<s>[INST] A congruence in a quadratic number field 
If $\displaystyle L = \frac{3+\sqrt{-3}}{2}$,
  and if $x\equiv 1\pmod{L}$,
  show that $x^3\equiv 1\pmod{L^4}$. 

I have already shown that if $x\equiv 1\pmod{L}$,
then $x^3\equiv 1\pmod{L^3}$. 
Thanks.
 [/INST] Let $w = \frac{-1 + \sqrt{-3}}{2}$. Your question is to show that if $x \equiv 1 \pmod{2+w}$, then $x^3 \equiv 1 \pmod{(2+w)^4}$.
By direct expansion, we see that $(2+w)^4 = 9w$. So it suffices to show $x^3 \equiv 1 \pmod{9}$. Now check the following,


*

*$x \equiv 1 \pmod{2+w}$ is equivalent to $x = a+bw$, such that $a+b \equiv 1 \pmod{3}$, where $a,b$ are integers.

*$(a+bw)^3 = a^3+b^3-3ab^2 + 3ab(a-b)w$.

*Check that each coefficient is divisible by 9, by looking at two cases: either a or b are divisible by 3, or that $a \equiv b \equiv -1 \pmod{3}$.

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