For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to...












0












$begingroup$


For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)equiv 0 mod(n)$ where $n$ is composite?



Is there a general way to determine the number of incongruent solutions modulo $n$?



My first idea is that we can of course break $n$ into its prime power factorization and look at $f(x)equiv 0 mod(p_{i}^{e_{i}})$ where $(p_{i}^{e_{i}})$ appears as a prime power factor in $n$.



Here's where I start to become confused, if $f(x)=x$ then the Chinese remainder theorem tells us that the solution is unique modulo $n$, but if $f(x)$ is non-constant and non-linear then we need to use the lifting method to solve $f(x)equiv 0$ for each $mod(p_{i}^{e_{i}})$ - but so far the method tells us nothing about the number of solutions.



I presume I am not incorrect in saying that the number of incongruent solutions to $f(x)equiv mod(p_{i}^{e_{i}})$ is at most $min(deg(f), p_{i}^{e_{i}})$, but is there a general way to determine precisely how many solutions are there?






abstract-algebra number-theory elementary-number-theory congruences







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $f(n)equiv 0pmod4$ has two solutions, and $f(n)equiv 0pmod{25}$ has six solutions, then $f(n)equiv 0pmod{100}$ has twelve solutions. Is that what you're asking?
    $endgroup$
    – Arthur
    May 5 '16 at 7:37










  • $begingroup$
    Essentially yes, but why's that?
    $endgroup$
    – user162089
    May 5 '16 at 7:42






  • 1




    $begingroup$
    @user162089 because of the Chinese Remainder Theorem
    $endgroup$
    – Hagen von Eitzen
    May 5 '16 at 7:48










  • $begingroup$
    Because every pair $(a,b)$ with $a$ being modulo four and $b$ being modulo $25$ gives a unique residue class $c$ modulo $100$. Also, this pairing respects polynomials, so $(f(a),f(b))$ corresponds to $f(c)$. There are two $a$ for which the polynomial is zero, and six $b$, which makes twelve pairs of solutions.
    $endgroup$
    – Arthur
    May 5 '16 at 7:55


















0












$begingroup$


For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)equiv 0 mod(n)$ where $n$ is composite?



Is there a general way to determine the number of incongruent solutions modulo $n$?



My first idea is that we can of course break $n$ into its prime power factorization and look at $f(x)equiv 0 mod(p_{i}^{e_{i}})$ where $(p_{i}^{e_{i}})$ appears as a prime power factor in $n$.



Here's where I start to become confused, if $f(x)=x$ then the Chinese remainder theorem tells us that the solution is unique modulo $n$, but if $f(x)$ is non-constant and non-linear then we need to use the lifting method to solve $f(x)equiv 0$ for each $mod(p_{i}^{e_{i}})$ - but so far the method tells us nothing about the number of solutions.



I presume I am not incorrect in saying that the number of incongruent solutions to $f(x)equiv mod(p_{i}^{e_{i}})$ is at most $min(deg(f), p_{i}^{e_{i}})$, but is there a general way to determine precisely how many solutions are there?






abstract-algebra number-theory elementary-number-theory congruences







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $f(n)equiv 0pmod4$ has two solutions, and $f(n)equiv 0pmod{25}$ has six solutions, then $f(n)equiv 0pmod{100}$ has twelve solutions. Is that what you're asking?
    $endgroup$
    – Arthur
    May 5 '16 at 7:37










  • $begingroup$
    Essentially yes, but why's that?
    $endgroup$
    – user162089
    May 5 '16 at 7:42






  • 1




    $begingroup$
    @user162089 because of the Chinese Remainder Theorem
    $endgroup$
    – Hagen von Eitzen
    May 5 '16 at 7:48










  • $begingroup$
    Because every pair $(a,b)$ with $a$ being modulo four and $b$ being modulo $25$ gives a unique residue class $c$ modulo $100$. Also, this pairing respects polynomials, so $(f(a),f(b))$ corresponds to $f(c)$. There are two $a$ for which the polynomial is zero, and six $b$, which makes twelve pairs of solutions.
    $endgroup$
    – Arthur
    May 5 '16 at 7:55
















0












0








0





$begingroup$


For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)equiv 0 mod(n)$ where $n$ is composite?



