Is element $5 + 2i$ irreducible in the ring $\mathbb{Z}[i]$? I want to prove that $5 + 2i$ is an irreducible in the ring $\mathbb{Z}[i]$. I was wondering how to do so without the use of the norm defined on this ring.
Let's suppose that it is reducible, i.e. $5 + 2i = (a + bi)(c + di)$ for some $a, b, c, d \in \mathbb{Z}$ and $a + bi, c + di$ are not units. Thus 
$$\begin{cases} 5 = ac - bd, \\ 2 = ad + bc.\end{cases}$$
How can I prove now that this system of equation does not have an integer solution?
EDIT
Extracting $c$ from the first equation, I get $c = \frac{5 + bd}{a}$, which substituted to the second one gives me $2a - 5b = d(a^2 + b^2)$ and therefore $d$ divides both $2a$ and $5b$. If $d \neq 1$, we've got $d |a$ and $d|b$, so the element $a + bi$ is invertible, which is contrary to the hypothesis?

Assume $5+2i = (a+ib)(c+id) = (ac - bd) + i(ad + bc)$.
Notice that
$$(a-ib)(c-id) = (ac - bd) - i(ad + bc) = 5-2i$$
Multiply these two equalities:
$$29 = 5^2 + 2^2 = (5+2i)(5-2i) = (a+ib)(c+id)(a-ib)(c-id) = (a^2+b^2)(c^2+d^2)$$
Since $29$ is prime, this implies $a^2+b^2 = 1$ or $c^2+d^2 = 1$. In the first case we have
$$(a+ib)(a-ib) = a^2 + b^2 = 1 \implies (a+ib)^{-1} = a-ib$$
and in the second case we have 
$$(c+id)(c-id) = c^2 + d^2 = 1 \implies (c+id)^{-1} = c-id$$
Hence, one of $a+ib$ and $c + id$ is invertible.
Therefore, $5+2i$ is irreducible. 
