How to solve $x^r [ x^n + 1] = x^{r[n]}$












1












$begingroup$


I send this message to have a piece of advice to solve my problem. Here is the statement:



Assume $k$ belongs to $N$ and $GF(2^k)[x]$ is a ring of polynomials with coefficients in the field $GF(2^k)$. Prove, that if $r$ belongs to $N$, $n$ belongs to $N$, $n geq 2$ and $x^r$ is a polynomial from the ring of polynomials
$GF(2^k )[x]$, then we have:



$$x^r bmod(x^n + 1) = x^{r bmod (n)}.$$



I wanted to prove this by a proposition. Showing that's true for $n = 2$ and then showing that it's true for all $n geq 2.$



I succeeded to prove this in case $r < 2$ because $x^r bmod(x^n+1) = x^r$ and $r[n] = r$. But then I encounter an issue for $n =2$ and for greater I don't know how to prove this.



for $r = 2$ we have $x^r [ x^n + 1] = -1$ and $x^0 = 1.$



Thank you in advance






finite-fields cryptography







share|cite|improve this question











$endgroup$












  • $begingroup$
    See here how to edit your question with mathjax
    $endgroup$
    – Jakobian
    Jan 10 at 21:04










  • $begingroup$
    @kelalaka Your edit distorted one of the formulas. If you don't understand, don't guess. Let it be.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:18












  • $begingroup$
    @MathLover Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @AdrianKeisler Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @JyrkiLahtonen my mistake, sorry.
    $endgroup$
    – kelalaka
    Jan 11 at 6:24
















1












$begingroup$


I send this message to have a piece of advice to solve my problem. Here is the statement:



Assume $k$ belongs to $N$ and $GF(2^k)[x]$ is a ring of polynomials with coefficients in the field $GF(2^k)$. Prove, that if $r$ belongs to $N$, $n$ belongs to $N$, $n geq 2$ and $x^r$ is a polynomial from the ring of polynomials
$GF(2^k )[x]$, then we have:



$$x^r bmod(x^n + 1) = x^{r bmod (n)}.$$



I wanted to prove this by a proposition. Showing that's true for $n = 2$ and then showing that it's true for all $n geq 2.$



I succeeded to prove this in case $r < 2$ because $x^r bmod(x^n+1) = x^r$ and $r[n] = r$. But then I encounter an issue for $n =2$ and for greater I don't know how to prove this.



for $r = 2$ we have $x^r [ x^n + 1] = -1$ and $x^0 = 1.$



Thank you in advance






finite-fields cryptography







share|cite|improve this question











$endgroup$












  • $begingroup$
    See here how to edit your question with mathjax
    $endgroup$
    – Jakobian
    Jan 10 at 21:04










  • $begingroup$
    @kelalaka Your edit distorted one of the formulas. If you don't understand, don't guess. Let it be.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:18












  • $begingroup$
    @MathLover Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @AdrianKeisler Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @JyrkiLahtonen my mistake, sorry.
    $endgroup$
    – kelalaka
    Jan 11 at 6:24














1












1








1





$begingroup$


I send this message to have a piece of advice to solve my problem. Here is the statement:



Assume $k$ belongs to $N$ and $GF(2^k)[x]$ is a ring of polynomials with coefficients in the field $GF(2^k)$. Prove, that if $r$ belongs to $N$, $n$ belongs to $N$, $n geq 2$ and $x^r$ is a polynomial from the ring of polynomials
$GF(2^k )[x]$, then we have:



$$x^r bmod(x^n + 1) = x^{r bmod (n)}.$$



I wanted to prove this by a proposition. Showing that's true for $n = 2$ and then showing that it's true for all $n geq 2.$



I succeeded to prove this in case $r < 2$ because $x^r bmod(x^n+1) = x^r$ and $r[n] = r$. But then I encounter an issue for $n =2$ and for greater I don't know how to prove this.



for $r = 2$ we have $x^r [ x^n + 1] = -1$ and $x^0 = 1.$



Thank you in advance






finite-fields cryptography







share|cite|improve this question











$endgroup$




I send this message to have a piece of advice to solve my problem. Here is the statement:



