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.






modular-arithmetic matrix-congruences







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  • 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.






modular-arithmetic matrix-congruences







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.






modular-arithmetic matrix-congruences







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




modular-arithmetic matrix-congruences






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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






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oldest

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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.






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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

    oldest

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    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

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