How many members $a$ in $Bbb{Z}^*$ have a number $b$, $b^3 = a pmod n$?
$begingroup$
Given two prime numbers $p$ and $q$ such that $3$ does not divide $p-1$ nor $q-1$, and let $pq = n$.
How many numbers in $mathbb{Z}_n^{*}$ (multiplicative group) are
equal to some $b^3$ where $b$ is a number ?
How does the fact that $3$ doesn't divide $(p-1)$ and $(q-1)$ help me find the answer ? All I know is that the size of $mathbb{Z}_n^{*}$ does not divide by $3$ (because its size is $(p-q)(q-1)$) but how does it help me ?
number-theory elementary-number-theory prime-numbers modular-arithmetic cryptography
asked Dec 18 '18 at 10:38
caffeincaffein
11
$endgroup$
|
show 7 more comments
$begingroup$
Given two prime numbers $p$ and $q$ such that $3$ does not divide $p-1$ nor $q-1$, and let $pq = n$.
How many numbers in $mathbb{Z}_n^{*}$ (multiplicative group) are
equal to some $b^3$ where $b$ is a number ?
How does the fact that $3$ doesn't divide $(p-1)$ and $(q-1)$ help me find the answer ? All I know is that the size of $mathbb{Z}_n^{*}$ does not divide by $3$ (because its size is $(p-q)(q-1)$) but how does it help me ?
number-theory elementary-number-theory prime-numbers modular-arithmetic cryptography
asked Dec 18 '18 at 10:38
caffeincaffein
11
$endgroup$
1
$begingroup$
Please use Mathjax
$endgroup$
– Math Girl
Dec 18 '18 at 10:39
$begingroup$
This equivalent to RSA with $e=3.$
$endgroup$
– gammatester
Dec 18 '18 at 10:53
$begingroup$
@gammatester But in RSA we perform b^3 mod n = a and not b^3 = a mod n. How is it still RSA ?
$endgroup$
– caffein
Dec 18 '18 at 11:08
$begingroup$
Huh? Do you know what a congruence is? Your notation is mathematically not appropriate (although it is often used in cryptography and programming). Both of your "equations" mean that $b^3 - a$ is a multiple of n.
$endgroup$
– gammatester
Dec 18 '18 at 11:14
$begingroup$
You are misunderstanding what $b^3equiv apmod n$ means. It doesn't mean you're applying the modulus operation only to $a$. You could say you're applying it to both $a$ and $b^3$, but usually $a<n$ already.
$endgroup$
– Matt Samuel
Dec 18 '18 at 11:19
|
show 7 more comments
$begingroup$
Given two prime numbers $p$ and $q$ such that $3$ does not divide $p-1$ nor $q-1$, and let $pq = n$.
How many numbers in $mathbb{Z}_n^{*}$ (multiplicative group) are
equal to some $b^3$ where $b$ is a number ?
How does the fact that $3$ doesn't divide $(p-1)$ and $(q-1)$ help me find the answer ? All I know is that the size of $mathbb{Z}_n^{*}$ does not divide by $3$ (because its size is $(p-q)(q-1)$) but how does it help me ?
number-theory elementary-number-theory prime-numbers modular-arithmetic cryptography
asked Dec 18 '18 at 10:38
caffeincaffein
11
$endgroup$
Given two prime numbers $p$ and $q$ such that $3$ does not divide $p-1$ nor $q-1$, and let $pq = n$.
How many numbers in $mathbb{Z}_n^{*}$ (multiplicative group) are
equal to some $b^3$ where $b$ is a number ?
How does the fact that $3$ doesn't divide $(p-1)$ and $(q-1)$ help me find the answer ? All I know is that the size of $mathbb{Z}_n^{*}$ does not divide by $3$ (because its size is $(p-q)(q-1)$) but how does it help me ?
number-theory elementary-number-theory prime-numbers modular-arithmetic cryptography
number-theory elementary-number-theory prime-numbers modular-arithmetic cryptography
asked Dec 18 '18 at 10:38
caffeincaffein
11
asked Dec 18 '18 at 10:38
caffeincaffein
11
edited Dec 18 '18 at 15:32
user593746
asked Dec 18 '18 at 10:38
caffeincaffein
11
asked Dec 18 '18 at 10:38
caffeincaffein
11
asked Dec 18 '18 at 10:38
caffeincaffein
11
11
1
$begingroup$
Please use Mathjax
$endgroup$
– Math Girl
Dec 18 '18 at 10:39
$begingroup$
This equivalent to RSA with $e=3.$
$endgroup$
– gammatester
Dec 18 '18 at 10:53
$begingroup$
@gammatester But in RSA we perform b^3 mod n = a and not b^3 = a mod n. How is it still RSA ?
