Existence of solution to linear congruence [closed]












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Given that $a$ and $b$ are relatively prime positive integers, and that $a$ is relatively prime to all the following primes: $p_1,p_2,...,p_n$ then the following congruence:$${a+bx}equiv 1pmod{ p_1cdot p_2cdot p_3 cdot...cdot p_n}$$
Has solution in x, why can we claim this?










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closed as off-topic by Carl Mummert, amWhy, Gibbs, Dave, Shailesh Dec 7 '18 at 2:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, amWhy, Gibbs, Dave, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.













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    Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product?
    $endgroup$
    – Bill Dubuque
    Dec 2 '18 at 17:38












  • $begingroup$
    @BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately!
    $endgroup$
    – Spasoje Durovic
    Dec 2 '18 at 20:46










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 6 '18 at 17:48
















0












$begingroup$


Given that $a$ and $b$ are relatively prime positive integers, and that $a$ is relatively prime to all the following primes: $p_1,p_2,...,p_n$ then the following congruence:$${a+bx}equiv 1pmod{ p_1cdot p_2cdot p_3 cdot...cdot p_n}$$
Has solution in x, why can we claim this?










share|cite|improve this question











$endgroup$



closed as off-topic by Carl Mummert, amWhy, Gibbs, Dave, Shailesh Dec 7 '18 at 2:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, amWhy, Gibbs, Dave, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product?
    $endgroup$
    – Bill Dubuque
    Dec 2 '18 at 17:38












  • $begingroup$
    @BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately!
    $endgroup$
    – Spasoje Durovic
    Dec 2 '18 at 20:46










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 6 '18 at 17:48














0












0








0





$begingroup$


Given that $a$ and $b$ are relatively prime positive integers, and that $a$ is relatively prime to all the following primes: $p_1,p_2,...,p_n$ then the following congruence:$${a+bx}equiv 1pmod{ p_1cdot p_2cdot p_3 cdot...cdot p_n}$$
Has solution in x, why can we claim this?










share|cite|improve this question











$endgroup$




Given that $a$ and $b$ are relatively prime positive integers, and that $a$ is relatively prime to all the following primes: $p_1,p_2,...,p_n$ then the following congruence:$${a+bx}equiv 1pmod{ p_1cdot p_2cdot p_3 cdot...cdot p_n}$$
Has solution in x, why can we claim this?







modular-arithmetic inverse






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share|cite|improve this question













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share|cite|improve this question








edited Dec 2 '18 at 20:47







Spasoje Durovic

















asked Dec 2 '18 at 12:03









Spasoje DurovicSpasoje Durovic

30610




30610




closed as off-topic by Carl Mummert, amWhy, Gibbs, Dave, Shailesh Dec 7 '18 at 2:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, amWhy, Gibbs, Dave, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Carl Mummert, amWhy, Gibbs, Dave, Shailesh Dec 7 '18 at 2:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, amWhy, Gibbs, Dave, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product?
    $endgroup$
    – Bill Dubuque
    Dec 2 '18 at 17:38












  • $begingroup$
    @BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately!
    $endgroup$
    – Spasoje Durovic
    Dec 2 '18 at 20:46










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 6 '18 at 17:48


















  • $begingroup$
    Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product?
    $endgroup$
    – Bill Dubuque
    Dec 2 '18 at 17:38












  • $begingroup$
    @BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately!
    $endgroup$
    – Spasoje Durovic
    Dec 2 '18 at 20:46










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Carl Mummert
    Dec 6 '18 at 17:48
















$begingroup$
Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product?
$endgroup$
– Bill Dubuque
Dec 2 '18 at 17:38






$begingroup$
Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product?
$endgroup$
– Bill Dubuque
Dec 2 '18 at 17:38














$begingroup$
@BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately!
$endgroup$
– Spasoje Durovic
Dec 2 '18 at 20:46




$begingroup$
@BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately!
$endgroup$
– Spasoje Durovic
Dec 2 '18 at 20:46












$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 6 '18 at 17:48




$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 6 '18 at 17:48










1 Answer
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$begingroup$

It is solvable $iff gcd(n,b)mid 1!-!a,,$ where $n = prod p_i.,$ Indeed



$qquadqquad begin{align}
!bmod n!: exists x!: bxequiv&, 1!-!a\[.3em]
iff exists x,y!: ny+bx =&, 1!-!a\[.3em]
iff gcd(n,,b), mid & 1!-!a {rm by Bezout}end{align}$






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    It is solvable $iff gcd(n,b)mid 1!-!a,,$ where $n = prod p_i.,$ Indeed



    $qquadqquad begin{align}
    !bmod n!: exists x!: bxequiv&, 1!-!a\[.3em]
    iff exists x,y!: ny+bx =&, 1!-!a\[.3em]
    iff gcd(n,,b), mid & 1!-!a {rm by Bezout}end{align}$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It is solvable $iff gcd(n,b)mid 1!-!a,,$ where $n = prod p_i.,$ Indeed



      $qquadqquad begin{align}
      !bmod n!: exists x!: bxequiv&, 1!-!a\[.3em]
      iff exists x,y!: ny+bx =&, 1!-!a\[.3em]
      iff gcd(n,,b), mid & 1!-!a {rm by Bezout}end{align}$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It is solvable $iff gcd(n,b)mid 1!-!a,,$ where $n = prod p_i.,$ Indeed



        $qquadqquad begin{align}
        !bmod n!: exists x!: bxequiv&, 1!-!a\[.3em]
        iff exists x,y!: ny+bx =&, 1!-!a\[.3em]
        iff gcd(n,,b), mid & 1!-!a {rm by Bezout}end{align}$






        share|cite|improve this answer









        $endgroup$



        It is solvable $iff gcd(n,b)mid 1!-!a,,$ where $n = prod p_i.,$ Indeed



        $qquadqquad begin{align}
        !bmod n!: exists x!: bxequiv&, 1!-!a\[.3em]
        iff exists x,y!: ny+bx =&, 1!-!a\[.3em]
        iff gcd(n,,b), mid & 1!-!a {rm by Bezout}end{align}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 22:14









        Bill DubuqueBill Dubuque

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        209k29190633















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