Are All Units in Z[sqrt(2)] of the Form +/- (1 +/- sqrt(2))^n?

[Phillips101]

How would I show that every unit in Z[sqrt(2)] is of the form +/- (1 +/- sqrt(2) )^n ?

I can show these are all units, but I can't show every unit is one of these. From some research, I'm aware this is a special case of Dirichlet's Unit Theorem, but that is far above the level I'm working at.

Any help would be appreciated :)

Thanks

 

[mrbohn1]

Every element of Z[\sqrt{2}] is of the form a+b\sqrt{2}, a,b integers.

Suppose a+b\sqrt{2} is a unit. Then: (a+b\sqrt{2})(c+d\sqrt{2})=1.

So:ac+(bc+ad)\sqrt{2}+2bd=1. So already you know that bc=-ad.

Now just try manipulating the expression you are left with: (ac+2bd=1) using this fact. You should be able to show that a,b,c,d must all be either 1 or -1.

 

[losiu99]

The problem is they don't have to. For example, 3+2\sqrt{2 is unit in \mathbb{Z}[\sqrt{2} ]. I'm also trying to solve this problem right now.

Edit: I think I got it. Let \nu(a+b\sqrt{2})=a^2-2b^2. It's multiplicative, so z is unit iff \nu(z)=\pm 1. It's easy to see that signs of a and b doesn't matter when it comes to invertibility, so we can assume z>0.

Lemma Smallest element of \mathbb{Z}[\sqrt{2}] with z>1 and \nu(z)=\pm 1 is 1+\sqrt{2}
Proof
Let a,b\in \mathbb{N}. If such element is of type a-b\sqrt{2}>1, then a+b\sqrt{2}>1 and so \nu({a-b\sqrt{2}})>1. Similarly if it's -a+b\sqrt{2}. It has to be a+b\sqrt{2}. Case with one of a, b equal zero is easily checked by hand. This proves the lemma.

Now suppose \nu(z)=1, z > 0. Let z_0 = 1+\sqrt{2}. There exists integer k (positive or negative) with z_0^k \leq z < z_0^{k+1}. Easily to check, 1\leq z z_0^{-k} < z_0. If z z_0^{-k}\neq 1, it contradicts minimality of z_0, and so z=z_0^k.

Hope I haven't messed up too badly.