Solving exercise 1.10 in Silverman's AEC Please note that although there is a very similarly titled question Exercise 1.10 from Silverman "The Arithmetic of Elliptic Curves" this question received no answers.
Let $p$ be an odd prime and $V_p\subseteq \mathbb P^2$ the variety $$V_p : X^2 + Y^2 = p Z^2 $$
I'm trying to show that $V_p \cong \mathbb P^1$ over $\mathbb Q$ if and only if $p\equiv 1\pmod 4$. The forward direction is easy: suppose $p\cong 3\pmod 4$. If $V_p$ is isomorphic to $\mathbb P^1$ over $\mathbb Q$ then $V_p (\mathbb Q) \cong \mathbb P^1 (\mathbb Q)$. But $X^2 + Y^2 = p Z^2$ has no solutions in integers and thus $V_p (\mathbb Q)=\varnothing$, and we know $\mathbb P^1 (\mathbb Q)\neq\varnothing$.
However I'm struggling to do the other direction because I can't cook up a good isomorphism $V_p \cong \mathbb P^1$ defined over $\mathbb Q$ when $p\equiv 1\pmod 4$. Can anyone help me out with this? Thank you!
 A: ‎Assume that $0\neq m = a^2‎+ ‎b^2$ for some $a‎, ‎b \in \mathbb{Z}$‎, 
‎and let $V_m \subset \mathbb{P}^2$ be the variety given by the equation‎: 
‎\begin{align*}‎ 
‎V_m : X^2‎ + ‎Y^2 = m Z^2‎. 
‎\end{align*}‎ 
‎We claim that $V_m$ is isomorphic to $\mathbb{P}^1$ over $\mathbb{Q}$‎. 


‎Consider the rational map‎ 
‎\begin{align*}‎ 
‎\phi‎ : ‎V_m \rightarrow \mathbb{P}^1‎, 
‎\qquad‎ 
‎\phi = [aX+bY+(a^2+b^2)Z‎, ‎aY-bX]‎. 
‎\end{align*}‎ 
‎Clearly $\phi$ is regular at every point of $V_m$‎, 
‎except possibly at $[a,b,-1]$‎, ‎i.e.‎, ‎at the points where $aX+bY+(a^2+b^2)Z=0=aY-bX$‎. 
‎However‎, ‎using‎ 
‎\begin{align*}‎ 
\Big( aX+bY+(a^2+b^2)Z \Big) \Big( aX+bY-(a^2+b^2)Z \Big) \equiv‎ ‎-(aY-bX)^2.‎ 
‎\qquad‎ 
‎(\mod I(V_m))‎, 
‎\end{align*}‎ 
‎we have‎ 
‎\begin{align*}‎ 
‎\phi &‎ 
‎= [aX+bY+(a^2+b^2)Z‎, ‎aY-bX]‎ 
‎\\ &‎ 
‎= [ (aX+bY+(a^2+b^2)Z) (aX+bY-(a^2+b^2)Z)‎, ‎(aY-bX) (aX+bY-(a^2+b^2)Z) ]‎ 
‎\\ &‎ 
‎= [‎ ‎-(aY-bX)^2,‎ ‎(aY-bX) (aX+bY-(a^2+b^2)Z) ]‎ 
‎\\ &‎ 
‎= [-(aY-bX)‎, ‎(aX+bY-(a^2+b^2)Z)]‎. 
‎\end{align*}‎ 
‎Thus $\phi([a‎, ‎b‎, -1])=[0, ‎2(a^2+b^2)]=[0,1]$‎, 
‎so $\phi$ is regular at every point of $V_m$‎, ‎i.e.‎, ‎$\phi$ is a morphism‎. 
‎One easily checks that the map‎: 
‎\begin{align*}‎ 
‎\psi‎ : ‎\mathbb{P}^1 \rightarrow V_m‎, 
‎\qquad‎ 
‎\psi = [a(S^2-T^2)‎ - ‎2bST‎, ‎2aST‎ + ‎b(S^2-T^2)‎, ‎S^2+T^2]‎, 
‎\end{align*}‎ 
‎is a morphism and provides an inverse for $\phi$‎, 
‎so $V_m$ and $\mathbb{P}^1$ are isomorphic‎.  
