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If you'll check out this answer, you'll get an idea how to prove that $\Bbb Z[i]$ is a Euclidean domain. Every Euclidean domain is a PID, and in a PID, the "prime" and "irreducible" elements are the same. Also, in a general ring $R$ with a non-$0$ ideal $I$, we have that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.

Since you know that $2\pm i$ are irreducible in $\Bbb Z[i]$, then in particular, $5=(2+i)(2-i)$ is not irreducible, so not prime. Thus, $\langle 5\rangle$ is not a prime ideal of $\Bbb Z[i]$, and so $\Bbb Z[i]/\langle 5\rangle$ is not an integral domain.

If you'll check out this answer, you'll get an idea how to prove that $\Bbb Z[i]$ is a Euclidean domain. Every Euclidean domain is a PID, and in a PID, the "prime" and "irreducible" elements are the same. Also, in a general ring $R$ with a non-$0$ ideal $I$, we have that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.

Since you know that $2\pm i$ are irreducible in $\Bbb Z[i]$, then in particular, $5=(2+i)(2-i)$ is not irreducible, so not prime. Thus, $\langle 5\rangle$ is not a prime ideal of $\Bbb Z[i]$, and so $\Bbb Z[i]/\langle 5\rangle$ is not an integral domain.

If you'll check out this answer, you'll get an idea how to prove that $\Bbb Z[i]$ is a Euclidean domain. Every Euclidean domain is a PID, and in a PID, the "prime" and "irredicible" elements are the same. Also, in a general ring $R$ with a non-$0$ ideal $I$, we have that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.

Since you know that $2\pm i$ are irreducible in $\Bbb Z[i]$, then in particular, $5=(2+i)(2-i)$ is not irreducible, so not prime. Thus, $\langle 5\rangle$ is not a prime ideal of $\Bbb Z[i]$, and so $\Bbb Z[i]/\langle 5\rangle$ is not an integral domain.

If you'll check out this answer, you'll get an idea how to prove that $\Bbb Z[i]$ is a Euclidean domain. Every Euclidean domain is a PID, and in a PID, the "prime" and "irreducible" elements are the same. Also, in a general ring $R$ with a non-$0$ ideal $I$, we have that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.

Since you know that $2\pm i$ are irreducible in $\Bbb Z[i]$, then in particular, $5=(2+i)(2-i)$ is not irreducible, so not prime. Thus, $\langle 5\rangle$ is not a prime ideal of $\Bbb Z[i]$, and so $\Bbb Z[i]/\langle 5\rangle$ is not an integral domain.

Oops. Just reread your post, and this answer doesn't do the job. Fixing it now.

If you'll check out this answer, you'll get an idea how to prove that $\Bbb Z[i]$ is a Euclidean domain. Every Euclidean domain is a PID, and in a PID, the "prime" and "irredicible" elements are the same. Also, in a general ring $R$ with a non-$0$ ideal $I$, we have that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.

Since you know that $2\pm i$ are irreducible in $\Bbb Z[i]$, then in particular, $5=(2+i)(2-i)$ is not irreducible, so not prime. Thus, $\langle 5\rangle$ is not a prime ideal of $\Bbb Z[i]$, and so $\Bbb Z[i]/\langle 5\rangle$ is not an integral domain.