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.
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.
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,