The variety $V_p$ given by $X^2 + Y^2 = pZ^2$ I'm doing exercise 1.10 in Silverman's Elliptic Curves. $p\ge 3$ is a prime. Part (a) asks for a proof that $p\equiv 1\mod{4}\iff V_p\cong \mathbb{P}^1$ over $\mathbb{Q}$. I proved this by (painfully) constructing an explicit bijection and inverse, using the fact that $x^2+y^2=p$ has a solution in rational numbers, by parametrizing solutions to the equation.
Part (b) asks for a proof that if $p, q\equiv 3\mod{4}$, $p\ne q$ then $V_p$ and $V_q$ are not isomorphic over $\mathbb{Q}$. I know that $V_p(\mathbb{Q}) = V_q(\mathbb{Q}) = \emptyset$, but I can't figure out how to use the fact that $p$ and $q$ are not expressible as a sum of two squares to show that no isomorphism exists. I constructed one proof that simply had to be wrong because it didn't use the congruence restrictions on $p$ and $q$. I'd appreciate a hint. Please don't post a complete solution; I want to figure this out myself.
Note: I interpret part (b) as asking: show that there is no isomorphism $\phi = [\phi_0,\phi_1,\phi_2]$ between
$$\{[a,b,c]\in\mathbb{P}^2(\overline{\mathbb{Q}})\,:\,a^2+b^2 = pc^2\}
\text{ and }
\{[d,e,f]\in\mathbb{P}^2(\overline{\mathbb{Q}})\,:\,d^2+e^2 = qf^2\}$$
where $\phi_i\in \mathbb{Q}[X]$ (or, if you prefer, $\mathbb{Q}[V_p]$).
 A: Here is a proof that the conics $x^2+y^2-pz^2$ are not isomorphic over $\mathbb Z$ for different $p$'s.
For every prime $p$ we  have a well defined reduction map $$r_p:\mathbb P^2=\mathbb P^2_{\mathbb Z}\to \mathbb P^2_{\mathbb F_p}$$ which in particular on rational points $A=[a:b:c]\in \mathbb P^2(\mathbb Q)$ (where $a,b,c$ are integers without any common prime factor) is given by  $r_p(A)=\bar A=[\bar a:\bar b:\bar c]\in \mathbb P^2(\mathbb F_p)$ .
(Elementary reference :  Silverman-Tate, Rational Points on Elliptic Curve, A5, page 302)
Given a subvariety $W\subset \mathbb P^2$ we'll call $\overline W=r_p(W)\subset \mathbb P^2_{\mathbb F_p}$ its image under $r_p$.
Now if $p, l \equiv 3 \operatorname {mod} 4$ are distinct primes we get $$\operatorname {card}[\overline {V_p}(\mathbb F_p)]=1 \quad  (\bullet) \quad\operatorname {but} \quad \operatorname {card}[\overline{V_l}(\mathbb F_p)]=p+1 \quad  (\bullet \bullet)$$ showing that $V_p, V_l\subset \mathbb P^2$ are not isomorphic since  their images $\overline {V_p},\overline {V_l}$ under $r_p$ are not isomorphic in $\mathbb P^2_{\mathbb F_p}$.
The reason that  $\overline {V_p},\overline {V_l}$ are not isomorphic in $\mathbb P^2_{\mathbb F_p}$ is that their $\mathbb F_p$ points are sets $\overline {V_p}(\mathbb F_p),\overline {V_l}(\mathbb F_p)$ of different cardinalities (namely $1$ and $p+1$).
Supplementary Explanations
$(\bullet )$ The equality $\operatorname {card}r_p(V_p)=1$ follows from $x^2+y^2=0$ having as solution in $\mathbb P^2(\mathbb F_p)$ only the point $[0:0:1]$.
 $(\bullet \bullet)$ The  equality $\operatorname {card}r_p(V_l)=p+1$ is due to the conic $r_p(V_l)\subset \mathbb P^2(\mathbb F_p)$  being isomorphic to $\mathbb P^1(\mathbb F_p)$, since   it  has  an $\mathbb F_p$-point (recall that in the finite field $\mathbb F_p$ we can solve $x^2+y^2=l\:$ for any $ l \in \mathbb F_p^*$) .  
Personal remark
The numerous comments pertaining to question and answer on this thread (not mentioning all the erased ones) prove that this exercise at the very beginning of Silverman's book is completely out of place. 
