Lines Matching refs:mod
26 \def\mod{{\mathit\ mod\ }}
27 \renewcommand{\pmod}[1]{\ ({\rm mod\ }{#1})}
793 C_{-1} = C_{-1} + 1\mbox{ }(\mbox{mod }2^W) \nonumber \\
3233 often simply denoted as $de \equiv 1\mbox{ }(\mbox{mod }\phi(N))$.
3236 $1 < M < N-2$ and computes the ciphertext $C = M^e\mbox{ }(\mbox{mod }N)$. Since finding the inverse exponent $d$
3238 $C^d \equiv \left (M^e \right)^d \equiv M^1 \equiv M\mbox{ }(\mbox{mod }N)$. Similarly the owner of the private key
3618 y^2 = x^3 - 3x + b\mbox{ }(\mbox{mod }p)
3965 -- check that g^q mod p == 1
3971 -- g^x mod p,
3972 -- check that y^q mod p == 1
3979 -- check that g^q mod p == 1
3985 -- g^x mod p,
3986 -- check that y^q mod p == 1
4035 \item $g = h^r \mbox{ (mod }p\mbox{)}$ a generator of order $q$ modulo $p$. $h$ can be any non-trivial random
4038 \item $y = g^x \mbox{ (mod }p\mbox{)}$ the public key.
4048 \item $(p-1) \equiv 0 \mbox{ (mod }q\mbox{)}$.
4049 \item $g^q \equiv 1 \mbox{ (mod }p\mbox{)}$.
4051 \item $y^q \equiv 1 \mbox{ (mod }p\mbox{)}$.
4898 as $\phi(pq)$ or $(p - 1)(q - 1)$. The decryption exponent $d$ is found as $de \equiv 1\mbox{ }(\mbox{mod } \phi(pq))$. If either $p$ or $q$ is composite the value of $d$ will be incorrect and the user
4918 to get a prime of the form $p \equiv 3\mbox{ }(\mbox{mod } 4)$. So if you want a 1024-bit prime of this sort pass
6219 @param d The destination (a*b mod c)
6227 @param c The destination (a*a mod b)
6235 @param c The destination (1/a mod b)
6466 /** The 1/q mod p CRT param */
6468 /** The d mod (p - 1) CRT param */
6470 /** The d mod (q - 1) CRT param */