Is there a general way to determine the number of incongruent solutions modulo $n$?



My first idea is that we can of course break $n$ into its prime power factorization and look at $f(x)equiv 0 mod(p_{i}^{e_{i}})$ where $(p_{i}^{e_{i}})$ appears as a prime power factor in $n$.



Here's where I start to become confused, if $f(x)=x$ then the Chinese remainder theorem tells us that the solution is unique modulo $n$, but if $f(x)$ is non-constant and non-linear then we need to use the lifting method to solve $f(x)equiv 0$ for each $mod(p_{i}^{e_{i}})$ - but so far the method tells us nothing about the number of solutions.



I presume I am not incorrect in saying that the number of incongruent solutions to $f(x)equiv mod(p_{i}^{e_{i}})$ is at most $min(deg(f), p_{i}^{e_{i}})$, but is there a general way to determine precisely how many solutions are there?






abstract-algebra number-theory elementary-number-theory congruences







share|cite|improve this question









$endgroup$




For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)equiv 0 mod(n)$ where $n$ is composite?



Is there a general way to determine the number of incongruent solutions modulo $n$?



My first idea is that we can of course break $n$ into its prime power factorization and look at $f(x)equiv 0 mod(p_{i}^{e_{i}})$ where $(p_{i}^{e_{i}})$ appears as a prime power factor in $n$.



Here's where I start to become confused, if $f(x)=x$ then the Chinese remainder theorem tells us that the solution is unique modulo $n$, but if $f(x)$ is non-constant and non-linear then we need to use the lifting method to solve $f(x)equiv 0$ for each $mod(p_{i}^{e_{i}})$ - but so far the method tells us nothing about the number of solutions.



I presume I am not incorrect in saying that the number of incongruent solutions to $f(x)equiv mod(p_{i}^{e_{i}})$ is at most $min(deg(f), p_{i}^{e_{i}})$, but is there a general way to determine precisely how many solutions are there?






abstract-algebra number-theory elementary-number-theory congruences




abstract-algebra number-theory elementary-number-theory congruences






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 5 '16 at 7:29









user162089user162089

211310




211310












  • $begingroup$
    If $f(n)equiv 0pmod4$ has two solutions, and $f(n)equiv 0pmod{25}$ has six solutions, then $f(n)equiv 0pmod{100}$ has twelve solutions. Is that what you're asking?
    $endgroup$
    – Arthur
    May 5 '16 at 7:37










  • $begingroup$
    Essentially yes, but why's that?
    $endgroup$
    – user162089
    May 5 '16 at 7:42






  • 1




    $begingroup$
    @user162089 because of the Chinese Remainder Theorem
    $endgroup$
    – Hagen von Eitzen
    May 5 '16 at 7:48










  • $begingroup$
    Because every pair $(a,b)$ with $a$ being modulo four and $b$ being modulo $25$ gives a unique residue class $c$ modulo $100$. Also, this pairing respects polynomials, so $(f(a),f(b))$ corresponds to $f(c)$. There are two $a$ for which the polynomial is zero, and six $b$, which makes twelve pairs of solutions.
    $endgroup$
    – Arthur
    May 5 '16 at 7:55




















  • $begingroup$
    If $f(n)equiv 0pmod4$ has two solutions, and $f(n)equiv 0pmod{25}$ has six solutions, then $f(n)equiv 0pmod{100}$ has twelve solutions. Is that what you're asking?
    $endgroup$
    – Arthur
    May 5 '16 at 7:37










  • $begingroup$
    Essentially yes, but why's that?
    $endgroup$
    – user162089
    May 5 '16 at 7:42






  • 1




    $begingroup$
    @user162089 because of the Chinese Remainder Theorem
    $endgroup$
    – Hagen von Eitzen
    May 5 '16 at 7:48










