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Adapting the Chinese Remainder Theorem (CRT) for integers to polynomials
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)mod Distributive Law, factoring $!!bmod!!:$ $ abbmod ac = a(bbmod c)$Remainder of polynomial by product of 2 polynomials$f$ dividing by $x + 1$ have remainder 4, when dividing with $x^2 + 1$ have remainder 2x+3. Find remainder dividing polynomial with($x+1$)($x^2+1$)chinese remainder theorem proofChinese Remainder Theorem InterpretationChinese Remainder Theorem clarificationI can't use Chinese Remainder Theorem.Chinese Remainder Theorem for $xequiv 0 pmod{y}$Comparing two statements of Chinese Remainder Theorem (Sun-Ze Theorem)Chinese remainder theorem methodChinese Remainder Theorem problem 7Chinese Remainder Theorem with 0 mod nSolve a system of congruences using the Chinese Remainder Theorem
$begingroup$
I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.
Here's an example with integers:
$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$
Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:
$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)
$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence
$bullet , M_i$ denotes $dfrac{M}{m_i}$
$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).
Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).
Now I want to apply the same technique to the following:
$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$
where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:
$M = (x^4)(x^2 + 1)$
$M_1 = x^4$
$M_2 = x^2 + 1$
Here's where I run into a problem. I need to find $y_1, y_2$ such that
$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$
But how does one find $y_1, y_2$?
abstract-algebra ring-theory chinese-remainder-theorem
asked 8 hours ago
JunglemathJunglemath
6017
$endgroup$
add a comment |
$begingroup$
I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.
Here's an example with integers:
$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$
Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:
$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)
$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence
$bullet , M_i$ denotes $dfrac{M}{m_i}$
$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).
Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).
Now I want to apply the same technique to the following:
$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$
where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:
$M = (x^4)(x^2 + 1)$
$M_1 = x^4$
$M_2 = x^2 + 1$
Here's where I run into a problem. I need to find $y_1, y_2$ such that
$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$
But how does one find $y_1, y_2$?
abstract-algebra ring-theory chinese-remainder-theorem
asked 8 hours ago
JunglemathJunglemath
6017
$endgroup$
1
$begingroup$
"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
$endgroup$
– kccu
8 hours ago
$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
$endgroup$
– Junglemath
8 hours ago
add a comment |
$begingroup$
I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.
Here's an example with integers:
$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$
Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:
$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)
$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence
$bullet , M_i$ denotes $dfrac{M}{m_i}$
$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).
Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).
Now I want to apply the same technique to the following:
$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$
where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:
$M = (x^4)(x^2 + 1)$
$M_1 = x^4$
$M_2 = x^2 + 1$
Here's where I run into a problem. I need to find $y_1, y_2$ such that
$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$
But how does one find $y_1, y_2$?
abstract-algebra ring-theory chinese-remainder-theorem
asked 8 hours ago
JunglemathJunglemath
6017
$endgroup$
I did a few examples using the CRT to solve congruences where everything was in terms of integers. I'm trying to use the same technique for polynomials over $mathbb{Q}$, but I'm getting stuck.
Here's an example with integers:
$begin{cases}x equiv 1 , (mathrm{mod} , 5) \
x equiv 2 , (mathrm{mod} , 7) \
x equiv 3 , (mathrm{mod} , 9) \
x equiv 4 , (mathrm{mod} , 11).
end{cases}$
Since all the moduli are pairwise relatively prime, we can use the CRT. Here's some notation I'm using:
$bullet , M$ denotes the product of the moduli (in this case, $M = 5 cdot7 cdot 9 cdot 11$)
$bullet , m_i $ denotes the modulus in the $i^{mathrm{th}}$ congruence
$bullet , M_i$ denotes $dfrac{M}{m_i}$
$bullet , y_i$ denotes the inverse of $M_i$ (mod $m_i$), i.e. $y_i$ satisfies $y_i M_i equiv 1$ (mod $m_i$).
Then $x = displaystyle sum_{i = 1}^n a_iM_iy_i$, and this solution is unique (mod $M$).
