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Fix a prime power \(q \equiv 3 \pmod4\). Define \(\chi : \mathbb {F}_q \to \mathbb {F}_q\) as the quadratic character of \(\mathbb {F}_q\), with its values \(0, \pm 1\) read inside \(\mathbb {F}_q\); equivalently
If \(a\) is a nonzero square then \(\chi (a) = 1\); if \(a\) is a non-square then \(\chi (a) = -1\); if \(a = 0\) then \(\chi (a) = 0\).
In the situation of Theorem 1, the decoding function for the complete Edwards curve \(E : x^2 + y^2 = 1 + d x^2 y^2\) is the function \(\varphi : \mathbb {F}_q \to E(\mathbb {F}_q)\) defined as follows:
if \(t \notin \{ \pm 1\} \) then \(\varphi (t) = (x, y)\).
The second of the three conditions characterizing \(\varphi (\mathbb {F}_q)\) inside \(E(\mathbb {F}_q)\) in Theorem 3: a point \((x, y)\) satisfies that
is a square, where \(\eta = (y - 1)/(2(y + 1))\).
The conjunction of the three conditions of Theorem 3 for a point \((x, y) \in E(\mathbb {F}_q)\): \(y + 1 \neq 0\); \((1 + \eta r)^2 - 1\) is a square, where \(\eta = (y - 1)/(2(y + 1))\); and if \(\eta r = -2\) then \(x = 2s(c - 1)\chi (c)/r\).
In the situation of Theorem 1, the decoding function for the complete Edwards curve \(E : x^2 + y^2 = 1 + d x^2 y^2\) is the function \(\varphi : \mathbb {F}_q \to E(\mathbb {F}_q)\) with
Here \(\varphi \) is regarded as a map \(\mathbb {F}_q \to \mathbb {F}_q \times \mathbb {F}_q\), forgetting the proof that the image lies on \(E\).
In the situation of Theorem 1, let \(t \in \mathbb {F}_q \setminus \{ \pm 1\} \) and let \(x\), \(y\) be as above. Then \((x, y)\) is a point of the complete Edwards curve \(E : x^2 + y^2 = 1 + d x^2 y^2\), i.e.
The reverse part of statement 2 of Theorem 3: every \((x, y) \in E(\mathbb {F}_q)\) such that \(y + 1 \neq 0\); \((1 + \eta r)^2 - 1\) is a square, where \(\eta = (y - 1)/(2(y + 1))\); and \(x = 2s(c - 1)\chi (c)/r\) whenever \(\eta r = -2\), lies in \(\varphi (\mathbb {F}_q)\).
The forward part of statement 2 of Theorem 3: every \((x, y) \in \varphi (\mathbb {F}_q)\) satisfies \(y + 1 \neq 0\); \((1 + \eta r)^2 - 1\) is a square, where \(\eta = (y - 1)/(2(y + 1))\); and if \(\eta r = -2\) then \(x = 2s(c - 1)\chi (c)/r\).
In the situation of Definition 2: if \(t \in \mathbb {F}_q\) then the set of preimages of \(\varphi (t)\) under \(\varphi \) is \(\{ t, -t\} \). Equivalently, \(\varphi (t) = \varphi (-t)\) if and only if no element of \(\mathbb {F}_q\) other than \(t\) and \(-t\) maps to \(\varphi (t)\).
In the situation of Definition 2: \(\varphi (\mathbb {F}_q)\) is the set of \((x, y) \in E(\mathbb {F}_q)\) such that
\(y + 1 \neq 0\);
\((1 + \eta r)^2 - 1\) is a square, where \(\eta = \frac{y - 1}{2(y + 1)}\); and
if \(\eta r = -2\) then \(x = 2s(c - 1)\chi (c)/r\).
In the situation of Definition 2: if \((x, y) \in \varphi (\mathbb {F}_q)\) then the following elements \(\bar X, z, \bar u, \bar t\) of \(\mathbb {F}_q\) are defined and \(\varphi (\bar t) = (x, y)\):