How many members $a$ in $Bbb{Z}^*$ have a number $b$, $b^3 = a pmod n$?












-1












$begingroup$


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 ?










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$endgroup$








  • 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
















-1












$begingroup$


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 ?










share|cite|improve this question











$endgroup$








  • 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














-1












-1








-1





$begingroup$


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 ?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 15:32







user593746

















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














  • 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










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$begingroup$

Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.






share|cite|improve this answer









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    $begingroup$

    Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.






      share|cite|improve this answer









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        0












        0








        0





        $begingroup$

        Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.






        share|cite|improve this answer









        $endgroup$



        Hint: If $3$ does not divide $p-1$, then $x mapsto x^3$ is a bijection in $mathbb Z_p$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 10:50









        lhflhf

        165k10171396




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