Now I want to apply the same technique to the following:
$begin{cases}
f(x) equiv 1 , (mathrm{ mod } , x^2 + 1) \
f(x) equiv x , (mathrm{mod} , x^4),
end{cases}$
where $f(x) in mathbb{Q}(x)$. Having checked that the moduli are relatively prime, we should be able to use the CRT. Using the notation above, I have the following:
$M = (x^4)(x^2 + 1)$
$M_1 = x^4$
$M_2 = x^2 + 1$
Here's where I run into a problem. I need to find $y_1, y_2$ such that
$begin{cases}
y_1 (x^4) equiv 1 , (mathrm{mod} , x^2 + 1) \
y_2 (x^2+1) equiv 1 , (mathrm{mod} , x^4).
end{cases}$
But how does one find $y_1, y_2$?
abstract-algebra ring-theory chinese-remainder-theorem
abstract-algebra ring-theory chinese-remainder-theorem
asked 8 hours ago
JunglemathJunglemath
6017
asked 8 hours ago
JunglemathJunglemath
6017
asked 8 hours ago
JunglemathJunglemath
6017
asked 8 hours ago
JunglemathJunglemath
6017
asked 8 hours ago
JunglemathJunglemath
6017
6017
1
$begingroup$
"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
$endgroup$
– kccu
8 hours ago
$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
$endgroup$
– Junglemath
8 hours ago
add a comment |
1
$begingroup$
"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
$endgroup$
– kccu
8 hours ago
$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
$endgroup$
– Junglemath
8 hours ago
1
1
$begingroup$
"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
$endgroup$
– kccu
8 hours ago
$begingroup$
"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
$endgroup$
– kccu
8 hours ago
$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
$endgroup$
– Junglemath
8 hours ago
$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
$endgroup$
– Junglemath
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To find $y_1$ and $y_2$ consider solving the problem
$$y_1x^4+y_2(x^2+1)=1.$$
This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
$$(x^2-1)(x^2+1)=x^4-1,$$
hence
$$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
This tells us that we can choose
$$y_1=1,$$
$$y_2=(1-x^2).$$
answered 8 hours ago
MelodyMelody
1,42212
$endgroup$
$begingroup$
Is there an algorithmic way of solving these, rather than relying on intuition?
$endgroup$
– Junglemath
8 hours ago
$begingroup$
@Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
$endgroup$
– Paolo
8 hours ago
2
$begingroup$
@Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
$endgroup$
– Melody
8 hours ago
2
$begingroup$
@Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
$endgroup$
– Melody
7 hours ago
2
$begingroup$
@Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
$endgroup$
– Bill Dubuque
6 hours ago
|
show 1 more comment
$begingroup$
Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:
$ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$
Remark $ $ Here are further examples done using MDL (an operational form of CRT).
You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
To find $y_1$ and $y_2$ consider solving the problem
$$y_1x^4+y_2(x^2+1)=1.$$
This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
$$(x^2-1)(x^2+1)=x^4-1,$$
hence
$$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
This tells us that we can choose
$$y_1=1,$$
$$y_2=(1-x^2).$$
answered 8 hours ago
MelodyMelody
1,42212
$endgroup$
$begingroup$
Is there an algorithmic way of solving these, rather than relying on intuition?
$endgroup$
– Junglemath
8 hours ago
$begingroup$
@Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
$endgroup$
– Paolo
8 hours ago
2
$begingroup$
@Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
$endgroup$
– Melody
8 hours ago
2
$begingroup$
@Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
$endgroup$
– Melody
7 hours ago
2
$begingroup$
@Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
$endgroup$
– Bill Dubuque
6 hours ago
|
show 1 more comment
$begingroup$
To find $y_1$ and $y_2$ consider solving the problem
$$y_1x^4+y_2(x^2+1)=1.$$
This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
$$(x^2-1)(x^2+1)=x^4-1,$$
hence
$$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
This tells us that we can choose
$$y_1=1,$$
$$y_2=(1-x^2).$$
answered 8 hours ago
MelodyMelody
1,42212
$endgroup$
$begingroup$
Is there an algorithmic way of solving these, rather than relying on intuition?
$endgroup$
– Junglemath
8 hours ago
$begingroup$
@Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
$endgroup$
– Paolo
8 hours ago
2
$begingroup$
@Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
$endgroup$
– Melody
8 hours ago
2
$begingroup$
@Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
$endgroup$
– Melody
7 hours ago
2
$begingroup$
@Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
$endgroup$
– Bill Dubuque
6 hours ago
|
show 1 more comment
$begingroup$
To find $y_1$ and $y_2$ consider solving the problem
$$y_1x^4+y_2(x^2+1)=1.$$
This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
$$(x^2-1)(x^2+1)=x^4-1,$$
hence
$$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
This tells us that we can choose
$$y_1=1,$$
$$y_2=(1-x^2).$$
answered 8 hours ago
MelodyMelody
1,42212
$endgroup$
To find $y_1$ and $y_2$ consider solving the problem
$$y_1x^4+y_2(x^2+1)=1.$$
This is not always easy to solve, but in this case a solution comes to mind. Note that by difference of squares
$$(x^2-1)(x^2+1)=x^4-1,$$
hence
$$x^4+[(-1)(x^2-1)](x^2+1)=1.$$
This tells us that we can choose
$$y_1=1,$$
$$y_2=(1-x^2).$$
answered 8 hours ago
MelodyMelody
1,42212
edited 8 hours ago
answered 8 hours ago
MelodyMelody
1,42212
answered 8 hours ago
MelodyMelody
1,42212
answered 8 hours ago
MelodyMelody
1,42212
1,42212
$begingroup$
Is there an algorithmic way of solving these, rather than relying on intuition?