$endgroup$
– caffein
Dec 18 '18 at 11:08
$begingroup$
Huh? Do you know what a congruence is? Your notation is mathematically not appropriate (although it is often used in cryptography and programming). Both of your "equations" mean that $b^3 - a$ is a multiple of n.
$endgroup$
– gammatester
Dec 18 '18 at 11:14
$begingroup$
You are misunderstanding what $b^3equiv apmod n$ means. It doesn't mean you're applying the modulus operation only to $a$. You could say you're applying it to both $a$ and $b^3$, but usually $a<n$ already.
$endgroup$
– Matt Samuel
Dec 18 '18 at 11:19
|
show 7 more comments
1
$begingroup$
Please use Mathjax
$endgroup$
– Math Girl
Dec 18 '18 at 10:39
$begingroup$
This equivalent to RSA with $e=3.$
$endgroup$
– gammatester
Dec 18 '18 at 10:53
$begingroup$
@gammatester But in RSA we perform b^3 mod n = a and not b^3 = a mod n. How is it still RSA ?
$endgroup$
– caffein
Dec 18 '18 at 11:08
$begingroup$
Huh? Do you know what a congruence is? Your notation is mathematically not appropriate (although it is often used in cryptography and programming). Both of your "equations" mean that $b^3 - a$ is a multiple of n.
$endgroup$
– gammatester
Dec 18 '18 at 11:14
$begingroup$
You are misunderstanding what $b^3equiv apmod n$ means. It doesn't mean you're applying the modulus operation only to $a$. You could say you're applying it to both $a$ and $b^3$, but usually $a<n$ already.
$endgroup$
– Matt Samuel
Dec 18 '18 at 11:19
1
1
$begingroup$
Please use Mathjax
$endgroup$
– Math Girl
Dec 18 '18 at 10:39
$begingroup$
Please use Mathjax
$endgroup$
– Math Girl
Dec 18 '18 at 10:39
$begingroup$
This equivalent to RSA with $e=3.$
$endgroup$
– gammatester
Dec 18 '18 at 10:53
$begingroup$
This equivalent to RSA with $e=3.$
$endgroup$
– gammatester
Dec 18 '18 at 10:53
$begingroup$
@gammatester But in RSA we perform b^3 mod n = a and not b^3 = a mod n. How is it still RSA ?
$endgroup$
– caffein
Dec 18 '18 at 11:08
$begingroup$
@gammatester But in RSA we perform b^3 mod n = a and not b^3 = a mod n. How is it still RSA ?
$endgroup$
– caffein
Dec 18 '18 at 11:08
$begingroup$
Huh? Do you know what a congruence is? Your notation is mathematically not appropriate (although it is often used in cryptography and programming). Both of your "equations" mean that $b^3 - a$ is a multiple of n.
$endgroup$
– gammatester
Dec 18 '18 at 11:14
$begingroup$
Huh? Do you know what a congruence is? Your notation is mathematically not appropriate (although it is often used in cryptography and programming). Both of your "equations" mean that $b^3 - a$ is a multiple of n.
$endgroup$
– gammatester
Dec 18 '18 at 11:14
$begingroup$
You are misunderstanding what $b^3equiv apmod n$ means. It doesn't mean you're applying the modulus operation only to $a$. You could say you're applying it to both $a$ and $b^3$, but usually $a<n$ already.
$endgroup$
– Matt Samuel
Dec 18 '18 at 11:19
$begingroup$
You are misunderstanding what $b^3equiv apmod n$ means. It doesn't mean you're applying the modulus operation only to $a$. You could say you're applying it to both $a$ and $b^3$, but usually $a<n$ already.
$endgroup$
– Matt Samuel
Dec 18 '18 at 11:19
|
show 7 more comments
1 Answer
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$begingroup$
Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
$endgroup$
add a comment |
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$begingroup$
Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
$endgroup$
Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
answered Dec 18 '18 at 10:50
lhflhf
167k11172403
167k11172403
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1
$begingroup$
Please use Mathjax
$endgroup$
– Math Girl
Dec 18 '18 at 10:39
$begingroup$
This equivalent to RSA with $e=3.$
$endgroup$
– gammatester
Dec 18 '18 at 10:53
$begingroup$
@gammatester But in RSA we perform b^3 mod n = a and not b^3 = a mod n. How is it still RSA ?
$endgroup$
– caffein
Dec 18 '18 at 11:08
$begingroup$
Huh? Do you know what a congruence is? Your notation is mathematically not appropriate (although it is often used in cryptography and programming). Both of your "equations" mean that $b^3 - a$ is a multiple of n.
$endgroup$
– gammatester
Dec 18 '18 at 11:14
$begingroup$
You are misunderstanding what $b^3equiv apmod n$ means. It doesn't mean you're applying the modulus operation only to $a$. You could say you're applying it to both $a$ and $b^3$, but usually $a<n$ already.
$endgroup$
– Matt Samuel
Dec 18 '18 at 11:19