# Thread: Find all elements a+b(-3)^0.5 in Z[-3^0.5] such that |a+b(-3)^0.5|^2=1

1. ## Find all elements a+b(-3)^0.5 in Z[-3^0.5] such that |a+b(-3)^0.5|^2=1

Hi,

I have the following problem:

"Find all elements $a+b(-3)^{0.5}$ in $Z[-3^{0.5}]$ such that $|a+b(-3)^{0.5}|^2=1$"

And I think I have the answer, but I don't know if it is right since it says "elements" and not "element."

Well, I know that $|a+b(-3)^{0.5}|^2=a+b(-3)^{0.5}$ (obvious), and that if $a+b(-3)^{0.5}$ is a unit, then $|a+b(-3)^{0.5}|^2=1$... is that correct?

Thanks!

2. $(a+b\;\sqrt{-3})^2=a^2-3b^2+2ab\;\sqrt{-3}=1\Leftrightarrow a^2-3b^2=1\;\wedge\;2ab=0$

Now, analyze the cases $a=0$ and $b=0$ .

3. Originally Posted by mbmstudent
Hi,

I have the following problem:

"Find all elements $a+b(-3)^{0.5}$ in $Z[-3^{0.5}]$ such that $|a+b(-3)^{0.5}|^2=1$"

And I think I have the answer, but I don't know if it is right since it says "elements" and not "element."

Well, I know that $|a+b(-3)^{0.5}|^2=a+b(-3)^{0.5}$ (obvious)

Not ony not obvious but even wrong: $|a+b\sqrt{-3}|^2=a^2+3b^2$ , since we're talking here about

the module (=absolute value) of a complex number

, and that if $a+b(-3)^{0.5}$ is a unit, then $|a+b(-3)^{0.5}|^2=1$... is that correct?

Thanks!

In fact, $a^2+3b^2=1\Longleftrightarrow a=\pm 1$ , so we have two different units here.

Tonio

4. Originally Posted by FernandoRevilla
$(a+b\;\sqrt{-3})^2=a^2-3b^2+2ab\;\sqrt{-3}=1\Leftrightarrow a^2-3b^2=1\;\wedge\;2ab=0$

Now, analyze the cases $a=0$ and $b=0$ .

I think you missed the absolute value (module) symbol the OP wrote in his mesage.

Tonio

5. Originally Posted by tonio
I think you missed the absolute value (module) symbol the OP wrote in his mesage.

Of course.