How to solve system of linear congruences with the same modulo?












0














I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$



I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.










share|cite|improve this question






















  • Search on Hermite(-Smith) normal form for the general ideas.
    – Bill Dubuque
    Nov 22 '18 at 19:20
















0














I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$



I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.










share|cite|improve this question






















  • Search on Hermite(-Smith) normal form for the general ideas.
    – Bill Dubuque
    Nov 22 '18 at 19:20














0












0








0







I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$



I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.










share|cite|improve this question













I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$



I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.







modular-arithmetic matrix-congruences






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 '18 at 12:13









Jacek Kurek

1




1












  • Search on Hermite(-Smith) normal form for the general ideas.
    – Bill Dubuque
    Nov 22 '18 at 19:20


















  • Search on Hermite(-Smith) normal form for the general ideas.
    – Bill Dubuque
    Nov 22 '18 at 19:20
















Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 '18 at 19:20




Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 '18 at 19:20










1 Answer
1






active

oldest

votes


















0














This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.



It seems you have to go the hard way and




  1. factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,

  2. you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.

  3. for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.

  4. reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$

  5. You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.

  6. Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$


Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009056%2fhow-to-solve-system-of-linear-congruences-with-the-same-modulo%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.



    It seems you have to go the hard way and




    1. factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,

    2. you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.

    3. for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.

    4. reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$

    5. You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.

    6. Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$


    Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.






    share|cite|improve this answer


























      0














      This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.



      It seems you have to go the hard way and




      1. factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,

      2. you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.

      3. for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.

      4. reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$

      5. You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.

      6. Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$


      Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.






      share|cite|improve this answer
























        0












        0








        0






        This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.



        It seems you have to go the hard way and




        1. factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,

        2. you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.

        3. for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.

        4. reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$

        5. You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.

        6. Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$


        Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.






        share|cite|improve this answer












        This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.



        It seems you have to go the hard way and




        1. factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,

        2. you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.

        3. for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.

        4. reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$

        5. You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.

        6. Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$


        Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 '18 at 15:27









        Ingix

        3,344145




        3,344145






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009056%2fhow-to-solve-system-of-linear-congruences-with-the-same-modulo%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Can I use Tabulator js library in my java Spring + Thymeleaf project?