In the set $\mathbb{Z}[i]:=\{a+bi:a,b\in\mathbb{Z}\}$ of Gaussian Integers, we define the norm to be $N:\mathbb{Z}[i]\to\mathbb{Z}$ by $$ N(\alpha)=\alpha\alpha^*=a^2+b^2,\ \forall\alpha=a+bi\in\mathbb{Z}[i]. $$ We then use this to determine factorisations of Gaussian integers. Since if $\alpha\mid\beta$ in $\mathbb{Z}[i]$, then $N(\alpha)\mid N(\beta)$ in $\mathbb{Z}$. Further, we know that if $\eta$ is a highest common factor of $\alpha$ and $\beta$ in $\mathbb{Z}[i]$, and $h$ is a highest common factor of $N(\alpha)$ and $N(\beta)$ in $\mathbb{Z}$, it follows that $N(\eta)\mid h$. My question is this:
When does $N(\eta)\neq h$? Is it possible to characterise such a case in terms of a prime factorisation?
I feel as though this is related to Fermat's theorem regarding the parity of the exponents of primes $q\equiv 3 \pmod 4$, i.e. an integer $n$ is a sum of squares (and therefore a norm of a Gaussian integer) iff in a prime factorisation $$ n=2^{a_0}p_1^{a_1}\dotsm p_r^{a_r}q_1^{b_1}\dotsm q_s^{b_s} \qquad (*)$$ where $p_i\equiv 1\pmod 4,\ q_i\equiv 3\pmod 4$ are prime and $a_i,b_i\in\mathbb{Z},\ \forall i$, has that all $b_i$ are even, but I've struggled to relate the two.
Thanks in advance.
Edit: I've been trying to construct examples with some level of difficulty, and I'll share my attempts here.
First of all, we know that since $N(\eta)\mid h$ and $N(\eta)\neq h$, we must have that $h\neq 1$. Then $N(\alpha)$ and $N(\beta)$ must not be coprime. Further, for such Gaussian integers to exist, they must have the form of $(*)$. Then pick primes congruent to $1\bmod 4$, $\ p_1,p_2,p_3,p_4$ (or equivalently, take primes congruent to $3\bmod 4$, but square them) and let $$ N(\alpha)=p_1p_2p_3, \quad N(\beta)=p_1p_2p_4. $$ Then $h=p_1p_2$, and we may have $N(\eta)=p_1$ or $N(\eta)=p_2$, in particular, $N(\eta)\neq h$. Does this seem to be a reasonable method? Or is there some easier way.
Alright, let's plug in some specific numbers into the formula you built: $p_1 = 2 - i$, $p_2 = 2 + i$, $p_3 = 3 - 2i$, $p_4 = 3 + 2i$. Then $\alpha = 15 - 10i$, $N(\alpha) = 325$, $\beta = 15 + 10i$ and $N(\beta) = 325$ as well.
Clearly $\gcd(15 - 10i, 15 + 10i) = 5$, which has a norm of $25$, not $325$. It checks out, it might be the most elegant example possible.
This is what I was groping towards, so I'm glad I helped you get there.