How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$?












3















Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?




Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?



(Please let me know if I should add some details to this problem)






calculus combinatorics number-theory elementary-number-theory discrete-mathematics







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  • 2




    This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
    – Dr. Mathva
    Nov 26 at 18:38










  • @Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
    – color
    Nov 26 at 18:43






  • 1




    Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
    – reuns
    Nov 27 at 0:56
















3















Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?




Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?



(Please let me know if I should add some details to this problem)






calculus combinatorics number-theory elementary-number-theory discrete-mathematics







share|cite|improve this question




















  • 2




    This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
    – Dr. Mathva
    Nov 26 at 18:38










  • @Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
    – color
    Nov 26 at 18:43






  • 1




    Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
    – reuns
    Nov 27 at 0:56














3












3








3


1






Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?




Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?



(Please let me know if I should add some details to this problem)






calculus combinatorics number-theory elementary-number-theory discrete-mathematics







share|cite|improve this question
















Let $p$ be an odd prime number and $a,b,c$ be integers coprime with $p$. How many different pairs of integers $(x,y)$ modulo $p$ such that $ax^2+by^2+c equiv 0pmod{p}$ ?




Until now I haven't had any specific way to approach this problem. How can I find the number of solutions of the equation $ax^2+by^2+c equiv 0pmod{p}$ ? Or can it only be counted with some given conditions of $a,b,c$ ?



(Please let me know if I should add some details to this problem)






calculus combinatorics number-theory elementary-number-theory discrete-mathematics




calculus combinatorics number-theory elementary-number-theory discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 0:44

























asked Nov 26 at 18:32









color

186




186








  • 2




    This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
    – Dr. Mathva
    Nov 26 at 18:38










  • @Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
    – color
    Nov 26 at 18:43






  • 1




    Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
    – reuns
    Nov 27 at 0:56














  • 2




    This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
    – Dr. Mathva
    Nov 26 at 18:38










  • @Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
    – color
    Nov 26 at 18:43






  • 1




    Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
    – reuns
    Nov 27 at 0:56








2




2




This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38




This question seems to be related with this one here: math.stackexchange.com/questions/1190423/…
– Dr. Mathva
Nov 26 at 18:38












@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43




@Dr.Mathva Yes, thank you. However I am looking for the number of solutions, not just if a solution exists.
– color
Nov 26 at 18:43




1




1




Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56




Let $chi(n) = n^{(p-1)/2}= pm 1, chi(0) = 0$. Then there are $sum_{x bmod p} chi(b^{-1})chi(-ax^2-c)+chi(b^{-1})^2chi(-ax^2-c)^2$ solutions with $y ne 0$. Can you go further assuming $-ac^{-1} = d^2$ ?
– reuns
Nov 27 at 0:56










1 Answer
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Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):



$d^{p-1}-1 ≡0mod p$



$ax^2+by^2=d^{p-1}$



If $y=0$ then:



$$x=frac{d^{(p-1)/2}}{a^{1/2}}$$



So



$x=<frac{d^{(p-1)/2}}{a^{1/2}}$



That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:



$$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$



That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:



$$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$



You can exclude 1 from solutions and find better formula.






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    Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):



    $d^{p-1}-1 ≡0mod p$



    $ax^2+by^2=d^{p-1}$



    If $y=0$ then:



    $$x=frac{d^{(p-1)/2}}{a^{1/2}}$$



    So



    $x=<frac{d^{(p-1)/2}}{a^{1/2}}$



    That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:



    $$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$



    That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:



    $$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$



    You can exclude 1 from solutions and find better formula.






    share|cite|improve this answer




























      0














      Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):



      $d^{p-1}-1 ≡0mod p$



      $ax^2+by^2=d^{p-1}$



      If $y=0$ then:



      $$x=frac{d^{(p-1)/2}}{a^{1/2}}$$



      So



      $x=<frac{d^{(p-1)/2}}{a^{1/2}}$



      That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:



      $$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$



      That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:



      $$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$



      You can exclude 1 from solutions and find better formula.






      share|cite|improve this answer


























        0












        0








        0






        Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):



        $d^{p-1}-1 ≡0mod p$



        $ax^2+by^2=d^{p-1}$



        If $y=0$ then:



        $$x=frac{d^{(p-1)/2}}{a^{1/2}}$$



        So



        $x=<frac{d^{(p-1)/2}}{a^{1/2}}$



        That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:



        $$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$



        That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:



        $$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$



        You can exclude 1 from solutions and find better formula.






        share|cite|improve this answer














        Suppose There is a number d which is prime to p such that we can write following relations (for a simple form let $c=-1$):



        $d^{p-1}-1 ≡0mod p$



        $ax^2+by^2=d^{p-1}$



        If $y=0$ then:



        $$x=frac{d^{(p-1)/2}}{a^{1/2}}$$



        So



        $x=<frac{d^{(p-1)/2}}{a^{1/2}}$



        That is $1,2,3, . . frac{d^{(p-1)/2}}{a^{1/2}}$ can be the solutions for x, so number of solutions for x is $$chi_x=frac{d^{(p-1)/2}}{a^{1/2}}$$ . Clearly there will be corresponding value for y for each value of x. The equation is symmetric and the same argument could be applied for y, therefore:



        $$y=<frac{d^{(p-1)/2}}{b^{1/2}}$$



        That is the number of solutions for y is $$chi_y=frac{d^{(p-1)/2}}{b^{1/2}}$$ So the number of solution must be equal to:



        $$chi=frac{d^{(p-1)/2}}{a^{1/2}}+frac{d^{(p-1)/2}}{b^{1/2}}$$



        You can exclude 1 from solutions and find better formula.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 at 17:31

























        answered Nov 29 at 17:26









        sirous

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