$x^2+y^2 = p$ and $x^2+y^2 = q$ are not $\mathbb{Q}$-isomorphic when p,q are primes 3 mod 4 This is a problem in Silverman's book "The arithmetic of Dynamical Systems". Concretely, problem 4.44.
If we call $X_p$ the curve with equation $x^2 + y^2 = p$, it turns out that $X_p$ is $\mathbb{Q}$-isomorphic to $X_1$ when $p$ is a prime 1 mod 4, but in the case of primes 3 mod 4, different primes 3 mod 4 will give different twists of $X_1$.
Any idea of how to prove it?
One approach that I tried was to prove that $X_q$ does not have solutions in $\mathbb{Q}(\sqrt{p})$ but I failed.
 A: Thanks Carlos Rivera for giving me the ideas. I post the solution maybe can help somebody else in the future.
Suppose $X_p$ and $X_q$ are $\mathbb{Q}$-isomorphic. This imply they are isomorphic in every extension of $\mathbb{Q}$. The trick is to prove that $X_q(\mathbb{Q}_q)$ is trivial whereas $X_p(\mathbb{Q}_q)$ is not.
$X_q(\mathbb{Q}_q)$ is trivial: Suppose $(u,v)$ is a solution in $\mathbb{Q}_q$. Using the power expansion of elements in $\mathbb{Q}_q$, there are unique $x, y \in \mathbb{Z}_q^\times$ and $a,b \in \mathbb{N}$ such that $(u,v) = (x/q^a, y/q^b)$. By symmetry, assume $a \geq b$. Putting this into the equation of $X_q$ we get $$x^2+q^{2(a-b)}y^2 = q^{2a+1}.$$
If $a > b$, $x \notin \mathbb{Z}_q^\times$, so $a = b$. Taking modulo $q$, $$x^2 \equiv -y^2 \pmod{q},$$ so $-1$ is an square modulo $q$. However, this is false when $q \equiv 3 \pmod{4}$.
$X_p(\mathbb{Q}_q)$ is not trivial. Observe that $\{x^2 : x \in \mathbb{F}_q \}$ and $\{ p - y^2: y \in \mathbb{F}_q \} $ have $(q+1)/2$ points. So, they meet in at least one point $(x_0, y_0)$. Since $q \neq p$, one of the coordinate is not $0 \pmod{q}$. Using a trick like Hensel's lemma, one can find a lift to $\mathbb{Z}_q$.
