Is it possible to find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i)=(\alpha)$?

Question

Is it possible to find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i)=(\alpha)$?

Is anyone could give me a full explication in ''Answer the question''?

Answer 1

$$\begin{align*}&\frac{3+5i}{1+3i}=\frac{-18-4i}{10}=-\frac95-\frac25i\cong-2\\{}\\ &\frac{1+3i}{-2}=-\frac12-\frac32i\cong-1-i\\{}\\ &\frac{-2}{-1-i}=\frac{2-2i}2=1-i\\{}\\ &-\frac{1+i}{1-i}=\frac{2i}2=i\end{align*}$$

and since $\;i\;$ is a unit we're finished, so $\gcd(3+2i,\,1+3i)=1-i\;$, so

$$\langle\,3+5i,\,1+3i\,\rangle=\langle\,1-i\rangle.$$

Answer 2

A quickish of doing this particular case is the following.

One can prove easily that $1 + i$ divides $a + i b$ iff $a, b$ have the same parity (both odd or both even). So $1 + i$ divides both $3 + 5 i$ and $1 + 3 i$.

Now looking at the euclidean norms $\lvert a + i b \rvert = a^{2} + b^{2}$, we see that $1 + i$ has norm $2$, and $ 1 + 3 i$ has norm $10$, so the Gaussian integer $\beta = (1 + 3 i)/(1 + i)$ has norm $5$, a prime integer, and thus $\beta$ is irreducible. Since $5$ does not divide the norm $34$ of $3 + 5 i$, the gcd is $1 + i$.