1. ## multiplicative inverse

Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].

2. Originally Posted by NikoBellic
Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].
$\displaystyle \frac{1}{1+\sqrt2} = \frac{1-\sqrt2}{(1+\sqrt2)(1-\sqrt2)} = \sqrt2-1$

3. Originally Posted by NikoBellic
Show that sqrt(2) + 1 has a multiplicative inverse in Z[sqrt(2)].
This is kind of overkill but here goes:

You can also do this using norms. $\displaystyle \mathbb{Z}[\sqrt2]$ has a norm $\displaystyle N(a+b\sqrt2) = a^2-2b^2$.

Suppose $\displaystyle (1+\sqrt2)^{-1} = a+b\sqrt2$, then $\displaystyle N(1+\sqrt2)=-1\implies N(a+b\sqrt2) = -1 \implies a^2-2b^2 = -1$.

So we're looking to solve $\displaystyle a^2-2b^2=-1$.

Now $\displaystyle (a^2-2b^2)^2=(-1)^2\implies (a^2+2b^2)^2-2(2ab)^2=1\implies (2a^2+1)^2-2(2ab)^2=1$ since $\displaystyle 2b^2=a^2+1$ for the last equality.

So as you can see to find a solution, we're left to solve a form of Pell's equation, which is solvable i.e. we can find $\displaystyle a$ and $\displaystyle b$. Thus $\displaystyle 1+\sqrt2$ has an inverse.

After Pell's equation is solved though, we'll have an answer in terms of $\displaystyle 2a^2+1$ and $\displaystyle 2ab$, so this proof needs some polishing at the end. I won't do this however since I've already given you another solution; this posting was just a thought I had.