Mixing
For a strong prime $p = kq+1$, $langle g^k rangle le mathbb{QR}_p$
$begingroup$
Context
In the context of DLP-based cryptography one has to choose a group whose order behaves well with respect to the chinese remainder theorem. Ideally want wants the order to be prime. This is why for instance, one may choose for a safe prime $p$ the groups $mathbb{Z}_p^{*}$ or $mathbb{QR}_p$. This problem studies a different candidate for DLP-based cryptography.
Problem
We call strong prime to a prime $p$ of the form $p = kq+1$ with $q$ prime and $k in mathbb{N}$. Consider $mathbb{Z}_p^{*} = langle g rangle$ and $mathbb{QR}_p = {x^2 ; mod ; p: x in mathbb{Z}_p^{*}} $ is the set of quadratic residues modulo $p$.
How can I show that $langle g^k rangle le mathbb{QR}_p$?
abstract-algebra group-theory number-theory cryptography
asked Jan 21 at 16:21
JavierJavier
2,06621234
$endgroup$
add a comment |
$begingroup$
Context
In the context of DLP-based cryptography one has to choose a group whose order behaves well with respect to the chinese remainder theorem. Ideally want wants the order to be prime. This is why for instance, one may choose for a safe prime $p$ the groups $mathbb{Z}_p^{*}$ or $mathbb{QR}_p$. This problem studies a different candidate for DLP-based cryptography.
Problem
We call strong prime to a prime $p$ of the form $p = kq+1$ with $q$ prime and $k in mathbb{N}$. Consider $mathbb{Z}_p^{*} = langle g rangle$ and $mathbb{QR}_p = {x^2 ; mod ; p: x in mathbb{Z}_p^{*}} $ is the set of quadratic residues modulo $p$.
How can I show that $langle g^k rangle le mathbb{QR}_p$?
abstract-algebra group-theory number-theory cryptography
asked Jan 21 at 16:21
JavierJavier
2,06621234
$endgroup$
$begingroup$
Maybe I’m missing something, but I am suspicious of the claimed result (i.e. the claim that if $p = kq+1$ with $q$ prime and $mathbb{Z}_p^* = langle g rangle$, then $langle g^k rangle$ is contained in the set of quadratic residues $pmod{p}$). What if we take $k = 3$ and $q =2$ to give $p = 7$? Then $3$ is a generator for $mathbb{Z}_7^*$. However, $3^k = 3^3$ is congruent to $6$, which is not a square $pmod{7}$.
$endgroup$
– Jordan Green
Jan 21 at 18:14
$begingroup$
It is true that $langle g^k rangle subset mathbb{Q} mathbb{R}_p$ if you add the additional hypothesis that $q$ is odd, because this forces $k$ to be even.
$endgroup$
– Jordan Green
Jan 21 at 18:22
add a comment |
$begingroup$
Context
In the context of DLP-based cryptography one has to choose a group whose order behaves well with respect to the chinese remainder theorem. Ideally want wants the order to be prime. This is why for instance, one may choose for a safe prime $p$ the groups $mathbb{Z}_p^{*}$ or $mathbb{QR}_p$. This problem studies a different candidate for DLP-based cryptography.
Problem
We call strong prime to a prime $p$ of the form $p = kq+1$ with $q$ prime and $k in mathbb{N}$. Consider $mathbb{Z}_p^{*} = langle g rangle$ and $mathbb{QR}_p = {x^2 ; mod ; p: x in mathbb{Z}_p^{*}} $ is the set of quadratic residues modulo $p$.
How can I show that $langle g^k rangle le mathbb{QR}_p$?
abstract-algebra group-theory number-theory cryptography
asked Jan 21 at 16:21
JavierJavier
2,06621234
$endgroup$
Context
In the context of DLP-based cryptography one has to choose a group whose order behaves well with respect to the chinese remainder theorem. Ideally want wants the order to be prime. This is why for instance, one may choose for a safe prime $p$ the groups $mathbb{Z}_p^{*}$ or $mathbb{QR}_p$. This problem studies a different candidate for DLP-based cryptography.
Problem
We call strong prime to a prime $p$ of the form $p = kq+1$ with $q$ prime and $k in mathbb{N}$. Consider $mathbb{Z}_p^{*} = langle g rangle$ and $mathbb{QR}_p = {x^2 ; mod ; p: x in mathbb{Z}_p^{*}} $ is the set of quadratic residues modulo $p$.
