Given $\displaystyle p \equiv 3 \mod{4} $ is prime in $\displaystyle \mathbb{Z} $, show $\displaystyle \mathbb{Z}[i]/(p) $ has order $\displaystyle p^2 $.
An odd prime p has a form either 1(mod 4) or 3(mod 4). We can see that 1(mod 4) is reducible in Z[i] (For example, 5=(1+2i)(1-2i)). An odd prime p is irreducible iff it has the form 3(mod 4).
We know that Z[i] is an integral domain and a PID. We also know that $\displaystyle p \equiv 3 (mod{4}) $ is irreducible. Thus <p> is maximal ideal in Z[i]. It follows that Z[i]/<p> is a field having p^2 elements.
An element of Z[i]/<p> has the form a+bi. The number of choices of a is p and the number of choices for b is p. So it has p^2 elements.
For example, let p =3. Then, Z[i]/<3>= {0, 1,2, i, 2i, 1+i, 1+2i, 2+i, 2+2i}. It has 9 elements. You can check other cases and verify them.