + 3 i)^2=( 6 + 6 i)^2$$ $$( 4 + 1 i)^2+( 7 + 4 i)^2=( 8 + 4 i)^2$$ $$( 5 + 5 i)^2+( 7 + 1 i)^2=( 8 + 4 i)^2$$ $$( 6 + 2 i)^2+( 2 + 3 i)^2=( 6 + 3 i)^2$$ $$( 6 + 3 i)^2+( 3 + 6 i)^2=( 6 + 6 i)^2$$ $$( 6 + 3 i)^2+( 8 + 4 i)^2=( 10 + 5 i)^2$$ $$( 7 + 1 i)^2+( 5 + 5 i)^2=( 8 + 4 i)^2$$ $$( 7 + 4 i)^2+( 4 + 1 i)^2=( 8 + 4 i)^2$$ $$( 7 + 6 i)^2+( 2 + 9 i)^2=( 6 + 10 i)^2$$ $$( 8 + 4 i)^2+( 6 + 3 i)^2=( 10 + 5 i)^2$$ $$( 8 + 9 i)^2+( 9 + 8 i)^2=( 12 + 12 i)^2$$ $$( 9 + 2 i)^2+( 2 + 9 i)^2=( 6 + 6 i)^2$$ $$( 9 + 3 i)^2+( 1 + 5 i)^2=( 8 + 4 i)^2$$ $$( 9 + 8 i)^2+( 8 + 9 i)^2=( 12 + 12 i)^2$$

Answer 2

Have you tried working through an algebraic derivation of the primitive Pythagorean triple parametric formula in $\mathbf Z$ to get an analogue in $\mathbf Z[i]$? Because $-1$ is a square, the terms in $\alpha^2 + \beta^2 = \gamma^2$ are essentially symmetric in $\alpha$, $\beta$, and $\gamma$, so we can assume without loss of generality that $\beta$ is divisible by $1+i$ while $\alpha$ and $\gamma$ are not. Then it turns out that $$ \alpha = u(z^2 - w^2), \ \beta = \pm 2uzw, \ \gamma = u(z^2+w^2) $$ where $u$ is a unit in the Gaussian integers, $z$ and $w$ are relatively prime in the Gaussian integers, and $z \not\equiv w \bmod 1+i$.