Let $α=a+bi \in\mathbb{Z}[i]$ with $\gcd(a, b)=1$. Show that there exists $c\in\mathbb{Z}$ such that $c+i$ is a multiple of $α$ in $\mathbb{Z}[i]$ (that is, $c + i = αδ$ for some $δ \in\mathbb{Z}[i]$).
I have got so far that $N(c+i) = c^2 + 1 = N(α)N(δ)$ and so $c^2 = -1 \pmod{a^2 + b^2}$ which has a solution if $a^2 + b^2 = p$ where $p$ is a prime and is equal to $1 \pmod{4}$.
However, $a^2 + b^2$ doesn't always equal a prime. How do I show such a c exists when this is the case?
You need to show that there is $m+ni\in Z[i]$ such that $an+bm=1$. Since $(a,b)=1$, we can apply Bezout's lemma to ensure that those integers exist, and use the Euclidean algorithm to construct them.
Once you have them you get that $(a+bi)(m+ni)=(am-bn)+(an+bm)i=(am-bn)+i$.
Therefore, we can take $c=am-bn$.