What is the set A = {$a + b\sqrt{3} : (a,b)\in \mathbb Z ^2, a^2 - 3b^2 = 1$}?

Question

I don't even understand the notation of this set. Does the b have be a multiple of $\sqrt 3$ and $a^2 - 3b^2 = 1$?

If I have a $(a,b) \in \mathbb R^2$, how do I know if this element is in A?

Again, I'm not sure how to process what the set is representing. Any insight is appreciated, thank you.

Answer 1

The notation means "$A$ is the set of all possible numbers which can be written as $a+b\sqrt3$, where $a$ and $b$ are integers such that $a^2-3b^2=1$." So, for instance, $1=1+0\sqrt3$ is an element of $A$, since $1$ and $0$ are integers and $1^2-3\cdot 0^2=1$.

Does the b have be a multiple of $\sqrt 3$ and $a^2 - 3b^2 = 1$?

No. $b$ must be an integer, as must $a$, and together they must satisfy $a^2-3b^2=1$. If this is satisfied, the number $a+b\sqrt3$ is an element of $A$.

If I have a $(a,b) \in \mathbb R^2$, how do I know if this element is in A?

It isn't. $A$ doesn't contain pairs of numbers, it just contains numbers. However, if you have an $x\in \Bbb R$, and you're wondering whether $x\in A$, then you check whether there are integers $a,b$ such that both $x=a+b\sqrt3$ and $a^2-3b^2=1$.

Answer 2

To check whether $x\in A$, write $x=(a+b\sqrt3)$ and observe that $$1=a^2-3b^2=(a+b\sqrt3)(a-b\sqrt3).$$ Therefore $$a-b\sqrt3=\frac1x.$$ Thus $x\in A$ iff both $$a=\frac12\left(x+\frac1x\right)\in\Bbb Z$$ and $$b=\frac1{2\sqrt3}\left(x-\frac1x\right)\in\Bbb Z.$$