How to solve $x^r [ x^n + 1] = x^{r[n]}$
$begingroup$
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
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
asked Jan 10 at 20:55
Hugo PeyronHugo Peyron
244
$endgroup$
add a comment |
$begingroup$
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
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
asked Jan 10 at 20:55
Hugo PeyronHugo Peyron
244
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
asked Jan 10 at 20:55
Hugo PeyronHugo Peyron
244
$endgroup$
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
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
asked Jan 10 at 20:55
Hugo PeyronHugo Peyron
244
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
asked Jan 10 at 20:55
Hugo PeyronHugo Peyron
244
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
edited Jan 11 at 6:17
Jyrki Lahtonen
110k13172390
110k13172390
asked Jan 10 at 20:55
Hugo PeyronHugo Peyron
244
asked Jan 10 at 20:55
Hugo PeyronHugo Peyron
244
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 $n$.
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $n$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
$endgroup$
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 $n$.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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