Over $\mathbb{Z}$, the congruence $e_1^4 + 2e_2^2 \equiv e_1^2 – 2e_2 \mod 4$ holds for any $e_1,e_2 \in \mathbb{Z}$, e.g. by consideration of residue classes for $e_1$ and $e_2$.
Is there a $D = \sqrt{d}$ if $d = 2,3 \mod 4$ and $D = \frac{\sqrt{d}+1}{2}$ if $d = 1 \mod 4$ such that the same congruence holds in the ring of algebraic integers of the quadratic number field $\mathbb{Z}[D] \subset \mathbb{Q}(\sqrt{d})$?
Edit: Also, can we find such a quadratic number field where the congruence holds, in which $2$ is prime in the corresponding ring of algebraic integers? Is this necessary for the congruence to hold?
Best Answer
If $d\equiv1\bmod8$ then the congruence certainly holds. In that case, $\frac12(1+\sqrt{d})$ reduces to a $2$-adic integer, and thus the integers in $\mathbb Q[\sqrt{d}]$ reduce to ordinary integers $\bmod 4$.
Note that this includes $d=-7$ where $2$ is not prime, for $2=[\frac12(1+\sqrt{-7})][\frac12(1-\sqrt{-7})]$. Any case where $d$ is eight less than an odd square similarly renders $2$ non-prime.