Prove that the group of the rational points on the conic $u^2-Av^2=1$ is not finitely generated. This is an exercise from Rational Points on Elliptic Curves by Silverman.

Let $H$ be the conic $u^2-Av^2=1$ where $\sqrt{A}\notin \mathbb{Q}$. If $(u_1,v_1), (u_2,v_2)$ are two points in $H(\mathbb{Q})$, we define their product by
  $$(u_1,v_1)*(u_2,v_2)=(u_1u_2+Av_1v_2, u_1v_2+u_2v_1).$$
  Prove that this group is not finitely generated. 

Approach 1:
Using a line passing through $(1,0)$ with rational slope $t$, we can find a one to one correspondence of this group with the rational numbers. But this is not an isomorphism. So it does not work.
Approach 2:
I can show that the map that sends $(u,v)$ on $H(\mathbb{Q})$ to $u+\sqrt{A}v$ in $\mathbb{Q}(\sqrt{A})^*$ is a homomorphism and injection. So the group is isomorphic to a subgroup of $\mathbb{Q}(\sqrt{A})^*$. I am not sure whether the following argument is enough to show that this group is not finitely generated.


*

*Suppose it is finitely generated. Then it is isomorphic to 
$$\mathbb{Z}/d_1\mathbb{Z}\oplus \cdots \mathbb{Z}/d_m\mathbb{Z}\oplus \underbrace{\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}_\text{n's}.$$

*Since $\mathbb{Q}(\sqrt{A})^*$ has characteristic $0$, $m$ must be $0$.

*Since $\mathbb{Q}(\sqrt{A})^*$ has no linearly independent elements, $n=1$. So $H(\mathbb{Q})$ has to be cyclic.

*All integer solutions are generated by one smallest element, but it cannot generate fraction points, so the generator has to be fraction.

*A fraction cannot generate integer solutions. Contradiction.


If someone can verify whether this is the correct approach, or point out my mistake, it would be greatly appreciated. 
 A: Your solution does not work. First, $m$ is always different from $0$ because $H(\mathbb Q)$ contains the cyclic subgroup $\{(1,0),(-1,0)\}$. Second, what does it mean that "$\mathbb Q(\sqrt{A})^*$ has no linearly independent elements"? Because actually $\mathbb Q(\sqrt{A})^*$ contains a lot of linearly independent elements!
To prove the claim, one can proceed in the following elementary way:
1) suppose that $H(\mathbb Q)$ is finitely generated. Then there exists a finite set of primes $P=\{p_1,\ldots,p_n\}$ such that if $(\frac{r}{s},\frac{u}{v})\in H(\mathbb Q)$ (with $(r,s)=(u,v)=1$) and $p$ is a prime dividing $sv$, then $p\in P$. To prove this, simply assume $(u_1,v_1),\ldots, (u_n,v_n)$ are generators for $H(\mathbb Q)$ and use the group law to see that primes dividing the denominators of any linear combination of them are primes dividing the denominators of the $u_i,v_i$'s and of $A$ (plus maybe $2$).
2) one proves that there are arbitrarily big primes dividing the denominator of a point $(u,v)\in H(\mathbb Q)$. This is in contradiction with 1), and therefore you have your claim. To see this, parametrize all the elements of $H(\mathbb Q)$ via the lines through $(1,0)$, as you suggested. Then you get $x=-\frac{At^2+1}{At^2-1}$, where $t\in\mathbb Q$. Up to multiplying $t$ by the denominator of $A$, you can assume that $A\in \mathbb Z$. Thus, for every $t\in\mathbb Z$, the only possible prime factor shared by $At^2+1$ and $At^2-1$ is $2$. Now you get the claim by observing that the set of primes dividing $At^2-1$, as $t$ runs over $\mathbb Z$, is infinite.
