highlevel.tex 24.5 KB
 Peter Schwabe committed Jun 20, 2019 1 \section{Proving that X25519 in Coq matches the mathematical model}  Benoit Viguier committed Jun 21, 2019 2 3 4 5 6 \label{sec3-maths} In this section we first present the work of Bartzia and Strub \cite{DBLP:conf/itp/BartziaS14} (\ref{Weierstrass}). We extend it to support Montgomery curves (\ref{montgomery}) with homogeneous coordinates (\ref{projective}) and prove the correctness of the ladder (\ref{ladder}).  Benoit Viguier committed Aug 01, 2019 7 We then prove the Montgomery ladder computes  Benoit Viguier committed Jun 21, 2019 8 9 10 11 12 13 14 15 16 17 the x-coordinate of scalar multiplication over $\F{p^2}$ (Theorem 2.1 by Bernstein \cite{Ber06}) where $p$ is the prime $\p$. \subsection{Formalization of Elliptic Curves} We consider elliptic curves over a field $\K$. We assume that the characteristic of $\K$ is neither 2 or 3. \begin{definition} Let a field $\K$, using an appropriate choice of coordinates, an elliptic curve $E$  Benoit Viguier committed Jul 22, 2019 18 is a plane cubic algebraic curve $E(x,y)$ defined by an equation of the form:  Benoit Viguier committed Jun 21, 2019 19 20 21 22 23 24 25 26 27 $$E : y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$$ where the $a_i$'s are in \K\ and the curve has no singular point (\ie no cusps or self-intersections). The set of points, written $E(\K)$, is formed by the solutions $(x,y)$ of $E$ augmented by a distinguished point $\Oinf$ (called point at infinity): $$E(\K) = \{(x,y) \in \K \times \K | E(x,y)\} \cup \{\Oinf\}$$ \end{definition} \subsubsection{Weierstra{\ss} Curves} \label{Weierstrass}  Benoit Viguier committed Jul 22, 2019 28 This equation $E(x,y)$ can be reduced into its short Weierstra{\ss} form.  Benoit Viguier committed Jun 21, 2019 29 30  \begin{definition}  Benoit Viguier committed Jul 22, 2019 31 32 Let $a \in \K$, and $b \in \K$ such that $$\Delta(a,b) = -16(4a^3 + 27b^2) \neq 0.$$ The \textit{elliptic curve} $E_{a,b}(\K)$ is the set of all points $(x,y) \in \K^2$ satisfying the equation:  Benoit Viguier committed Jun 21, 2019 33 34 35 36 37 $$y^2 = x^3 + ax + b,$$ along with an additional formal point $\Oinf$, at infinity''. Such curve does not present any singularity. \end{definition} In this setting, Bartzia and Strub defined the parametric type \texttt{ec} which  Benoit Viguier committed Aug 01, 2019 38 represent the points on a specific curve. It is parameterized by  Benoit Viguier committed Jun 21, 2019 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 a \texttt{K : ecuFieldType} -- the type of fields which characteristic is not 2 or 3 -- and \texttt{E : ecuType} -- a record that packs the curve parameters $a$ and $b$ along with the proof that $\Delta(a,b) \neq 0$. \begin{lstlisting}[language=Coq] Inductive point := EC_Inf | EC_In of K * K. Notation "(| x, y |)" := (EC_In x y). Notation "\infty" := (EC_Inf). Record ecuType := { A : K; B : K; _ : 4 * A^3 + 27 * B^2 != 0}. Definition oncurve (p : point) := if p is (| x, y |) then y^2 == x^3 + A * x + B else true. Inductive ec : Type := EC p of oncurve p. \end{lstlisting}  Benoit Viguier committed Aug 01, 2019 56 Points of an elliptic curve can be equipped with a structure of an abelian group.  Benoit Viguier committed Jun 21, 2019 57 \begin{itemize}  Benoit Viguier committed Aug 01, 2019 58  \item The negation of a point $P = (x,y)$ by taking the symmetric with respect to the x axis $-P = (x, -y)$.  Benoit Viguier committed Jun 21, 2019 59 60  \item The addition of two points $P$ and $Q$ is defined by the negation of third intersection of the line passing by $P$ and $Q$ or tangent to $P$ if $P = Q$.  