How to calculate mod inverse
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Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?
I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.
What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.
I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$
I am not sure if it is any different since I am doing a mod inverse.
PS. I am a programmer, not a mathematician so please elaborate.
number-theory
edited Nov 19 at 21:02
Mason
1,9161426
asked Nov 19 at 20:29
User
1
|
show 3 more comments
up vote
0
down vote
favorite
Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?
I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.
What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.
I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$
I am not sure if it is any different since I am doing a mod inverse.
PS. I am a programmer, not a mathematician so please elaborate.
number-theory
edited Nov 19 at 21:02
Mason
1,9161426
asked Nov 19 at 20:29
User
1
2
Use the Extended Euclidean Algorithm, e.g. see here
– Bill Dubuque
Nov 19 at 20:36
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
– User
Nov 19 at 20:39
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
– Bill Dubuque
Nov 19 at 20:58
So what if I have a number * an inverse mod p -1. How would I break that down?
– User
Nov 19 at 21:01
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
– Bill Dubuque
Nov 19 at 21:05
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?
I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.
What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.
I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$
I am not sure if it is any different since I am doing a mod inverse.
PS. I am a programmer, not a mathematician so please elaborate.
number-theory
edited Nov 19 at 21:02
Mason
1,9161426
asked Nov 19 at 20:29
User
1
Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?
I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.
What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.
I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$
I am not sure if it is any different since I am doing a mod inverse.
PS. I am a programmer, not a mathematician so please elaborate.
number-theory
number-theory
edited Nov 19 at 21:02
Mason
1,9161426
asked Nov 19 at 20:29
User
1
edited Nov 19 at 21:02
Mason
1,9161426
asked Nov 19 at 20:29
User
1
edited Nov 19 at 21:02
Mason
1,9161426
edited Nov 19 at 21:02
Mason
1,9161426
edited Nov 19 at 21:02
Mason
1,9161426
1,9161426
asked Nov 19 at 20:29
User
1
asked Nov 19 at 20:29
User
1
asked Nov 19 at 20:29
User
1
1
2
Use the Extended Euclidean Algorithm, e.g. see here
– Bill Dubuque
Nov 19 at 20:36
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
– User
Nov 19 at 20:39
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
– Bill Dubuque
Nov 19 at 20:58
So what if I have a number * an inverse mod p -1. How would I break that down?
– User
Nov 19 at 21:01
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
– Bill Dubuque
Nov 19 at 21:05
|
show 3 more comments
2
Use the Extended Euclidean Algorithm, e.g. see here
– Bill Dubuque
Nov 19 at 20:36
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
– User
Nov 19 at 20:39
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
– Bill Dubuque
Nov 19 at 20:58
So what if I have a number * an inverse mod p -1. How would I break that down?
– User
Nov 19 at 21:01
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
– Bill Dubuque
Nov 19 at 21:05
2
2
Use the Extended Euclidean Algorithm, e.g. see here
– Bill Dubuque
Nov 19 at 20:36
Use the Extended Euclidean Algorithm, e.g. see here
– Bill Dubuque
Nov 19 at 20:36
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
– User
Nov 19 at 20:39
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
– User
Nov 19 at 20:39
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
– Bill Dubuque
Nov 19 at 20:58
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
– Bill Dubuque
Nov 19 at 20:58
So what if I have a number * an inverse mod p -1. How would I break that down?
– User
Nov 19 at 21:01
So what if I have a number * an inverse mod p -1. How would I break that down?
– User
Nov 19 at 21:01
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
– Bill Dubuque
Nov 19 at 21:05
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
– Bill Dubuque
Nov 19 at 21:05
|
show 3 more comments
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Use the Extended Euclidean Algorithm, e.g. see here
– Bill Dubuque
Nov 19 at 20:36
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
– User
Nov 19 at 20:39
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
– Bill Dubuque
Nov 19 at 20:58
So what if I have a number * an inverse mod p -1. How would I break that down?
– User
Nov 19 at 21:01
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
– Bill Dubuque
Nov 19 at 21:05