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












2















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.










share|cite|improve this question




















  • 2




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


















2















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.










share|cite|improve this question




















  • 2




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
















2












2








2


1






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.










share|cite|improve this question
















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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 22:05









greedoid

38.6k114797




38.6k114797










asked Jan 4 at 22:05









John John

394




394








  • 2




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
















  • 2




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










2




2




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






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












1 Answer
1






active

oldest

votes


















3














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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062168%2ffor-1-leq-a-leq-p-1-and-5-leq-p-show-that-sum-a-p-1-a-equiv-0-pm%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    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


























      3














      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
























        3












        3








        3






        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












        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

        1,633414




        1,633414






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062168%2ffor-1-leq-a-leq-p-1-and-5-leq-p-show-that-sum-a-p-1-a-equiv-0-pm%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            1300-talet

            1300-talet

            Display a custom attribute below product name in the front-end Magento 1.9.3.8