Find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i) = (\alpha)$ 
Let $\mathbb{Z}[i] = \{a+bi : a,b \in \mathbb{Z}\}$ and $\mathbb{Q}(i) = \{a+bi : a,b \in \mathbb{Q}\}$.
Find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i) = (\alpha)$

Definition : If $A$ is a ring et $S \subset A$, then we note $(S)$ the smallest ideal containing S.
I took I lot of time to try solving this problem. Is anyone could help me for this question?
 A: Hint : $3+5i-2(1+3i)=1-i$. Try use this and fact that $(3+5i,1+3i)$ is an ideal.
A: Since $\mathbb{Z}[i]$ is a Euclidean domain, every ideal is principal, so $\alpha$ exists and it is a greatest common divisor of $3+5i$ and $1+3i$. You could find it with the Euclidean algorithm, but also factorization is possible: a greatest common divisor can be computed from the prime factorization as usual for the integers.
Suppose you know a prime factor $z$; then $3+5i=zw$ for some $w$ and therefore
$$
(3+5i)(3-5i)=34=z\bar{z}w\bar{w}
$$
Since $34=2\cdot 17$ we know that the prime factors of $3+5i$ are among $1+i$, $4+i$ and $4-i$.
Similarly, if $1+3i=zw$ for a prime $z$,
$$
(1+3i)(1-3i)=10=z\bar{z}w\bar{w}
$$
so the prime factors of $1+3i$ are among $1+i$, $2+i$ and $2-i$.
Thus the candidate is $1+i$ and, indeed
$$
\frac{3+5i}{1+i}=\frac{(3+5i)(1-i)}{2}=4+i,
\qquad
\frac{1+3i}{1+i}=\frac{(1+3i)(1-i)}{2}=2+i
$$
so a greatest common divisor is $1-i$.
Note that $4+i$ and $2+i$ are coprime, because $17$ and $5$ are so in $\mathbb{Z}$.

How did I find the candidate primes in the decomposition of $3+5i$? The primes in $\mathbb{Z}[i]$ are, up to multiplication by units, $1+i$, the couples $a+bi$, $a-bi$ where $a^2+b^2$ is an odd prime integer, and the prime integers $p$ such that $p\equiv3\pmod{4}$.
If $3+5i=zw$, where $z$ is a prime, then $z\bar{z}w\bar{w}$ is a factor of $(3+5i)(3-5i)=34$. But we know from the classification above that $z\bar{z}$ must be $2=(1+i)(1-i)$, an odd prime of the form $a^2+b^2$ or $p^2$ (with $p$ prime and $p\equiv3\pmod{4}$). As the factorization of $3+5i$ is $2\cdot 17$ we know where to look for.
Note that we do know that $3+5i$ is divisible by $1+i$, but the method doesn't tell which one among $4+i$ or $4-i$ is a factor; this of course can be checked with a division.
Of course the Euclidean algorithm is faster if the numbers have large modulus, but it's not as easy to perform as in the integers (see Gaussian integers & division theorem).
