Around the property $(X+a)^n = X^n +a pmod{X^r-1,p}$.
$begingroup$
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
edited Dec 27 '18 at 20:06
Bill Dubuque
212k29195654
asked Dec 27 '18 at 17:26
I want to make gamesI want to make games
866511
$endgroup$
add a comment |
$begingroup$
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
edited Dec 27 '18 at 20:06
Bill Dubuque
212k29195654
asked Dec 27 '18 at 17:26
I want to make gamesI want to make games
866511
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 27 '18 at 20:06
Bill Dubuque
212k29195654
asked Dec 27 '18 at 17:26
I want to make gamesI want to make games
866511
$endgroup$
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
edited Dec 27 '18 at 20:06
Bill Dubuque
212k29195654
asked Dec 27 '18 at 17:26
I want to make gamesI want to make games
866511
edited Dec 27 '18 at 20:06
Bill Dubuque
212k29195654
asked Dec 27 '18 at 17:26
I want to make gamesI want to make games
866511
edited Dec 27 '18 at 20:06
Bill Dubuque
212k29195654
edited Dec 27 '18 at 20:06
Bill Dubuque
212k29195654
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
asked Dec 27 '18 at 17:26
I want to make gamesI want to make games
866511
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $B=X^{n/p}+a$.
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
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1 Answer
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1 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 $B=X^{n/p}+a$.
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
$begingroup$
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 $B=X^{n/p}+a$.
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
$begingroup$
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 $B=X^{n/p}+a$.
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
$endgroup$
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 $B=X^{n/p}+a$.
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
edited Dec 27 '18 at 20:24
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
answered Dec 27 '18 at 19:15
metamorphymetamorphy
3,7021621
3,7021621
add a comment |
add a comment |
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$begingroup$
Appears to be asked already.
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– 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