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.
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.