Benoit Viguier committed Aug 01, 2019 61  \item $\Oinf$ is the neutral element under this law: if 3 points are collinear, their sum is equal to $\Oinf$.  Benoit Viguier committed Jun 21, 2019 62 63 \end{itemize}  Benoit Viguier committed Aug 01, 2019 64 This operation are defined in Coq as follow:  Benoit Viguier committed Jun 21, 2019 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 \begin{lstlisting}[language=Coq] Definition neg (p : point) := if p is (| x, y |) then (| x, -y |) else EC_Inf. Definition add (p1 p2 : point) := match p1, p2 with | \infty , _ => p2 | _ , \infty => p1 | (| x1, y1 |), (| x2, y2 |) => if x1 == x2 then ... else let s := (y2 - y1) / (x2 - x1) in let xs := s^2 - x1 - x2 in (| xs, - s * (xs - x1 ) - y1 |) end. \end{lstlisting} And is proven internal to the curve (with coercion): \begin{lstlisting}[language=Coq] Lemma addO (p q : ec): oncurve (add p q). Definition addec (p1 p2 : ec) : ec := EC p1 p2 (addO p1 p2) \end{lstlisting} \subsubsection{Montgomery Curves} \label{montgomery} Computation over elliptic curves are hard. Speedups can be obtained by using homogeneous coordinates and other forms than the Weierstra{\ss} form. We consider the Montgomery form \cite{MontgomerySpeeding}. \begin{definition}  Benoit Viguier committed Jul 22, 2019 96 97  Let $a \in \K \backslash \{-2, 2\}$, and $b \in \K \backslash \{ 0\}$. The \textit{Montgomery curve} $M_{a,b}(\K)$ is the set of all points $(x,y) \in \K^2$ satisfying the equation:  Benoit Viguier committed Jun 21, 2019 98 99 100 101  $$by^2 = x^3 + ax^2 + x,$$ along with an additional formal point $\Oinf$, at infinity''. \end{definition} Using a similar representation, we defined the parametric type \texttt{mc} which  Benoit Viguier committed Aug 01, 2019 102 represent the points on a specific Montgomery curve. It is parameterized by  Benoit Viguier committed Jun 21, 2019 103 a \texttt{K : ecuFieldType} -- the type of fields which characteristic is not 2 or 3 --  Benoit Viguier committed Aug 01, 2019 104 and \texttt{M : mcuType} -- a record that packs the curve parameters $a$ and $b$  Benoit Viguier committed Jul 22, 2019 105 along with the proofs that $b \neq 0$ and $a^2 \neq 4$.  Benoit Viguier committed Jun 21, 2019 106 107 108 109 110 111 112 113 114 115 116 \begin{lstlisting}[language=Coq] Record mcuType := { cA : K; cB : K; _ : cB != 0; _ : cA^2 != 4}. Definition oncurve (p : point K) := if p is (| x, y |) then cB * y^+2 == x^+3 + cA * x^+2 + x else true. Inductive mc : Type := MC p of oncurve p. Lemma oncurve_mc: forall p : mc, oncurve p. \end{lstlisting}  Benoit Viguier committed Jul 22, 2019 117 We define the addition on Montgomery curves the same way as it is in the Weierstra{\ss} form,  Benoit Viguier committed Jun 21, 2019 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 however the actual computations will be slightly different. \begin{lstlisting}[language=Coq] Definition add (p1 p2 : point K) := match p1, p2 with | \infty, _ => p2 | _, \infty => p1 | (|x1, y1|), (|x2, y2|) => if x1 == x2 then if (y1 == y2) && (y1 != 0) then ... else \infty else let s := (y2 - y1) / (x2 - x1) in let xs := s^+2 * cB - cA - x1 - x2 in (| xs, - s * (xs - x1) - y1 |) end. \end{lstlisting} But we prove it is internal to the curve (again with coercion): \begin{lstlisting}[language=Coq] Lemma addO (p q : mc) : oncurve (add p q). Definition addmc (p1 p2 : mc) : mc := MC p1 p2 (addO p1 p2) \end{lstlisting} We then prove a bijection between a Montgomery curve and its Weierstra{\ss} equation. \begin{lemma} Let $M_{a,b}(\K)$ be a Montgomery curve, define $$a' = \frac{3-a^2}{3b^2} \text{\ \ \ \ and\ \ \ \ } b' = \frac{2a^3 - 9a}{27b^3}.