1. ## Unit

Let R={m+n(2)^(1/2)|m,n are integers}
Show that m+n(2)^(1/2) is a unit in R if and only if m^2-2n^2=+-1

I know a we have a unit a if ab=1.
I was given a hint to try (m+n(2)^(1/2))(x+y(2)^(1/2))=1 then(m-n(2)^(1/2))(x-y(2)^(1/2))=1 and multiply the 2 equations, but I don't understand the reasoning for this hint.

2. The function $f(m+n\sqrt2)=m^2-2n^2$ is a multiplicative norm, that is, a function from R into the integers such that for $r,s\in R$, $f(rs)=f(r)f(s)$. What is the norm of 1? Does that help?

3. I don't think I've really done anything norms in this section.

4. Well, the fact that it's called a "norm" isn't really important--the only feature of it that's important to the problem is the one I described, which is straightforward to verify, and the result falls out immediately.

5. Well f(rs)=f(r)f(s)=m^2-2n^2=1(1)=1

6. Originally Posted by kathrynmath
Well f(rs)=f(r)f(s)=m^2-2n^2=1(1)=1
Since f(1)=1, what do you know about f(a) and f(b) if ab=1?

7. Oh f(a)=f(b)=+-1

8. You got it!

9. Ok it's an iff statement, so I would have to go backwards, too right?
Assume m^2-2n^2=+-1
Then (m^2-2^(1/2)n)(m^2+2^(1/2)n)=+-1

10. Let $m^2-2n^2=\pm 1$. Then $m=\pm\sqrt{n^2\pm 1}$, which means

$m+n\sqrt{2}=\pm\sqrt{n^2\pm 1}+n\sqrt{2}$ or $\mp\sqrt{n^2\pm 1}+n\sqrt{2}$.

So $\sqrt{2n^2\pm1}\mp n\sqrt{2},-\sqrt{2n^2\pm1}\mp n\sqrt{2}$ are the respective inverses.