Hello All,

I have 2 questions regarding quadratic integers.

1a. Assume that $\lambda$= $(3+Sqrt(-3))/2$ exists in Q[Sqrt(3)]. How can we show that if $x$= $1(mod \lambda)^3$ that $x^3$= $1(mod \lambda)^3$? Also, how can we show that if $x$= $-1(mod \lambda$), then $x^3$= $-1(mod \lambda)^3$? Lastly, if $x$= $0(mod \lambda)$, how can we show that $x^3$= $0(mod \lambda)^3$ ?

I know that $\lambda$ is prime and that there are three congruence classes (mod $\lambda$): one with -1, 0, and 1.

1b. How can it be proven that if $x^3$+ $y^3$= $z^3$ and x,y,z are quadratic integers in Q[Sqrt(-3)], then $\lambda$ (as described in 1a) must divide of one x, y, or z?

My first guess at this one is to reduce the equation to modulo $\lamda^3$.

All help is appreciated!

2. Originally Posted by Samson
Hello All,

I have 2 questions regarding quadratic integers.

1a. Assume that $\lambda$= $(3+Sqrt(-3))/2$ exists in Q[Sqrt(3)]. How can we show that if $x$= $1(mod \lambda)^3$ that $x^3$= $1(mod \lambda)^3$? Also, how can we show that if $x$= $-1(mod \lambda$), then $x^3$= $-1(mod \lambda)^3$? Lastly, if $x$= $0(mod \lambda)$, how can we show that $x^3$= $0(mod \lambda)^3$ ?

I know that $\lambda$ is prime and that there are three congruence classes (mod $\lambda$): one with -1, 0, and 1.

1b. How can it be proven that if $x^3$+ $y^3$= $z^3$ and x,y,z are quadratic integers in Q[Sqrt(-3)], then $\lambda$ (as described in 1a) must divide of one x, y, or z?

My first guess at this one is to reduce the equation to modulo $\lamda^3$.

All help is appreciated!
Does reducing this to modulo 3 do me any good? Can anyone offer some pointers on this?