Quadratic Integers

**Samson** · August 26th 2010, 08:15 AM

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!

**Samson** · September 5th 2010, 11:02 AM

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?

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