$$ then $E_{a',b'}(\K)$ is an elliptic curve, and the mapping $\varphi : M_{a,b}(\K) \mapsto E_{a',b'}(\K)$ defined as: \begin{align*} \varphi(\Oinf_M) &= \Oinf_E\\ \varphi( (x , y) ) &= ( \frac{x}{b} + \frac{a}{3b} , \frac{y}{b} ) \end{align*} is an isomorphism between groups. \end{lemma} \begin{lstlisting}[language=Coq] Definition ec_of_mc_point p := match p with | \infty => \infty | (|x, y|) => (|x/(M#b) + (M#a)/(3%:R * (M#b)), y/(M#b)|) end. Lemma ec_of_mc_point_ok p : oncurve M p -> ec.oncurve E (ec_of_mc_point p). Definition ec_of_mc p := EC (ec_of_mc_point_ok (oncurve_mc p)). Lemma ec_of_mc_bij : bijective ec_of_mc. \end{lstlisting} \subsubsection{Projective Coordinates} \label{projective}  Benoit Viguier committed Jul 22, 2019 169 170 171 172 Points on a projective plane are represented with a triple $(X:Y:Z)$. Any points except $(0:0:0)$ defines a point on a projective plane. A scalar multiple of a point defines the same point, \ie for all $\alpha \neq 0$, $(X:Y:Z)$ and $(\alpha X:\alpha Y:\alpha Z)$ defines the same point. For $Z\neq 0$, the projective point $(X:Y:Z)$ corresponds to the  Benoit Viguier committed Aug 01, 2019 173 point $(X/Z,Y/Z)$ on the Euclidean plane, likewise the point $(X,Y)$ on the Euclidean plane corresponds to $(X:Y:1)$ on the projective plane.  Benoit Viguier committed Jun 21, 2019 174 175 176 177 178 179 180 181 182  We write the equation for a Montgomery curve $M_{a,b}(\K)$ as such: b \bigg(\frac{Y}{Z}\bigg)^2 = \bigg(\frac{X}{Z}\bigg)^3 + a \bigg(\frac{X}{Z}\bigg)^2 + \bigg(\frac{X}{Z}\bigg) Multiplying both sides by $Z^3$ yields: b Y^2Z = X^3 + a X^2Z + XZ^2  Benoit Viguier committed Jul 22, 2019 183 184 With this equation we can additionally represent the point at infinity''. By setting $Z=0$, we derive $X=0$, giving us the infinite points'' $(0:Y:0)$ with $Y\neq 0$.  Benoit Viguier committed Jun 21, 2019 185   Benoit Viguier committed Aug 01, 2019 186 By restricting the parameter $a$ of $M_{a,b}(\K)$ such that $a^2-4$ is not a  Benoit Viguier committed Jul 22, 2019 187 square in \K, we ensure that $(0,0)$ is the only point with a $y$-coordinate of $0$.  Benoit Viguier committed Jun 21, 2019 188 189 190 191 192 193 194 195 196 197 198 199 200 \begin{lstlisting}[language=Coq] Hypothesis mcu_no_square : forall x : K, x^+2 != (M#a)^+2 - 4%:R. \end{lstlisting} With those coordinates we prove the following lemmas for the addition of two points. \begin{definition}Define the functions $\chi$ and $\chi_0$:\\ -- $\chi : M_{a,b}(\K) \to \K \cup \{\infty\}$\\ such that $\chi(\Oinf) = \infty$ and $\chi((x,y)) = x$.\\ -- $\chi_0 : M_{a,b}(\K) \to \K$\\ such that $\chi_0(\Oinf) = 0$ and $\chi_0((x,y)) = x$. \end{definition} \begin{lemma} \label{lemma-add}  Benoit Viguier committed Jul 22, 2019 201 202 Let $M_{a,b}(\K)$ be a Montgomery curve such that $a^2-4$ is not a square, and let $X_1, Z_1, X_2, Z_2, X_3, Z_3 \in \K$, such that $(X_1,Z_1) \neq (0,0)$, $(X_2,Z_2) \neq (0,0)$, $X_4 \neq 0$ and $Z_4 \neq 0$.  Benoit Viguier committed Jun 21, 2019 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 Define \begin{align*} X_3 &= Z_4((X_1 - Z_1)(X_2+Z_2) + (X_1+Z_1)(X_2-Z_2))^2\\ Z_3 &= X_4((X_1 - Z_1)(X_2+Z_2) - (X_1+Z_1)(X_2-Z_2))^2, \end{align*} then for any point $P_1$ and $P_2$ on $M_{a,b}(\K)$ such that $X_1/Z_1 = \chi(P_1), X_2/Z_2 = \chi(P_2)$, and $X_4/Z_4 = \chi(P_1 - P_2)$, we have $X_3/Z_3 = \chi(P_1+P_2)$.