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.

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