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author | Niels Möller <nisse@lysator.liu.se> | 2014-08-02 21:41:03 +0200 |
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committer | Niels Möller <nisse@lysator.liu.se> | 2014-08-02 21:41:03 +0200 |
commit | b6c445639015fb8dbe4058006dac7a7affcc7437 (patch) | |
tree | 036bdfbb5e95ef636a78a72f4a7dc225c73ad348 /misc | |
parent | 131d068d4fdbe41e13b4d04cb14ff3c764464552 (diff) | |
download | nettle-b6c445639015fb8dbe4058006dac7a7affcc7437.tar.gz |
Fixed equations for Montgomery->Edwards transformation.
Diffstat (limited to 'misc')
-rw-r--r-- | misc/ecc-formulas.tex | 4 |
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 |