How can I show that $langle g^k rangle le mathbb{QR}_p$?
abstract-algebra group-theory number-theory cryptography
abstract-algebra group-theory number-theory cryptography
asked Jan 21 at 16:21
JavierJavier
2,06621234
asked Jan 21 at 16:21
JavierJavier
2,06621234
edited Jan 25 at 14:55
Javier
asked Jan 21 at 16:21
JavierJavier
2,06621234
asked Jan 21 at 16:21
JavierJavier
2,06621234
asked Jan 21 at 16:21
JavierJavier
2,06621234
2,06621234
$begingroup$
Maybe I’m missing something, but I am suspicious of the claimed result (i.e. the claim that if $p = kq+1$ with $q$ prime and $mathbb{Z}_p^* = langle g rangle$, then $langle g^k rangle$ is contained in the set of quadratic residues $pmod{p}$). What if we take $k = 3$ and $q =2$ to give $p = 7$? Then $3$ is a generator for $mathbb{Z}_7^*$. However, $3^k = 3^3$ is congruent to $6$, which is not a square $pmod{7}$.
$endgroup$
– Jordan Green
Jan 21 at 18:14
$begingroup$
It is true that $langle g^k rangle subset mathbb{Q} mathbb{R}_p$ if you add the additional hypothesis that $q$ is odd, because this forces $k$ to be even.
$endgroup$
– Jordan Green
Jan 21 at 18:22
add a comment |
$begingroup$
Maybe I’m missing something, but I am suspicious of the claimed result (i.e. the claim that if $p = kq+1$ with $q$ prime and $mathbb{Z}_p^* = langle g rangle$, then $langle g^k rangle$ is contained in the set of quadratic residues $pmod{p}$). What if we take $k = 3$ and $q =2$ to give $p = 7$? Then $3$ is a generator for $mathbb{Z}_7^*$. However, $3^k = 3^3$ is congruent to $6$, which is not a square $pmod{7}$.
$endgroup$
– Jordan Green
Jan 21 at 18:14
$begingroup$
It is true that $langle g^k rangle subset mathbb{Q} mathbb{R}_p$ if you add the additional hypothesis that $q$ is odd, because this forces $k$ to be even.
$endgroup$
– Jordan Green
Jan 21 at 18:22
$begingroup$
Maybe I’m missing something, but I am suspicious of the claimed result (i.e. the claim that if $p = kq+1$ with $q$ prime and $mathbb{Z}_p^* = langle g rangle$, then $langle g^k rangle$ is contained in the set of quadratic residues $pmod{p}$). What if we take $k = 3$ and $q =2$ to give $p = 7$? Then $3$ is a generator for $mathbb{Z}_7^*$. However, $3^k = 3^3$ is congruent to $6$, which is not a square $pmod{7}$.
$endgroup$
– Jordan Green
Jan 21 at 18:14
$begingroup$
Maybe I’m missing something, but I am suspicious of the claimed result (i.e. the claim that if $p = kq+1$ with $q$ prime and $mathbb{Z}_p^* = langle g rangle$, then $langle g^k rangle$ is contained in the set of quadratic residues $pmod{p}$). What if we take $k = 3$ and $q =2$ to give $p = 7$? Then $3$ is a generator for $mathbb{Z}_7^*$. However, $3^k = 3^3$ is congruent to $6$, which is not a square $pmod{7}$.
$endgroup$
– Jordan Green
Jan 21 at 18:14
$begingroup$
It is true that $langle g^k rangle subset mathbb{Q} mathbb{R}_p$ if you add the additional hypothesis that $q$ is odd, because this forces $k$ to be even.
$endgroup$
– Jordan Green
Jan 21 at 18:22
$begingroup$
It is true that $langle g^k rangle subset mathbb{Q} mathbb{R}_p$ if you add the additional hypothesis that $q$ is odd, because this forces $k$ to be even.
$endgroup$
– Jordan Green
Jan 21 at 18:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assume $p$ odd.
Either $q$ is odd and then $k$ is even. Hence $g^k$ is trivially a quadratic residue.
Or $q=2$ and $p=2k+1.$
But since $g^{2k}=1,$ we must have $g^k=-1,$ otherwise $g^k=1,$ contradicting the primitivity of $g.$ Now $-1$ is a quadratic residue iff $p equiv 1 pmod{4}.$ by Euler criterion
https://en.wikipedia.org/wiki/Euler%27s_criterion
Thus if $q=2$ the statement must assume that condition.
A counterexample is obtained for $p=7=2times 3+1$ and $g=5.$
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
$endgroup$
$begingroup$
Please see how I edited just adding a few$. And what did you mean with your last part
$endgroup$
– reuns
Jan 21 at 18:58
$begingroup$
I also do not understand the last sentence of this post.
$endgroup$
– Jordan Green
Jan 21 at 19:37
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Assume $p$ odd.
Either $q$ is odd and then $k$ is even. Hence $g^k$ is trivially a quadratic residue.
Or $q=2$ and $p=2k+1.$
But since $g^{2k}=1,$ we must have $g^k=-1,$ otherwise $g^k=1,$ contradicting the primitivity of $g.$ Now $-1$ is a quadratic residue iff $p equiv 1 pmod{4}.$ by Euler criterion
https://en.wikipedia.org/wiki/Euler%27s_criterion
Thus if $q=2$ the statement must assume that condition.