\\ \textbf{Remark:} For any $x \in \K \backslash\{0\}, x/0$ should be understood as $\infty$. \end{lemma} % This can be formalized as follow: % \begin{lstlisting}[language=Coq] % Inductive K_infty := % | K_Inf : K_infty % | K_Fin : K -> K_infty. % % Definition point_x (p : point K) := % if p is (|x, _|) then K_Fin x else K_Inf. % Local Notation "p '#x'" := (point_x p) (at level 30). % Definition point_x0 (p : point K) := % if p is (|x, _|) then x else 0. % Local Notation "p '#x0'" := (point_x0 p) (at level 30). % % Definition inf_div (x z : K) := % if z == 0 then K_Inf else K_Fin (x / z). % Definition hom_ok (x z : K) := (x != 0) || (z != 0). % Lemma montgomery_hom_neq : % forall x1 x2 x4 z1 z2 z4 : K, % hom_ok x1 z1 -> hom_ok x2 z2 -> % (x4 != 0) && (z4 != 0) -> % let x3 := z4 * ((x1 - z1)*(x2 + z2) % + (x1 + z1)*(x2 - z2))^+2 in % let z3 := x4 * ((x1 - z1)*(x2 + z2) % - (x1 + z1)*(x2 - z2))^+2 in % forall p1 p2 : point K, % oncurve M p1 -> oncurve M p2 -> % p1#x = inf_div x1 z1 -> % p2#x = inf_div x2 z2 -> % (p1 \- p2)#x = inf_div x4 z4 -> % hom_ok x3 z3 && ((p1 \+ p2)#x == inf_div x3 z3). % \end{lstlisting} With those coordinates we also prove a similar lemma for point doubling. \begin{lemma} \label{lemma-double}  Benoit Viguier committed Jul 22, 2019 246 247 Let $M_{a,b}(\K)$ be a Montgomery curve such that $a^2-4$ is not a square, and let $X_1, Z_1, X_2, Z_2, X_3, Z_3 \in \K$, such that $(X_1,Z_1) \neq (0,0)$. Define  Benoit Viguier committed Jun 21, 2019 248 249 250 251 252 \begin{align*} c &= (X_1 + Z_1)^2 - (X_1 - Z_1)^2\\ X_3 &= (X_1 + Z_1)^2(X_1-Z_1)^2\\ Z_3 &= c\Big((X_1 + Z_1)^2+\frac{a-2}{4}\times c\Big), \end{align*}  Benoit Viguier committed Jul 22, 2019 253 254 then for any point $P_1$ on $M_{a,b}(\K)$ such that $X_1/Z_1 = \chi(P_1)$, we have $X_3/Z_3 = \chi(2P_1)$.  Benoit Viguier committed Jun 21, 2019 255 256 257 258 259 260 261 262 263 264 265 266 267 268 \end{lemma} % Which is formalized as follow: % \begin{lstlisting}[language=Coq] % Lemma montgomery_hom_eq : % forall x1 z1 : K, % hom_ok x1 z1 -> % let c := (x1 + z1)^+2 - (x1 - z1)^+2 in % let x3 := (x1 + z1)^+2 * (x1 - z1)^+2 in % let z3 := c * ((x1 + z1)^+2 + (((M#a) - 2%:R)/4%:R) * c) in % forall p : point K, oncurve M p -> % p#x = inf_div x1 z1 -> % (p \+ p)#x = inf_div x3 z3. % \end{lstlisting}  Benoit Viguier committed Jul 22, 2019 269 270 With these two lemmas (\ref{lemma-add} and \ref{lemma-double}), we have the basic tools to compute efficiently additions and point doubling on projective coordinates.  Benoit Viguier committed Jun 21, 2019 271 272 273 274  \subsubsection{Scalar Multiplication Algorithms} \label{ladder}  Benoit Viguier committed Jul 22, 2019 275 276 277 278 Suppose we have a scalar $n$ and a point $P$ on some curve. The most straightforward way to compute $n \cdot P$ is to repetitively add $P$ \ie computing $P + \ldots + P$. However there is an more efficient algorithm which makes use of the binary representation of $n$ and by combining doubling and adding and starting from $\Oinf$.  Benoit Viguier committed Jun 21, 2019 279 280 \eg for $n=11$, we compute $2(2(2(2\Oinf + P)) + P)+ P$.  Benoit Viguier committed Jul 22, 2019 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 % \begin{algorithm} % \caption{Double-and-add for scalar mult.