Mixing
How to remove power in a modular exponent expression?
$begingroup$
I am implementing a cryptography scheme which involves verifiying some data through the following process:
Suppose party A wants to verify data held by party B
Party A has: $a^x mod N$
Party B has: $x^{-1}$ i.e. modular inverse of $x$ with respect
to some $p$ such that $xx^{-1} equiv 1 mod p$
To carry out the verification, B has to use $a^x bmod N$ and $x^{-1}$ to attempt creating $abmod N$
If $x^{-1}$ is correctly calculated, then the data held by B is verified.
My question is, how can I, using $a^x bmod N$ and modular inverse $x^{-1}$ attempt to generate $abmod N$
modular-arithmetic exponentiation
edited Jan 28 at 9:38
Bernard
123k741117
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
$endgroup$
add a comment |
$begingroup$
I am implementing a cryptography scheme which involves verifiying some data through the following process:
Suppose party A wants to verify data held by party B
Party A has: $a^x mod N$
Party B has: $x^{-1}$ i.e. modular inverse of $x$ with respect
to some $p$ such that $xx^{-1} equiv 1 mod p$
To carry out the verification, B has to use $a^x bmod N$ and $x^{-1}$ to attempt creating $abmod N$
If $x^{-1}$ is correctly calculated, then the data held by B is verified.
My question is, how can I, using $a^x bmod N$ and modular inverse $x^{-1}$ attempt to generate $abmod N$
modular-arithmetic exponentiation
edited Jan 28 at 9:38
Bernard
123k741117
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
$endgroup$
$begingroup$
Based on the cryptgraphic scheme detailed here: link.springer.com/chapter/10.1007/0-387-34805-0_20
$endgroup$
– Tabish Mir
Jan 28 at 7:20
2
$begingroup$
Given y, it's quick to get y inverse mod p by the euclidean algorithm on p, y. So take x inverse, compute x, then you know the exponent used
$endgroup$
– Artimis Fowl
Jan 28 at 7:25
$begingroup$
x is a confidential value only to be known by B. As such, A attempts to regenerate a mod N using his version of x, a^x mod N and the corresponding x inverse. If 'x' held by B is valid, he should be able to successfully generate a mod N
$endgroup$
– Tabish Mir
Jan 28 at 7:32
add a comment |
$begingroup$
I am implementing a cryptography scheme which involves verifiying some data through the following process:
Suppose party A wants to verify data held by party B
Party A has: $a^x mod N$
Party B has: $x^{-1}$ i.e. modular inverse of $x$ with respect
to some $p$ such that $xx^{-1} equiv 1 mod p$
To carry out the verification, B has to use $a^x bmod N$ and $x^{-1}$ to attempt creating $abmod N$
If $x^{-1}$ is correctly calculated, then the data held by B is verified.
My question is, how can I, using $a^x bmod N$ and modular inverse $x^{-1}$ attempt to generate $abmod N$
modular-arithmetic exponentiation
edited Jan 28 at 9:38
Bernard
123k741117
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
$endgroup$
I am implementing a cryptography scheme which involves verifiying some data through the following process:
Suppose party A wants to verify data held by party B
Party A has: $a^x mod N$
Party B has: $x^{-1}$ i.e. modular inverse of $x$ with respect
to some $p$ such that $xx^{-1} equiv 1 mod p$
To carry out the verification, B has to use $a^x bmod N$ and $x^{-1}$ to attempt creating $abmod N$
If $x^{-1}$ is correctly calculated, then the data held by B is verified.
My question is, how can I, using $a^x bmod N$ and modular inverse $x^{-1}$ attempt to generate $abmod N$
modular-arithmetic exponentiation
modular-arithmetic exponentiation
edited Jan 28 at 9:38
Bernard
123k741117
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
edited Jan 28 at 9:38
Bernard
123k741117
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
edited Jan 28 at 9:38
Bernard
123k741117
edited Jan 28 at 9:38
Bernard
123k741117
edited Jan 28 at 9:38
Bernard
123k741117
123k741117
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
asked Jan 28 at 7:20
Tabish MirTabish Mir
12617
12617
$begingroup$
Based on the cryptgraphic scheme detailed here: link.springer.com/chapter/10.1007/0-387-34805-0_20
$endgroup$
– Tabish Mir
Jan 28 at 7:20
2
$begingroup$
Given y, it's quick to get y inverse mod p by the euclidean algorithm on p, y. So take x inverse, compute x, then you know the exponent used
$endgroup$
– Artimis Fowl
Jan 28 at 7:25
$begingroup$
x is a confidential value only to be known by B. As such, A attempts to regenerate a mod N using his version of x, a^x mod N and the corresponding x inverse. If 'x' held by B is valid, he should be able to successfully generate a mod N
$endgroup$
– Tabish Mir
Jan 28 at 7:32
add a comment |
$begingroup$
Based on the cryptgraphic scheme detailed here: link.springer.com/chapter/10.1007/0-387-34805-0_20
$endgroup$
– Tabish Mir
Jan 28 at 7:20
2
$begingroup$
Given y, it's quick to get y inverse mod p by the euclidean algorithm on p, y. So take x inverse, compute x, then you know the exponent used
$endgroup$
– Artimis Fowl
Jan 28 at 7:25
$begingroup$
x is a confidential value only to be known by B. As such, A attempts to regenerate a mod N using his version of x, a^x mod N and the corresponding x inverse. If 'x' held by B is valid, he should be able to successfully generate a mod N
$endgroup$
– Tabish Mir
Jan 28 at 7:32
$begingroup$
Based on the cryptgraphic scheme detailed here: link.springer.com/chapter/10.1007/0-387-34805-0_20
$endgroup$
– Tabish Mir
Jan 28 at 7:20
$begingroup$
Based on the cryptgraphic scheme detailed here: link.springer.com/chapter/10.1007/0-387-34805-0_20
$endgroup$
– Tabish Mir
Jan 28 at 7:20
2
2
$begingroup$
Given y, it's quick to get y inverse mod p by the euclidean algorithm on p, y. So take x inverse, compute x, then you know the exponent used
$endgroup$
– Artimis Fowl
Jan 28 at 7:25
$begingroup$
Given y, it's quick to get y inverse mod p by the euclidean algorithm on p, y. So take x inverse, compute x, then you know the exponent used
$endgroup$
– Artimis Fowl
Jan 28 at 7:25
$begingroup$
x is a confidential value only to be known by B. As such, A attempts to regenerate a mod N using his version of x, a^x mod N and the corresponding x inverse. If 'x' held by B is valid, he should be able to successfully generate a mod N
$endgroup$
– Tabish Mir
Jan 28 at 7:32
$begingroup$
x is a confidential value only to be known by B. As such, A attempts to regenerate a mod N using his version of x, a^x mod N and the corresponding x inverse. If 'x' held by B is valid, he should be able to successfully generate a mod N
$endgroup$
– Tabish Mir
Jan 28 at 7:32
add a comment |
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$begingroup$
Based on the cryptgraphic scheme detailed here: link.springer.com/chapter/10.1007/0-387-34805-0_20
$endgroup$
– Tabish Mir
Jan 28 at 7:20
2
$begingroup$
Given y, it's quick to get y inverse mod p by the euclidean algorithm on p, y. So take x inverse, compute x, then you know the exponent used
$endgroup$
– Artimis Fowl
Jan 28 at 7:25
$begingroup$
x is a confidential value only to be known by B. As such, A attempts to regenerate a mod N using his version of x, a^x mod N and the corresponding x inverse. If 'x' held by B is valid, he should be able to successfully generate a mod N
$endgroup$
– Tabish Mir
Jan 28 at 7:32