Assume $k$ belongs to $N$ and $GF(2^k)[x]$ is a ring of polynomials with coefficients in the field $GF(2^k)$. Prove, that if $r$ belongs to $N$, $n$ belongs to $N$, $n geq 2$ and $x^r$ is a polynomial from the ring of polynomials
$GF(2^k )[x]$, then we have:



$$x^r bmod(x^n + 1) = x^{r bmod (n)}.$$



I wanted to prove this by a proposition. Showing that's true for $n = 2$ and then showing that it's true for all $n geq 2.$



I succeeded to prove this in case $r < 2$ because $x^r bmod(x^n+1) = x^r$ and $r[n] = r$. But then I encounter an issue for $n =2$ and for greater I don't know how to prove this.



for $r = 2$ we have $x^r [ x^n + 1] = -1$ and $x^0 = 1.$



Thank you in advance






finite-fields cryptography




finite-fields cryptography






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 6:17









Jyrki Lahtonen

110k13172390




110k13172390










asked Jan 10 at 20:55









Hugo PeyronHugo Peyron

244




244












  • $begingroup$
    See here how to edit your question with mathjax
    $endgroup$
    – Jakobian
    Jan 10 at 21:04










  • $begingroup$
    @kelalaka Your edit distorted one of the formulas. If you don't understand, don't guess. Let it be.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:18












  • $begingroup$
    @MathLover Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @AdrianKeisler Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @JyrkiLahtonen my mistake, sorry.
    $endgroup$
    – kelalaka
    Jan 11 at 6:24


















  • $begingroup$
    See here how to edit your question with mathjax
    $endgroup$
    – Jakobian
    Jan 10 at 21:04










  • $begingroup$
    @kelalaka Your edit distorted one of the formulas. If you don't understand, don't guess. Let it be.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:18












  • $begingroup$
    @MathLover Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @AdrianKeisler Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 6:20










  • $begingroup$
    @JyrkiLahtonen my mistake, sorry.
    $endgroup$
    – kelalaka
    Jan 11 at 6:24
















$begingroup$
See here how to edit your question with mathjax
$endgroup$
– Jakobian
Jan 10 at 21:04




$begingroup$
See here how to edit your question with mathjax
$endgroup$
– Jakobian
Jan 10 at 21:04












$begingroup$
@kelalaka Your edit distorted one of the formulas. If you don't understand, don't guess. Let it be.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 6:18






$begingroup$
@kelalaka Your edit distorted one of the formulas. If you don't understand, don't guess. Let it be.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 6:18














$begingroup$
@MathLover Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 6:20




$begingroup$
@MathLover Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 6:20












$begingroup$
@AdrianKeisler Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 6:20




$begingroup$
@AdrianKeisler Careful when approving edits. One of the remainder operations was in the exponent, and you approved an edit turning the question into non-sense.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 6:20












$begingroup$
@JyrkiLahtonen my mistake, sorry.
$endgroup$
– kelalaka
Jan 11 at 6:24




$begingroup$
@JyrkiLahtonen my mistake, sorry.
$endgroup$
– kelalaka
Jan 11 at 6:24










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hints:




  • Show that for all positive integers $q$ the polynomial $x^{qn}-1$ is divisible by the polynomial $x^n-1$. Observe that we are in characteristic two, so $x^n-1$ and $x^n+1$ mean the same thing.

  • Show that $x^{qn}equiv1pmod{x^n+1}$ for all positive integers $q$.

  • Show that $x^{qn+b}equiv x^bpmod{x^n+1}$ for all positive integers $q,b$.

  • We can write any integer $r$ in the form $r=qn+b$ where $q$ is an integer, and $b$ is the remainder of $r$ when divided modulo $n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Essentially all the calculations are done here and many other places. I was a bit baffled in not finding this as a question themed on remainders. Only then it dawned on me to search using gcd, where the same calculation is needed. Luckily I had the sense to CW this right away.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:46










  • $begingroup$
    See here for a polynomial variant.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:47












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

Hints:




  • Show that for all positive integers $q$ the polynomial $x^{qn}-1$ is divisible by the polynomial $x^n-1$. Observe that we are in characteristic two, so $x^n-1$ and $x^n+1$ mean the same thing.