} % \label{double-add} % \begin{algorithmic} % \REQUIRE{Point $P$, scalars $n$ and $m$, $n < 2^m$} % \ENSURE{$Q = n \cdot P$} % \STATE $Q \leftarrow \Oinf$ % \FOR{$k$ := $m$ downto $1$} % \STATE $Q \leftarrow 2Q$ % \IF{$k^{\text{th}}$ bit of $n$ is $1$} % \STATE $Q \leftarrow Q + P$ % \ENDIF % \ENDFOR % \end{algorithmic} % \end{algorithm}  Benoit Viguier committed Jun 21, 2019 296   Benoit Viguier committed Jul 22, 2019 297 298 299 300 % \begin{lemma} % \label{lemma-double-add} % Algorithm \ref{double-add} is correct, \ie it respects its output conditions given the input conditions. % \end{lemma}  Benoit Viguier committed Jun 21, 2019 301   Benoit Viguier committed Jul 22, 2019 302 303 304 305 % We prove Lemma \ref{lemma-double-add}. However With a simple double-and-add algorithm, with careful timing, an attacker could reconstruct $n$. In the case of X25519, $n$ is the private key. With the Montgomery's ladder, while it provides slightly more computations and an extra variable, we can prevent such weakness.  Benoit Viguier committed Jun 21, 2019 306 307 308 309 310 311 312 See Algorithm \ref{montgomery-ladder}. \begin{algorithm} \caption{Montgomery ladder for scalar mult.} \label{montgomery-ladder} \begin{algorithmic} \REQUIRE{Point $P$, scalars $n$ and $m$, $n < 2^m$}  Benoit Viguier committed Jul 22, 2019 313 \ENSURE{$Q = n \cdot P$}  Benoit Viguier committed Jun 21, 2019 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 \STATE $Q \leftarrow \Oinf$ \STATE $R \leftarrow P$ \FOR{$k$ := $m$ downto $1$} \IF{$k^{\text{th}}$ bit of $n$ is $0$} \STATE $R \leftarrow Q + R$ \STATE $Q \leftarrow 2Q$ \ELSE \STATE $Q \leftarrow Q + R$ \STATE $R \leftarrow 2R$ \ENDIF \ENDFOR \end{algorithmic} \end{algorithm} \begin{lemma} \label{lemma-montgomery-ladder} Algorithm \ref{montgomery-ladder} is correct, \ie it respects its output conditions given the input conditions. \end{lemma}  Benoit Viguier committed Jul 22, 2019 333 334 335 336 In Curve25519 we are only interested in the $x$ coordinate of points, using Lemmas \ref{lemma-add} and \ref{lemma-double}, and replacing the if statements with conditional swapping we can define a ladder similar to the one used in TweetNaCl. See Algorithm \ref{montgomery-double-add}  Benoit Viguier committed Jun 21, 2019 337 338 339 340 341 342  \begin{algorithm} \caption{Montgomery ladder for scalar multiplication on $M_{a,b}(\K)$ with optimizations} \label{montgomery-double-add} \begin{algorithmic} \REQUIRE{$x \in \K\backslash \{0\}$, scalars $n$ and $m$, $n < 2^m$}  Benoit Viguier committed Jul 22, 2019 343 \ENSURE{$a/c = \chi_0(n \cdot P)$ for any $P$ such that $\chi_0(P) = x$}  Benoit Viguier committed Jun 21, 2019 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 \STATE $(a,b,c,d) \leftarrow (1,x,0,1)$ \FOR{$k$ := $m$ downto $1$} \IF{$k^{\text{th}}$ bit of $n$ is $1$} \STATE $(a,b) \leftarrow (b,a)$ \STATE $(c,d) \leftarrow (d,c)$ \ENDIF \STATE $e \leftarrow a + c$ \STATE $a \leftarrow a - c$ \STATE $c \leftarrow b + d$ \STATE $b \leftarrow b - d$ \STATE $d \leftarrow e^2$ \STATE $f \leftarrow a^2$ \STATE $a \leftarrow c \times a$ \STATE $c \leftarrow b \times e$ \STATE $e \leftarrow a + c$ \STATE $a \leftarrow a - c$ \STATE $b \leftarrow a^2$ \STATE $c \leftarrow d-f$ \STATE $a \leftarrow c\times\frac{A - 2}{4}$ \STATE $a \leftarrow a + d$ \STATE $c \leftarrow c \times a$ \STATE $a \leftarrow d \times f$ \STATE $d \leftarrow b \times x$ \STATE $b \leftarrow e^2$ \IF{$k^{\text{th}}$ bit of $n$ is $1$} \STATE $(a,b) \leftarrow (b,a)$ \STATE $(c,d) \leftarrow (d,c)$ \ENDIF \ENDFOR \end{algorithmic} \end{algorithm} \begin{lemma} \label{lemma-montgomery-double-add}  Benoit Viguier committed Jul 22, 2019 378 379 Algorithm \ref{montgomery-double-add} is correct, \ie it respects its output conditions given the input conditions.  