## Solve a linear congruence over the Gaussian integers

Solve the following congruence over $\mathbb{Z}[i]$: $(3-2i)x \equiv 1 \mod (1-2i)$.

As a consequence of Fermat’s Little Theorem for $\mathbb{Z}[i]$, we have $x \equiv (3-2i)^{N(1-2i)-2} \mod (1-2i)$. Now $N(1-2i) = 5$, and $(3-2i)^3 = -9-46i \equiv i \mod (1-2i)$. Indeed, $(3-2i)i - 1 = (1-2i)(-1+i)$.

The complete set of solutions in $\mathbb{Z}[i]$ is $\{ (1-2i)\delta + i \ |\ \delta \in \mathbb{Z}[i] \}$.