TITLE: Factorize $999999$ into a product of irreducibles in $\mathbb{Z}[i]$ (clarification)
QUESTION [5 upvotes]: Factorize $999999$ into a product of irreducibles in $\mathbb{Z}[i]$.
First I factorized $999999$ into a product of primes in $\mathbb{Z}$ as follows
Factorize $999999$ into a product of irreducibles in $\mathbb{Z}[i]$.
First I factorized $999999$ into a product of primes in $\mathbb{Z}$ as follows
$999999 = 3^3\cdot7\cdot11\cdot13\cdot37$
From here I see that $37 = (6+i)(6-i)$
Now we have $999999 = 3^3\cdot7\cdot11\cdot13\cdot(6+i)(6-i)$
Then $13 = (2-3i)(2+3i)$
So $999999 = 3^3\cdot7\cdot11\cdot(2-3i)(2+3i)\cdot(6+i)(6-i)$
I see that $3, 7,$ and $11$ are congruent to $3 \pmod 4$ so these integers are prime in $\mathbb{Z}[i]$. 
So I want to know if I have completed the problem because $3, 7,$ and $11$ are irreducible; or if there is more work to do.

REPLY [2 votes]: You are finished. The factors are irreducible and the factorization is even unique (up to the order of the elements and units)  because the Gaussian integers are a unique factorization domain. A characterization of the irreducible elements in $\mathbb{Z}[i]$ can be found here.