Need help for finding a generator $I=\{a+bi \in R\mid a \equiv b\pmod 2\}$ is an ideal of $R=\mathbb Z[i]=\{a+bi\mid a,b \in \mathbb Z\}$. Can somebody help me to find the generator of $I$?
 A: Claim: $I = \langle 1 + i \rangle$.
Proof: Notice that $1 + i \in I$ as $1 \equiv 1 \pmod 2$ which means $\langle 1 + i \rangle \subseteq I$.
For the reverse containment, let $a + bi \in I$, and $a, b \in \mathbb Z$ with $a \equiv b \mod 2$. Since $\mathbb Z[i]$ is a Euclidean Domain, we can use the division algorithm and so there exist $r, q \in \mathbb Z[i]$ with $r = c + di$, $q = e + fi$ such that $$a + bi = (1 + i)q + r$$ where $c^2 + d^2 = N(r) < N(1 + i) = 2$ or $r = 0$. Notice that $$(1 + i)q = (1 + i)(e + fi) = e + fi + ei - f = (e - f) + (e + f)i$$ and so $e - f \equiv e + f \pmod 2$ since $-f \equiv f \pmod 2$. Since $r = c + di = (a + bi) - (1 - i)q$ and since $a \equiv b \pmod 2$ and $e - f \equiv e + f \pmod 2$, then $$c = a + e - f \equiv b + e + f = d \pmod 2.$$
For $c^2 + d^2 < N(1 + i) = 2$ we must have $c = d = 0$ and hence $r = 0$. Conclude that $a + bi = (1 + i)q$ and so $a + bi \in \langle 1 + i \rangle$ which means $I \subseteq \langle 1 + i \rangle$.
Conclude $I = \langle 1 + i \rangle$.
