## Solve a linear congruence over a quotient of an algebraic integer ring

Solve the congruence $(1+\sqrt{-5})\xi \equiv 3$ mod $(7,1+2\sqrt{-5})$ in the ring $\mathbb{Z}[\sqrt{-5}]$.

Let $Q = (7,1+2\sqrt{-5})$. We claim that $Q$ is prime. To that end, we claim that $(7,1+2\sqrt{-5})(7,1-2\sqrt{-5}) = (7)$. Indeed, $7 = 7^2 - 2(1+2\sqrt{-5})(1-2\sqrt{-5})$. Next we claim that $Q = (7,1+2\sqrt{-5})$ and $Q_2 = (7,1-2\sqrt{-5})$ are proper. If to the contrary $1 \in Q_1$, then $1 = 7(a+b\sqrt{-5}) + (1+2\sqrt{-5})(h+k\sqrt{-5})$ for some integers $a$, $b$, $h$, and $k$. Comparing coefficients mod 7 we have $1 \equiv 0$, a contradiction. Similarly, $Q_2$ is proper. Now $N((7)) = 49$, so that $N((7,1+2\sqrt{-5})) = 7$. By Corollary 9.15, $(7,1+2\sqrt{-5})$ is prime.

Now we claim that $1+\sqrt{-5} \notin Q$. Indeed, note that $1+\sqrt{-5} \equiv 8+8\sqrt{-5} \equiv 4$ mod $Q$. As we saw in a previous exercise, since $N(Q)$ is prime, $0 \not\equiv 4$ mod $Q$. By Fermat’s Theorem, $(1+\sqrt{-5})^6 \equiv 1$ mod $Q$. In particular, $(1+\sqrt{-5})^{-1} \equiv (1+\sqrt{-5})^5$ mod $Q$. So we have $\xi \equiv 3(1+\sqrt{-5})^5 \equiv 228-60\sqrt{-5}$ mod $Q$. Reducing coefficients mod 7, we have $\xi \equiv 4+3\sqrt{-5}$ mod $Q$, and so $\xi \equiv 6$ mod $Q$.

Indeed, $6(1+\sqrt{-5}) - 3 = 3(1+2\sqrt{-5}) \in Q$.