Adapting the Chinese Remainder Theorem (CRT) for integers to polynomials












2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
    Apr 21 at 19:23










  • $begingroup$
    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
    Apr 21 at 19:56
















2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
    Apr 21 at 19:23










  • $begingroup$
    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
    Apr 21 at 19:56














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 21 at 19:16









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
    Apr 21 at 19:23










  • $begingroup$
    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
    Apr 21 at 19:56














  • 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
    Apr 21 at 19:23










  • $begingroup$
    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
    Apr 21 at 19:56








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
Apr 21 at 19:23




$begingroup$
"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
Apr 21 at 19:23












$begingroup$
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
Apr 21 at 19:56




$begingroup$
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
Apr 21 at 19:56










2 Answers
2






active

oldest

votes


















3












$begingroup$

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
$$y_1=1,$$
$$y_2=(1-x^2).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there an algorithmic way of solving these, rather than relying on intuition?
    $endgroup$
    – Junglemath
    Apr 21 at 19:59










  • $begingroup$
    @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
    Apr 21 at 20:09








  • 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
    Apr 21 at 20:10








  • 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
    Apr 21 at 20:22








  • 2




    $begingroup$
    @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
    $endgroup$
    – Bill Dubuque
    Apr 21 at 21:28





















2












$begingroup$

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).






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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
    $$y_1=1,$$
    $$y_2=(1-x^2).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @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
      Apr 21 at 20:09








    • 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
      Apr 21 at 20:10








    • 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
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28


















    3












    $begingroup$

    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
    $$y_1=1,$$
    $$y_2=(1-x^2).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @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
      Apr 21 at 20:09








    • 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
      Apr 21 at 20:10








    • 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
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28
















    3












    3








    3





    $begingroup$

    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
    $$y_1=1,$$
    $$y_2=(1-x^2).$$






    share|cite|improve this answer











    $endgroup$



    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
    $$y_1=1,$$
    $$y_2=(1-x^2).$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 21 at 19:53

























    answered Apr 21 at 19:26









    MelodyMelody

    1,46112




    1,46112












    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @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
      Apr 21 at 20:09








    • 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
      Apr 21 at 20:10








    • 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
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28




















    • $begingroup$
      Is there an algorithmic way of solving these, rather than relying on intuition?
      $endgroup$
      – Junglemath
      Apr 21 at 19:59










    • $begingroup$
      @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
      Apr 21 at 20:09








    • 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
      Apr 21 at 20:10








    • 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
      Apr 21 at 20:22








    • 2




      $begingroup$
      @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
      $endgroup$
      – Bill Dubuque
      Apr 21 at 21:28


















    $begingroup$
    Is there an algorithmic way of solving these, rather than relying on intuition?
    $endgroup$
    – Junglemath
    Apr 21 at 19:59




    $begingroup$
    Is there an algorithmic way of solving these, rather than relying on intuition?
    $endgroup$
    – Junglemath
    Apr 21 at 19:59












    $begingroup$
    @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
    Apr 21 at 20:09






    $begingroup$
    @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
    Apr 21 at 20:09






    2




    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
    Apr 21 at 20:10






    $begingroup$
    @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
    Apr 21 at 20:10






    2




    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
    Apr 21 at 20:22






    $begingroup$
    @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
    Apr 21 at 20:22






    2




    2




    $begingroup$
    @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
    $endgroup$
    – Bill Dubuque
    Apr 21 at 21:28






    $begingroup$
    @Junglemath I describe here at length this method of scaling the Bezout equation into a CRT solution.
    $endgroup$
    – Bill Dubuque
    Apr 21 at 21:28













    2












    $begingroup$

    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).






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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).






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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).






        share|cite|improve this answer











        $endgroup$



        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).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 21 at 21:41

























        answered Apr 21 at 21:25









        Bill DubuqueBill Dubuque

        214k29198660




        214k29198660






























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