Benoit Viguier committed Jun 21, 2019 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 \end{lemma} %% here we have \chi and \chi_0 ... We formalized this lemma (\ref{lemma-montgomery-double-add}): \begin{lstlisting}[language=Coq] Lemma opt_montgomery_x : forall (n m : nat) (x : K), n < 2^m -> x != 0 -> forall (p : mc M), p#x0 = x -> opt_montgomery n m x = (p *+ n)#x0. \end{lstlisting} We can remark that for an input $x = 0$, the ladder returns $0$. \begin{lstlisting}[language=Coq] Lemma opt_montgomery_0: forall (n m : nat), opt_montgomery n m 0 = 0. \end{lstlisting} Also \Oinf\ is the neutral element over $M_{a,b}(\K)$, we have: $$\forall P, P + \Oinf\ = P$$ thus we derive the following lemma. % \begin{lemma} % \label{lemma-montgomery-double-add} % Algorithm \ref{montgomery-double-add} is correct even if $x=0$, \ie it respects its output conditions given the input conditions or $x=0$. % \end{lemma} \begin{lstlisting}[language=Coq] Lemma p_x0_0_eq_0 : forall (n : nat) (p : mc M), p #x0 = 0%:R -> (p *+ n) #x0 = 0%R. \end{lstlisting} And thus the theorem of the correctness of the Montgomery ladder. \begin{theorem} \label{montgomery-ladder-correct} For all $n, m \in \N$, $x \in \K$, $P \in M_{a,b}(\K)$,  Benoit Viguier committed Jul 22, 2019 413 if $\chi_0(P) = x$ then Algorithm \ref{montgomery-double-add} returns $\chi_0(n \cdot P)$  Benoit Viguier committed Jun 21, 2019 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 \end{theorem} \begin{lstlisting}[language=Coq] Theorem opt_montgomery_ok (n m: nat) (x : K) : n < 2^m -> forall (p : mc M), p#x0 = x -> opt_montgomery n m x = (p *+ n)#x0. \end{lstlisting} \subsection{Curves, Twists and Extension Fields} One hypothesis to be able to use the above theorem is that $a^2-4$ is not a square: $$\forall x \in \K,\ x^2 \neq a^2-4$$ As Curve25519 is defined over the field $\K = \F{p^2}$, there exists $x$ such that $x^2 = a^2-4$. We first study Curve25519 and one of the quadratic twist Twist25519, first defined over \F{p}. \subsubsection{Curves and Twists} We define $\F{p}$ as the numbers between $0$ and $p = \p$. We create a \coqe{Zmodp} module to encapsulate those definitions. \begin{lstlisting}[language=Coq] Module Zmodp. Definition betweenb x y z := (x <=? z) && (z 0. Inductive type := Zmodp x of betweenb 0 p x. Lemma Z_mod_betweenb (x y : Z) : y > 0 -> betweenb 0 y (x mod y). Definition pi (x : Z) : type := Zmodp (Z_mod_betweenb x Hp_gt0). Coercion repr (x : type) : Z := let: @Zmodp x _ := x in x. End Zmodp. \end{lstlisting} We define the basic operations ($+, -, \times$) with their respective neutral elements ($0, 1$). \begin{lemma} $\F{p}$ is a commutative ring. \end{lemma} % \begin{lstlisting}[language=Coq] % Definition zero : type := pi 0. % Definition one : type := pi 1. % Definition opp (x : type) : type := pi (p - x). % Definition add (x y : type) : type := pi (x + y). % Definition sub (x y : type) : type := pi (x - y). % Definition mul (x y : type) : type := pi (x * y). % % Lemma Zmodp_ring : % ring_theory zero one add mul sub opp eq. % \end{lstlisting} And finally for $a = 486662$, by using the Legendre symbol we prove that $a^2 - 4$ and $2$ are not squares in $\F{p}$. \begin{lstlisting}[language=Coq] Lemma a_not_square : forall x: Zmodp.type, x^+2 != (Zmodp.pi 486662)^+2 - 4%:R. \end{lstlisting} \begin{lstlisting}[language=Coq,label=two_not_square] Lemma two_not_square : forall x : Zmodp.type, x^+2 != 2%:R. \end{lstlisting} We consider $M_{486662,1}(\F{p})$ and $M_{486662,2}(\F{p})$, one of its quadratic twist. % $M_{486662,1}(\F{p})$ has the same equation as $M_{486662,1}(\F{p^2})$ while $M_{486662,2}(\F{p})$ is one of its quadratic twist.  