  • $begingroup$
    Because every pair $(a,b)$ with $a$ being modulo four and $b$ being modulo $25$ gives a unique residue class $c$ modulo $100$. Also, this pairing respects polynomials, so $(f(a),f(b))$ corresponds to $f(c)$. There are two $a$ for which the polynomial is zero, and six $b$, which makes twelve pairs of solutions.
    $endgroup$
    – Arthur
    May 5 '16 at 7:55


















$begingroup$
If $f(n)equiv 0pmod4$ has two solutions, and $f(n)equiv 0pmod{25}$ has six solutions, then $f(n)equiv 0pmod{100}$ has twelve solutions. Is that what you're asking?
$endgroup$
– Arthur
May 5 '16 at 7:37




$begingroup$
If $f(n)equiv 0pmod4$ has two solutions, and $f(n)equiv 0pmod{25}$ has six solutions, then $f(n)equiv 0pmod{100}$ has twelve solutions. Is that what you're asking?
$endgroup$
– Arthur
May 5 '16 at 7:37












$begingroup$
Essentially yes, but why's that?
$endgroup$
– user162089
May 5 '16 at 7:42




$begingroup$
Essentially yes, but why's that?
$endgroup$
– user162089
May 5 '16 at 7:42




1




1




$begingroup$
@user162089 because of the Chinese Remainder Theorem
$endgroup$
– Hagen von Eitzen
May 5 '16 at 7:48




$begingroup$
@user162089 because of the Chinese Remainder Theorem
$endgroup$
– Hagen von Eitzen
May 5 '16 at 7:48












$begingroup$
Because every pair $(a,b)$ with $a$ being modulo four and $b$ being modulo $25$ gives a unique residue class $c$ modulo $100$. Also, this pairing respects polynomials, so $(f(a),f(b))$ corresponds to $f(c)$. There are two $a$ for which the polynomial is zero, and six $b$, which makes twelve pairs of solutions.
$endgroup$
– Arthur
May 5 '16 at 7:55






$begingroup$
Because every pair $(a,b)$ with $a$ being modulo four and $b$ being modulo $25$ gives a unique residue class $c$ modulo $100$. Also, this pairing respects polynomials, so $(f(a),f(b))$ corresponds to $f(c)$. There are two $a$ for which the polynomial is zero, and six $b$, which makes twelve pairs of solutions.
$endgroup$
– Arthur
May 5 '16 at 7:55












1 Answer
1






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$begingroup$

First factor $m$ to powers of primes. $$N(m) = PI N(p^a)$$ $N$ is number of roots. This is by CRT.



Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $<= p$. If $P'(x)$ is a factor of $(x^p - x) mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.



From $p$ to $p^a$, it's Hensel's Lemma.






share|cite|improve this answer











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    $begingroup$

    First factor $m$ to powers of primes. $$N(m) = PI N(p^a)$$ $N$ is number of roots. This is by CRT.



    Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $<= p$. If $P'(x)$ is a factor of $(x^p - x) mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.



    From $p$ to $p^a$, it's Hensel's Lemma.






    share|cite|improve this answer











    $endgroup$


















      -1












      $begingroup$

      First factor $m$ to powers of primes. $$N(m) = PI N(p^a)$$ $N$ is number of roots. This is by CRT.



      Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $<= p$. If $P'(x)$ is a factor of $(x^p - x) mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.



      From $p$ to $p^a$, it's Hensel's Lemma.






      share|cite|improve this answer











      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        First factor $m$ to powers of primes. $$N(m) = PI N(p^a)$$ $N$ is number of roots. This is by CRT.



        Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $<= p$. If $P'(x)$ is a factor of $(x^p - x) mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.



        From $p$ to $p^a$, it's Hensel's Lemma.






        share|cite|improve this answer











        $endgroup$



        First factor $m$ to powers of primes. $$N(m) = PI N(p^a)$$ $N$ is number of roots. This is by CRT.



        Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $<= p$. If $P'(x)$ is a factor of $(x^p - x) mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.



        From $p$ to $p^a$, it's Hensel's Lemma.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 14:44









        dantopa

        6,44932142




        6,44932142










        answered Dec 6 '18 at 9:26









        james0910james0910

        111




        111






























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