Around the property $(X+a)^n = X^n +a pmod{X^r-1,p}$.












3












$begingroup$


There's an unexplained step in a paper (AKS: Agrawal, Kayal, Saxena: PRIMES is in P) I'm reading that I don't understand. (equation (5))



Let $n$ be an integer divisible by a prime $p$.
Additionally, $r<n$ is coprime with $n$.



$a$ is a fixed arbitrary constant in $mathbb{Z}_p$.



Then the following two equalities
$$(X+a)^n = X^n + a pmod{X^r-1, p}$$
$$(X+a)^p = X^p + a pmod{X^r-1, p}$$



imply the third one:
$$(X+a)^{n/p} = X^{n/p} + a pmod{X^r-1, p}$$



I do not expect that the$pmod{X^r-1}$ will be relevant but I included it just in case it is.



Explanation of notation: $a = b pmod{X^r-1, p}$ means $a = b$ in the ring $mathbb{Z}_p[X]/(X^r-1)$.




I don't know how to approach proving this. I expect this is easy as it was not explained in any way in the paper. Thanks for any help.






polynomials prime-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Appears to be asked already.
    $endgroup$
    – metamorphy
    Dec 27 '18 at 19:22










  • $begingroup$
    Indeed I can't understand their answer so I'm happy you provided yours.
    $endgroup$
    – I want to make games
    Dec 27 '18 at 19:27
















3












$begingroup$


There's an unexplained step in a paper (AKS: Agrawal, Kayal, Saxena: PRIMES is in P) I'm reading that I don't understand. (equation (5))



Let $n$ be an integer divisible by a prime $p$.
Additionally, $r<n$ is coprime with $n$.



$a$ is a fixed arbitrary constant in $mathbb{Z}_p$.



Then the following two equalities
$$(X+a)^n = X^n + a pmod{X^r-1, p}$$
$$(X+a)^p = X^p + a pmod{X^r-1, p}$$



imply the third one:
$$(X+a)^{n/p} = X^{n/p} + a pmod{X^r-1, p}$$



I do not expect that the$pmod{X^r-1}$ will be relevant but I included it just in case it is.



Explanation of notation: $a = b pmod{X^r-1, p}$ means $a = b$ in the ring $mathbb{Z}_p[X]/(X^r-1)$.




I don't know how to approach proving this. I expect this is easy as it was not explained in any way in the paper. Thanks for any help.






polynomials prime-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Appears to be asked already.
    $endgroup$
    – metamorphy
    Dec 27 '18 at 19:22










  • $begingroup$
    Indeed I can't understand their answer so I'm happy you provided yours.
    $endgroup$
    – I want to make games
    Dec 27 '18 at 19:27














3












3








3


3



$begingroup$


There's an unexplained step in a paper (AKS: Agrawal, Kayal, Saxena: PRIMES is in P) I'm reading that I don't understand. (equation (5))



Let $n$ be an integer divisible by a prime $p$.
Additionally, $r<n$ is coprime with $n$.



$a$ is a fixed arbitrary constant in $mathbb{Z}_p$.



Then the following two equalities
$$(X+a)^n = X^n + a pmod{X^r-1, p}$$
$$(X+a)^p = X^p + a pmod{X^r-1, p}$$



imply the third one:
$$(X+a)^{n/p} = X^{n/p} + a pmod{X^r-1, p}$$



I do not expect that the$pmod{X^r-1}$ will be relevant but I included it just in case it is.



Explanation of notation: $a = b pmod{X^r-1, p}$ means $a = b$ in the ring $mathbb{Z}_p[X]/(X^r-1)$.




I don't know how to approach proving this. I expect this is easy as it was not explained in any way in the paper. Thanks for any help.






polynomials prime-numbers







share|cite|improve this question











$endgroup$




There's an unexplained step in a paper (AKS: Agrawal, Kayal, Saxena: PRIMES is in P) I'm reading that I don't understand. (equation (5))



Let $n$ be an integer divisible by a prime $p$.
Additionally, $r<n$ is coprime with $n$.



$a$ is a fixed arbitrary constant in $mathbb{Z}_p$.



Then the following two equalities
$$(X+a)^n = X^n + a pmod{X^r-1, p}$$
$$(X+a)^p = X^p + a pmod{X^r-1, p}$$



imply the third one:
$$(X+a)^{n/p} = X^{n/p} + a pmod{X^r-1, p}$$



I do not expect that the$pmod{X^r-1}$ will be relevant but I included it just in case it is.



Explanation of notation: $a = b pmod{X^r-1, p}$ means $a = b$ in the ring $mathbb{Z}_p[X]/(X^r-1)$.




