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authorNiels Möller <nisse@lysator.liu.se>2014-08-02 21:41:03 +0200
committerNiels Möller <nisse@lysator.liu.se>2014-08-02 21:41:03 +0200
commitb6c445639015fb8dbe4058006dac7a7affcc7437 (patch)
tree036bdfbb5e95ef636a78a72f4a7dc225c73ad348 /misc
parent131d068d4fdbe41e13b4d04cb14ff3c764464552 (diff)
downloadnettle-b6c445639015fb8dbe4058006dac7a7affcc7437.tar.gz
Fixed equations for Montgomery->Edwards transformation.
Diffstat (limited to 'misc')
-rw-r--r--misc/ecc-formulas.tex4
1 files changed, 2 insertions, 2 deletions
diff --git a/misc/ecc-formulas.tex b/misc/ecc-formulas.tex
index 36c15227..46225066 100644
--- a/misc/ecc-formulas.tex
+++ b/misc/ecc-formulas.tex
@@ -127,7 +127,7 @@ mapping $P = (x,y)$ to $P' = (u, v)$, as follows.
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}
@@ -135,7 +135,7 @@ mapping $P = (x,y)$ to $P' = (u, v)$, as follows.
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