Mathematics Problem solving

Solving in Gaussian integers Z[i] = {a + bi : a, b, in Z}.
All Gaussian integer are 0, units, or irreducible.
If not 0, nor unit, nor irreducible, then a=bc, whereby b nor c is not a unit.
N(a) = N(b)N(c), whereby N(b) is less than N(a) and N(c), which is also less than N(a).
We therefore factorize both b and c to acquire the factorization of a.
{an} is considered a sequence in real numbers.
An = bn + cni, which represents the sequence of complex numbers.
In this case, when {an} is bounded, all the rest {bn} and {cn} are bounded too.
There always exists subsequences {bnk} of {bn} converges at b.
(ai) for all i in N and (bi) for all i in N. The sequence therefore needs lemma.
Recalling all elements of a, including the communicative R ring, it is irreducible given that a = bc and a isn’t a unit. This then implies that c or b is a unit (Ho, 2016).
(1 + i)(1 – i) = 2. This case, 2 is not irreducible in Z[i].
The Z[i], closed under multiplication and addition, also contains Z. It proves to be a subring from C, which is not a field. Then, division of Gaussian integer by other results in Q(i) elements. This are assumed to have imaginary and real parts.
In this case 2 is not a prime number in Z[i]. Neither is another number as 5, given that 5 = (1 + 2i)(1-2i). For a number like 3 = (-1)(-3) = i(-3i)
Then, all elements in α in Z[i] is then an invertible element with units existing as β in Z[i] in that α . β = 1.

Reference
Ho, H. (2016). Gaussian integers. Pre-Print, paper edition.