I don't know how to approach proving this. I expect this is easy as it was not explained in any way in the paper. Thanks for any help.






polynomials prime-numbers




polynomials prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 20:06









Bill Dubuque

212k29195654




212k29195654










asked Dec 27 '18 at 17:26









I want to make gamesI want to make games

866511




866511








  • 1




    $begingroup$
    Appears to be asked already.
    $endgroup$
    – metamorphy
    Dec 27 '18 at 19:22










  • $begingroup$
    Indeed I can't understand their answer so I'm happy you provided yours.
    $endgroup$
    – I want to make games
    Dec 27 '18 at 19:27














  • 1




    $begingroup$
    Appears to be asked already.
    $endgroup$
    – metamorphy
    Dec 27 '18 at 19:22










  • $begingroup$
    Indeed I can't understand their answer so I'm happy you provided yours.
    $endgroup$
    – I want to make games
    Dec 27 '18 at 19:27








1




1




$begingroup$
Appears to be asked already.
$endgroup$
– metamorphy
Dec 27 '18 at 19:22




$begingroup$
Appears to be asked already.
$endgroup$
– metamorphy
Dec 27 '18 at 19:22












$begingroup$
Indeed I can't understand their answer so I'm happy you provided yours.
$endgroup$
– I want to make games
Dec 27 '18 at 19:27




$begingroup$
Indeed I can't understand their answer so I'm happy you provided yours.
$endgroup$
– I want to make games
Dec 27 '18 at 19:27










1 Answer
1






active

oldest

votes


















3












$begingroup$

In the ring $mathbb{Z}_p[X]/(X^r-1)$ we have $A^p=B^pimplies A=B$.



This follows from $(Apm B)^p=A^ppm B^p$ (which holds already in $mathbb{Z}_p[X]$) and $A^p=0implies A=0$ (which can be deduced from $r$ being coprime to $n$ and thus to $p$; for an alternate way to go, if $A=sum_{k=0}^{r-1}a_k X^k$, then $A^p=sum_{k=0}^{r-1}a_k X^{pkbmod r}$, and $kmapsto pkbmod r$ is a bijection).



It remains to put $A=(X+a)^{n/p}$ and $B=X^{n/p}+a$.






share|cite|improve this answer











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    $begingroup$

    In the ring $mathbb{Z}_p[X]/(X^r-1)$ we have $A^p=B^pimplies A=B$.



    This follows from $(Apm B)^p=A^ppm B^p$ (which holds already in $mathbb{Z}_p[X]$) and $A^p=0implies A=0$ (which can be deduced from $r$ being coprime to $n$ and thus to $p$; for an alternate way to go, if $A=sum_{k=0}^{r-1}a_k X^k$, then $A^p=sum_{k=0}^{r-1}a_k X^{pkbmod r}$, and $kmapsto pkbmod r$ is a bijection).



    It remains to put $A=(X+a)^{n/p}$ and $B=X^{n/p}+a$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      In the ring $mathbb{Z}_p[X]/(X^r-1)$ we have $A^p=B^pimplies A=B$.



      This follows from $(Apm B)^p=A^ppm B^p$ (which holds already in $mathbb{Z}_p[X]$) and $A^p=0implies A=0$ (which can be deduced from $r$ being coprime to $n$ and thus to $p$; for an alternate way to go, if $A=sum_{k=0}^{r-1}a_k X^k$, then $A^p=sum_{k=0}^{r-1}a_k X^{pkbmod r}$, and $kmapsto pkbmod r$ is a bijection).



      It remains to put $A=(X+a)^{n/p}$ and $B=X^{n/p}+a$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        In the ring $mathbb{Z}_p[X]/(X^r-1)$ we have $A^p=B^pimplies A=B$.



        This follows from $(Apm B)^p=A^ppm B^p$ (which holds already in $mathbb{Z}_p[X]$) and $A^p=0implies A=0$ (which can be deduced from $r$ being coprime to $n$ and thus to $p$; for an alternate way to go, if $A=sum_{k=0}^{r-1}a_k X^k$, then $A^p=sum_{k=0}^{r-1}a_k X^{pkbmod r}$, and $kmapsto pkbmod r$ is a bijection).



        It remains to put $A=(X+a)^{n/p}$ and $B=X^{n/p}+a$.






        share|cite|improve this answer











        $endgroup$



        In the ring $mathbb{Z}_p[X]/(X^r-1)$ we have $A^p=B^pimplies A=B$.



        This follows from $(Apm B)^p=A^ppm B^p$ (which holds already in $mathbb{Z}_p[X]$) and $A^p=0implies A=0$ (which can be deduced from $r$ being coprime to $n$ and thus to $p$; for an alternate way to go, if $A=sum_{k=0}^{r-1}a_k X^k$, then $A^p=sum_{k=0}^{r-1}a_k X^{pkbmod r}$, and $kmapsto pkbmod r$ is a bijection).



        It remains to put $A=(X+a)^{n/p}$ and $B=X^{n/p}+a$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 27 '18 at 20:24

























        answered Dec 27 '18 at 19:15









        metamorphymetamorphy

        3,7021621




        3,7021621






























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