I have the next question.
Let be $F=\mathbb{Q}[i]$. Also let $\alpha=a+bi$ and $\beta=c+di$ be non associate primes in $\mathbb{Z}[i]$ such that $N(\alpha),N(\beta)\equiv\;1\;(\mathrm{mod}\;4)$ and $N(\alpha)=N(\beta)$
Let $\alpha\mathbb{Z}[\alpha]=N(\alpha)\mathbb{Z}+\alpha\mathbb{Z}$ and $\beta\mathbb{Z}[\beta]=N(\beta)\mathbb{Z}+\beta\mathbb{Z}$
I am trying to figure out to which set is equal $\alpha\mathbb{Z}[\alpha]+\beta\mathbb{Z}[\beta]$.
Since $N(\alpha)=N(\beta)$, one solution is $\beta=\overline{\alpha}=a-bi$ therefore
$$\alpha\mathbb{Z}[\alpha]+\beta\mathbb{Z}[\beta]=N(\alpha)\mathbb{Z}+\alpha\mathbb{Z}+ N(\alpha)\mathbb{Z}+\overline{\alpha}\mathbb{Z}=N(\alpha)\mathbb{Z}+\alpha\mathbb{Z}+\overline{\alpha}\mathbb{Z}$$
For the part $K=\alpha\mathbb{Z}+\overline{\alpha}\mathbb{Z}$ I have that $x\in K$ if and only if $x=a(m+n)+b(m-n)i$ for $m,n\in \mathbb{Z}$ since $m$ and $n$ can be any integer $x$ can be written as $x=as+b(s+2t)i$ where $s, t\in\mathbb{Z}$.
Therefore
$\alpha\mathbb{Z}[\alpha]+\beta\mathbb{Z}[\beta]=\{Nr+as+b(s+2t)i\;:\; r,s,t\in\mathbb{Z}\}$
Since $(N,a)=1$ we have that there exist integers $r,s$ such that $1=Nr+as$ and hence we have freedom on that part however, in the case of the imaginary part we do not have freedom since depends of the integer $s$ so I am not sure how to proceed.