A counterexample is obtained for $p=7=2times 3+1$ and $g=5.$
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
$endgroup$
$begingroup$
Please see how I edited just adding a few$. And what did you mean with your last part
$endgroup$
– reuns
Jan 21 at 18:58
$begingroup$
I also do not understand the last sentence of this post.
$endgroup$
– Jordan Green
Jan 21 at 19:37
add a comment |
$begingroup$
Assume $p$ odd.
Either $q$ is odd and then $k$ is even. Hence $g^k$ is trivially a quadratic residue.
Or $q=2$ and $p=2k+1.$
But since $g^{2k}=1,$ we must have $g^k=-1,$ otherwise $g^k=1,$ contradicting the primitivity of $g.$ Now $-1$ is a quadratic residue iff $p equiv 1 pmod{4}.$ by Euler criterion
https://en.wikipedia.org/wiki/Euler%27s_criterion
Thus if $q=2$ the statement must assume that condition.
A counterexample is obtained for $p=7=2times 3+1$ and $g=5.$
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
$endgroup$
$begingroup$
Please see how I edited just adding a few$. And what did you mean with your last part
$endgroup$
– reuns
Jan 21 at 18:58
$begingroup$
I also do not understand the last sentence of this post.
$endgroup$
– Jordan Green
Jan 21 at 19:37
add a comment |
$begingroup$
Assume $p$ odd.
Either $q$ is odd and then $k$ is even. Hence $g^k$ is trivially a quadratic residue.
Or $q=2$ and $p=2k+1.$
But since $g^{2k}=1,$ we must have $g^k=-1,$ otherwise $g^k=1,$ contradicting the primitivity of $g.$ Now $-1$ is a quadratic residue iff $p equiv 1 pmod{4}.$ by Euler criterion
https://en.wikipedia.org/wiki/Euler%27s_criterion
Thus if $q=2$ the statement must assume that condition.
A counterexample is obtained for $p=7=2times 3+1$ and $g=5.$
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
$endgroup$
Assume $p$ odd.
Either $q$ is odd and then $k$ is even. Hence $g^k$ is trivially a quadratic residue.
Or $q=2$ and $p=2k+1.$
But since $g^{2k}=1,$ we must have $g^k=-1,$ otherwise $g^k=1,$ contradicting the primitivity of $g.$ Now $-1$ is a quadratic residue iff $p equiv 1 pmod{4}.$ by Euler criterion
https://en.wikipedia.org/wiki/Euler%27s_criterion
Thus if $q=2$ the statement must assume that condition.
A counterexample is obtained for $p=7=2times 3+1$ and $g=5.$
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
edited Jan 21 at 20:46
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
answered Jan 21 at 18:52
Patrick SolePatrick Sole
1227
1227
$begingroup$
Please see how I edited just adding a few$. And what did you mean with your last part
$endgroup$
– reuns
Jan 21 at 18:58
$begingroup$
I also do not understand the last sentence of this post.
$endgroup$
– Jordan Green
Jan 21 at 19:37
add a comment |
$begingroup$
Please see how I edited just adding a few$. And what did you mean with your last part
$endgroup$
– reuns
Jan 21 at 18:58
$begingroup$
I also do not understand the last sentence of this post.
$endgroup$
– Jordan Green
Jan 21 at 19:37
$begingroup$
Please see how I edited just adding a few
$. And what did you mean with your last part$endgroup$
– reuns
Jan 21 at 18:58
$begingroup$
Please see how I edited just adding a few
$. And what did you mean with your last part$endgroup$
– reuns
Jan 21 at 18:58
$begingroup$
I also do not understand the last sentence of this post.
$endgroup$
– Jordan Green
Jan 21 at 19:37
$begingroup$
I also do not understand the last sentence of this post.
$endgroup$
– Jordan Green
Jan 21 at 19:37
add a comment |
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$begingroup$
Maybe I’m missing something, but I am suspicious of the claimed result (i.e. the claim that if $p = kq+1$ with $q$ prime and $mathbb{Z}_p^* = langle g rangle$, then $langle g^k rangle$ is contained in the set of quadratic residues $pmod{p}$). What if we take $k = 3$ and $q =2$ to give $p = 7$? Then $3$ is a generator for $mathbb{Z}_7^*$. However, $3^k = 3^3$ is congruent to $6$, which is not a square $pmod{7}$.
$endgroup$
– Jordan Green
Jan 21 at 18:14
$begingroup$
It is true that $langle g^k rangle subset mathbb{Q} mathbb{R}_p$ if you add the additional hypothesis that $q$ is odd, because this forces $k$ to be even.
$endgroup$
– Jordan Green
Jan 21 at 18:22