# 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.

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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,
1. $x \equiv 1 \pmod{2+w}$ is equivalent to $x = a+bw$, such that $a+b \equiv 1 \pmod{3}$, where $a,b$ are integers.
2. $(a+bw)^3 = a^3+b^3-3ab^2 + 3ab(a-b)w$.
3. Check that each coefficient is divisible by 9, by looking at two cases: either a or b are divisible by 3, or that $a \equiv b \equiv -1 \pmod{3}$.