$endgroup$
– Junglemath
8 hours ago
$begingroup$
@Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
$endgroup$
– Paolo
8 hours ago
2
$begingroup$
@Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
$endgroup$
– Melody
8 hours ago
2
$begingroup$
@Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
$endgroup$
– Melody
7 hours ago
2
$begingroup$
@Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
$endgroup$
– Bill Dubuque
6 hours ago
|
show 1 more comment
$begingroup$
Is there an algorithmic way of solving these, rather than relying on intuition?
$endgroup$
– Junglemath
8 hours ago
$begingroup$
@Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
$endgroup$
– Paolo
8 hours ago
2
$begingroup$
@Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
$endgroup$
– Melody
8 hours ago
2
$begingroup$
@Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
$endgroup$
– Melody
7 hours ago
2
$begingroup$
@Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
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– Bill Dubuque
6 hours ago
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Is there an algorithmic way of solving these, rather than relying on intuition?
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– Junglemath
8 hours ago
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Is there an algorithmic way of solving these, rather than relying on intuition?
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– Junglemath
8 hours ago
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@Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
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– Paolo
8 hours ago
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@Junglemath By Euclidean algorithm you can find two polynomials $p(x), q(x) in mathbb{Q}[x]$ such that $p(x) x^4 + q(x) (x^2 + 1) = 1$.
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– Paolo
8 hours ago
2
2
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@Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
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– Melody
8 hours ago
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@Junglemath The polynomials over a field form a Euclidean Domain, so yes, there is. If the gcd of $f_1,f_2inmathbb{Q}[x]$ is a unit, then you can perform the Euclidean Algorithm to find their gcd. Now, you can reverse the algorithm to write them as a linear combination of their gcd the same way you would for integers. I said not easy, because the process can be time consuming and very tedious.
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– Melody
8 hours ago
2
2
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@Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
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– Melody
7 hours ago
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@Junglemath I didn't really convert it. I knew in advanced the single equation had a solution. This is because $x^4$ and $x^2+1$ have no common roots, hence no common irreducible factors. This means they are relatively prime, in which case we can write $1$ as a linear combination. Knowing that, I knew solving the single equation would give rise to a solution to the congruence equations. This is completely analogous to how you can solve everything over the integers.
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– Melody
7 hours ago
2
2
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@Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
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– Bill Dubuque
6 hours ago
$begingroup$
@Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
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– Bill Dubuque
6 hours ago
|
show 1 more comment
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Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:
$ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$
Remark $ $ Here are further examples done using MDL (an operational form of CRT).
You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
$endgroup$
add a comment |
$begingroup$
Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:
$ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$
Remark $ $ Here are further examples done using MDL (an operational form of CRT).
You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
$endgroup$
add a comment |
$begingroup$
Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:
$ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$
Remark $ $ Here are further examples done using MDL (an operational form of CRT).
You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
$endgroup$
Bu applying $ abbmod ac, =, a(bbmod c) $ [Mod Distributive Law] $ $ it is a bit simpler:
$ f-x,bmod, {x^{large 4}(x^{large 2}!+!1)}, =, x^{large 4}underbrace{{left[dfrac{color{#c00}f-x}{color{#0a0}{x^{large 4}}}bmod {x^{large 2}!+!1}right]}}_{large color{#0a0}{x^{Large 4}} equiv 1 {rm by} x^{Large 2} equiv -1 } =, x^{large 4}[1-x], $ by $,color{#c00}fequiv 1pmod{!x^{large 2}!+!1}$
Remark $ $ Here are further examples done using MDL (an operational form of CRT).
You can find further details here on transforming the Bezout equation into a CRT solution (the method sketched in Melody's answer).
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
edited 6 hours ago
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
answered 6 hours ago
Bill DubuqueBill Dubuque
214k29198660
214k29198660
add a comment |
add a comment |
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"Having checked that the moduli are relatively prime..." but that means precisely that there exist $p_1(x)$ and $p_2(x)$ such that $p_1(x)x^4 + p_2(x)(x^2+1)=1$.
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– kccu
8 hours ago
$begingroup$
Are you saying that I can find $p_1(x)$ and $p_2(x)$ in general by using the extended Euclidean algorithm, and that $p_1(x)$ and $p_2(x)$ are precisely my $y_1$ and $y_2$?
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– Junglemath
8 hours ago