Benoit Viguier committed Aug 01, 2019 478 By instantiating theorem \ref{montgomery-ladder-correct} we derive the following two lemmas:  Benoit Viguier committed Jun 21, 2019 479 480 \begin{lemma} For all $x \in \F{p},\ n \in \N,\ P \in \F{p} \times \F{p}$,\\ such that $P \in M_{486662,1}(\F{p})$ and $\chi_0(P) = x$.  Benoit Viguier committed Jul 22, 2019 481 Given $n$ and $x$, $Curve25519\_Fp(n,x) = \chi_0(n \cdot P)$.  Benoit Viguier committed Jun 21, 2019 482 483 484 \end{lemma} \begin{lemma} For all $x \in \F{p},\ n \in \N,\ P \in \F{p} \times \F{p}$\\ such that $P \in M_{486662,2}(\F{p})$ and $\chi_0(P) = x$.  Benoit Viguier committed Jul 22, 2019 485 Given $n$ and $x$, $Twist25519\_Fp(n,x) = \chi_0(n \cdot P)$.  Benoit Viguier committed Jun 21, 2019 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 \end{lemma} As the Montgomery ladder defined above does not depends on $b$, it is trivial to see that the computations done for points of $M_{486662,1}(\F{p})$ and of $M_{486662,2}(\F{p})$ are the same. \begin{lstlisting}[language=Coq] Theorem curve_twist_eq: forall n x, curve25519_Fp_ladder n x = twist25519_Fp_ladder n x. \end{lstlisting} Because $2$ is not a square in $\F{p}$, it allows us split $\F{p}$ into two sets. \begin{lemma} \label{square-or-2square} For all $x$ in $\F{p}$, there exists $y$ in $\F{p}$ such that $$y^2 = x\ \ \ \lor\ \ 2y^2 = x$$ \end{lemma} For all $x \in \F{p}$, we can compute $x^3 + ax^2 + x$. Using Lemma \ref{square-or-2square} we can find a $y$ such that $(x,y)$ is either on the curve or on the quadratic twist: \begin{lemma} \label{curve-or-twist} For all $x \in \F{p}$, there exists a point $P$ over $M_{486662,1}(\F{p})$ or over $M_{486662,2}(\F{p})$ such that the $x$-coordinate of $P$ is $x$. \end{lemma} \begin{lstlisting}[language=Coq] Theorem x_is_on_curve_or_twist: forall x : Zmodp.type, (exists (p : mc curve25519_mcuType), p#x0 = x) \/ (exists (p' : mc twist25519_mcuType), p'#x0 = x). \end{lstlisting} \subsubsection{Curve25519 over \F{p^2}} We use the same definitions as in \cite{Ber06}. We consider the extension field $\F{p^2}$ as the set $\F{p} \times \F{p}$ with $\delta = 2$, in other words, the polynomial with coefficients in $\F{p}$ modulo $X^2 - 2$. In a similar way as for $\F{p}$ we use Module in Coq. \begin{lstlisting}[language=Coq] Module Zmodp. Inductive type := Zmodp2 (x: Zmodp.type) (y:Zmodp.type). Definition pi (x : Zmodp.type * Zmodp.type) : type := Zmodp2 x.1 x.2. Coercion repr (x : type) : Zmodp.type*Zmodp.type := let: Zmodp2 u v := x in (u, v). Definition zero : type := pi ( 0%:R, 0%:R ). Definition one : type := pi ( 1, 0%:R ). Definition opp (x : type) : type := pi (- x.1 , - x.2). Definition add (x y : type) : type := pi (x.1 + y.1, x.2 + y.2). Definition sub (x y : type) : type := pi (x.1 - y.1, x.2 - y.2). Definition mul (x y : type) : type := pi ((x.1 * y.1) + (2%:R * (x.2 * y.2)), (x.1 * y.2) + (x.2 * y.1)). \end{lstlisting} We define the basic operations ($+, -, \times$) with their respective neutral elements ($0, 1$). Additionally we verify that for each element of in $\F{p^2}\backslash\{0\}$, there exists a multiplicative inverse. \begin{lemma} For all $x \in \F{p^2}\backslash\{0\}$ and $a,b \in \F{p}$ such that $x = (a,b)$, $$x^{-1} = \Big(\frac{a}{a^2-2b^2}\ , \frac{-b}{a^2-2b^2}\Big)$$ \end{lemma}  Benoit Viguier committed Aug 01, 2019 543 Similarly as in $\F{p}$, we define $0^{-1} = 0$.  