that $x^2 + bx + 1 = 0$, or $(x + b/2)^2 = (b/2)^2 - 1$, which also
isn't a quadratic residue). The correspondence is then given by
\begin{align*}
- u &= \sqrt{b} \, x / y \\
+ u &= \sqrt{b+2} \, x / y \\
v &= (x-1) / (x+1)
\end{align*}
\end{itemize}
The inverse transformation is
\begin{align*}
x &= (1+v) / (1-v) \\
- y &= \sqrt{b} x / u
+ y &= \sqrt{b+2} x / u
\end{align*}
If the Edwards coordinates are represented using homogeneous
coordinates, $u = U/W$ and $v = V/W$, then