EDIT: When I wrote this post originally, I was not careful about the distinction between $\displaystyle \mathbb{Q}(\sqrt{-1})$ and $\displaystyle \mathbb{Z}[i]$. Here's what I wrote, slightly edited:

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tonio's first post (post #2 of this thread) was not "generic" but was a direct and concrete answer to your questions. To spell it out more explicitly,

(1) The answer is, there does not exist any such Gaussian prime.

Further reading:

Gaussian Prime -- from Wolfram MathWorld Gaussian integer - Wikipedia, the free encyclopedia (subheading: As a unique factorization domain)

Note: I don't know why tonio mentions $\displaystyle 1+i$ separately, as it fits the description $\displaystyle a+bi$ where $\displaystyle N(a+bi)=a^2+b^2=p$ with $\displaystyle \displaystyle p$ a regular prime in $\displaystyle \mathbb{Z}$.

Correct. I was just thinking of primes which are 1 modulo 4 when I wrote $\displaystyle a^2+b^2=p$ , and tried to distinguish the first prime 2 which isn't 1 modulo 4 and still fulfills the condition: $\displaystyle 1^2+1^2=2$ , but I didn't write down the distinction.

All the above was thought to convey the fact (theorem) that a prime is a sum of squares iff it is 2 or it is 1 mod 4 , and from here to deduce, together with a little algebra, that an integer is a sum of squares iff any prime equal to 3 mod 4 that appears in its prime decompostion appears there to an even power.

Thanx

Tonio
(2) tonio gave you the prime factorisation of 6 in $\displaystyle \mathbb{Z}[i]$. Note that 2*3 is not a prime factorisation here because 2 is not a Gaussian prime.

Further reading:

Unique factorization domain - Wikipedia, the free encyclopedia Unique Factorization Domain -- from Wolfram MathWorld
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So, the above applies to $\displaystyle \mathbb{Z}[i]$. Question to the OP (Samson): Are you sure you want to know about primes in $\displaystyle \mathbb{Q}(\sqrt{-1})$ as opposed to $\displaystyle \mathbb{Z}[i]$? If so, I'll have to think/research more, or someone else will have to add some info..