Benoit Viguier committed Jun 21, 2019 544 545 546 \begin{lemma} $\F{p^2}$ is a commutative ring. \end{lemma}  Benoit Viguier committed Aug 01, 2019 547 We can then specialize the basic operations in order to speed up the verification of formulas by using rewrite rules:  Benoit Viguier committed Jun 21, 2019 548 549 550 551 552 553 554 555 556 557 558 \begin{align*} (a,0) + (b,0) &= (a+b, 0)\\ (a,0) \cdot (b,0) &= (a \cdot b, 0)\\ (a, 0)^n &= (a^n, 0)\\ (a, 0)^{-1} &= (a^{-1}, 0)\\ (a, 0)\cdot (0,b) &= (0, a\cdot b)\\ (0, a)\cdot (0,b) &= (2\cdot a\cdot b, 0)\\ (0,a)^{-1} &= ((2\cdot a)^{-1},0) \end{align*} The injection $a \mapsto (a,0)$ from $\F{p}$ to $\F{p^2}$ preserves $0, 1, +, -, \times$. Thus $(a,0)$ can be abbreviated as $a$ without confusions.  Benoit Viguier committed Aug 01, 2019 559 We define $M_{486662,1}(\F{p^2})$. With the rewrite rule above, it is straightforward to prove that any point on the curve $M_{486662,1}(\F{p})$ is also on the curve $M_{486662,1}(\F{p^2})$. Similarly, any point on the quadratic twist $M_{486662,2}(\F{p})$ is also on the curve $M_{486662,1}(\F{p^2})$.  Benoit Viguier committed Jun 21, 2019 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 As direct consequence, using lemma \ref{curve-or-twist}, we prove that for all $x \in \F{p}$, there exists a point $P \in \F{p^2}\times\F{p^2}$ on $M_{486662,2}(\F{p})$ such that $\chi_0(P)$ is $(x,0)$ \begin{lstlisting}[language=Coq] Theorem x_is_on_curve_or_twist_implies_x_in_Fp2: forall (x:Zmodp.type), exists (p: mc curve25519_Fp2_mcuType), p#x0 = Zmodp2.Zmodp2 x 0. \end{lstlisting} We now study the case of the scalar multiplication and show similar proofs. \begin{definition} Define the functions $\varphi_c$, $\varphi_t$ and $\psi$\\ -- $\varphi_c: M_{486662,1}(\F{p}) \mapsto M_{486662,1}(\F{p^2})$\\ such that $\varphi((x,y)) = ((x,0), (y,0))$.\\ -- $\varphi_t: M_{486662,2}(\F{p}) \mapsto M_{486662,1}(\F{p^2})$\\ such that $\varphi((x,y)) = ((x,0), (0,y))$.\\ -- $\psi: \F{p^2} \mapsto \F{p}$\\ such that $\psi(x,y) = (x)$. \end{definition} \begin{lemma} For all $n \in \N$, for all point $P\in\F{p}\times\F{p}$ on the curve $M_{486662,1}(\F{p})$ (respectively on the quadratic twist $M_{486662,2}(\F{p})$), we have: \begin{align*} P \in M_{486662,1}(\F{p}) &\implies \varphi_c(n \cdot P) = n \cdot \varphi_c(P)\\ P \in M_{486662,2}(\F{p}) &\implies \varphi_t(n \cdot P) = n \cdot \varphi_t(P) \end{align*} \end{lemma} Notice that: \begin{align*} \forall P \in M_{486662,1}(\F{p}),\ \ \psi(\chi_0(\varphi_c(P))) = \chi_0(P)\\ \forall P \in M_{486662,2}(\F{p}),\ \ \psi(\chi_0(\varphi_t(P))) = \chi_0(P) \end{align*}  Benoit Viguier committed Jul 22, 2019 594 In summary for all $n \in \N,\ n < 2^{255}$, for any given point $P\in\F{p}\times\F{p}$ on $M_{486662,1}(\F{p})$ or $M_{486662,2}(\F{p})$ \texttt{curve25519\_Fp\_ladder} computes the $\chi_0(n \cdot P)$.  Benoit Viguier committed Jun 21, 2019 595 596 597 We have proved that for all $P \in \F{p^2}\times\F{p^2}$ such that $\chi_0(P) \in \F{p}$ there exists a corresponding point on the curve or the twist over $\F{p}$. We have proved that for any point, on the curve or the twist we can compute the scalar multiplication by $n$ and yield to the same result as if we did the computation in $\F{p^2}$. As a result we have proved theorem 2.1 of \cite{Ber06}: \begin{theorem}  Benoit Viguier committed Jul 22, 2019 598 For all $n \in \N$, $x \in \F{P}$, $P \in M_{486662,1}(\F{p^2})$, such that $n < 2^{255}$ and $\chi_0(P) = \varphi(x)$, \texttt{curve25519\_Fp\_ladder}$(n, x)$ computes $\psi(\chi_0(n \cdot P))$.  Benoit Viguier committed Jun 21, 2019 599 600 601 602 603 604 605 606 607 \end{theorem} which can be formalized in Coq as: \begin{lstlisting}[language=Coq] Lemma curve25519_Fp2_ladder_ok (n : nat) (x:Zmodp.type) : (n < 2^255)%nat -> forall (p : mc curve25519_Fp2_mcuType), p #x0 = Zmodp2.Zmodp2 x 0 -> curve25519_Fp_ladder n x = (p *+ n)#x0 /p. \end{lstlisting}