For $1leq aleq p-1$ and $5leq p$, show that $sum_{(a/p)=1}^{} a equiv 0 pmod{p}$












2












$begingroup$



For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.




I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.



Hints or complete solutions would be appreciated.






number-theory







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  • 2




    $begingroup$
    Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
    $endgroup$
    – MisterRiemann
    Jan 4 at 22:09


















2












$begingroup$



For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.




I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.



Hints or complete solutions would be appreciated.






number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
    $endgroup$
    – MisterRiemann
    Jan 4 at 22:09
















2












2








2


1



$begingroup$



For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.




I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.



Hints or complete solutions would be appreciated.






number-theory







share|cite|improve this question











$endgroup$





For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.




I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.



Hints or complete solutions would be appreciated.






number-theory




number-theory






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share|cite|improve this question













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edited Jan 4 at 22:05









Maria Mazur

49.9k1361124




49.9k1361124










asked Jan 4 at 22:05









John John

444




444








  • 2




    $begingroup$
    Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
    $endgroup$
    – MisterRiemann
    Jan 4 at 22:09
















  • 2




    $begingroup$
    Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
    $endgroup$
    – MisterRiemann
    Jan 4 at 22:09










2




2




$begingroup$
Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09






$begingroup$
Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09












1 Answer
1






active

oldest

votes


















3












$begingroup$

Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.



Now,
$$
sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
$$
Since $pneq 2,3$, the result follows immediately.






share|cite|improve this answer









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    3












    $begingroup$

    Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.



    Now,
    $$
    sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
    $$
    Since $pneq 2,3$, the result follows immediately.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.



      Now,
      $$
      sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
      $$
      Since $pneq 2,3$, the result follows immediately.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.



        Now,
        $$
        sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
        $$
        Since $pneq 2,3$, the result follows immediately.






        share|cite|improve this answer









        $endgroup$



        Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.



        Now,
        $$
        sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
        $$
        Since $pneq 2,3$, the result follows immediately.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 22:14









        AaronAaron

        2,010415




        2,010415






























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