  • Show that $x^{qn}equiv1pmod{x^n+1}$ for all positive integers $q$.

  • Show that $x^{qn+b}equiv x^bpmod{x^n+1}$ for all positive integers $q,b$.

  • We can write any integer $r$ in the form $r=qn+b$ where $q$ is an integer, and $b$ is the remainder of $r$ when divided modulo $n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Essentially all the calculations are done here and many other places. I was a bit baffled in not finding this as a question themed on remainders. Only then it dawned on me to search using gcd, where the same calculation is needed. Luckily I had the sense to CW this right away.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:46










  • $begingroup$
    See here for a polynomial variant.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:47
















2












$begingroup$

Hints:




  • Show that for all positive integers $q$ the polynomial $x^{qn}-1$ is divisible by the polynomial $x^n-1$. Observe that we are in characteristic two, so $x^n-1$ and $x^n+1$ mean the same thing.

  • Show that $x^{qn}equiv1pmod{x^n+1}$ for all positive integers $q$.

  • Show that $x^{qn+b}equiv x^bpmod{x^n+1}$ for all positive integers $q,b$.

  • We can write any integer $r$ in the form $r=qn+b$ where $q$ is an integer, and $b$ is the remainder of $r$ when divided modulo $n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Essentially all the calculations are done here and many other places. I was a bit baffled in not finding this as a question themed on remainders. Only then it dawned on me to search using gcd, where the same calculation is needed. Luckily I had the sense to CW this right away.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:46










  • $begingroup$
    See here for a polynomial variant.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:47














2












2








2





$begingroup$

Hints:




  • Show that for all positive integers $q$ the polynomial $x^{qn}-1$ is divisible by the polynomial $x^n-1$. Observe that we are in characteristic two, so $x^n-1$ and $x^n+1$ mean the same thing.

  • Show that $x^{qn}equiv1pmod{x^n+1}$ for all positive integers $q$.

  • Show that $x^{qn+b}equiv x^bpmod{x^n+1}$ for all positive integers $q,b$.

  • We can write any integer $r$ in the form $r=qn+b$ where $q$ is an integer, and $b$ is the remainder of $r$ when divided modulo $n$.






share|cite|improve this answer











$endgroup$



Hints:




  • Show that for all positive integers $q$ the polynomial $x^{qn}-1$ is divisible by the polynomial $x^n-1$. Observe that we are in characteristic two, so $x^n-1$ and $x^n+1$ mean the same thing.

  • Show that $x^{qn}equiv1pmod{x^n+1}$ for all positive integers $q$.

  • Show that $x^{qn+b}equiv x^bpmod{x^n+1}$ for all positive integers $q,b$.

  • We can write any integer $r$ in the form $r=qn+b$ where $q$ is an integer, and $b$ is the remainder of $r$ when divided modulo $n$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered Jan 11 at 6:27


























community wiki





Jyrki Lahtonen













  • $begingroup$
    Essentially all the calculations are done here and many other places. I was a bit baffled in not finding this as a question themed on remainders. Only then it dawned on me to search using gcd, where the same calculation is needed. Luckily I had the sense to CW this right away.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:46










  • $begingroup$
    See here for a polynomial variant.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:47


















  • $begingroup$
    Essentially all the calculations are done here and many other places. I was a bit baffled in not finding this as a question themed on remainders. Only then it dawned on me to search using gcd, where the same calculation is needed. Luckily I had the sense to CW this right away.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:46










  • $begingroup$
    See here for a polynomial variant.
    $endgroup$
    – Jyrki Lahtonen
    Jan 11 at 17:47
















$begingroup$
Essentially all the calculations are done here and many other places. I was a bit baffled in not finding this as a question themed on remainders. Only then it dawned on me to search using gcd, where the same calculation is needed. Luckily I had the sense to CW this right away.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:46




$begingroup$
Essentially all the calculations are done here and many other places. I was a bit baffled in not finding this as a question themed on remainders. Only then it dawned on me to search using gcd, where the same calculation is needed. Luckily I had the sense to CW this right away.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:46












$begingroup$
See here for a polynomial variant.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:47




$begingroup$
See here for a polynomial variant.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:47


















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