Proof that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime iif $p equiv pm 1 mod 10$












3












$begingroup$


Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28
















3












$begingroup$


Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28














3












3








3





$begingroup$


Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol







share|cite|improve this question











$endgroup$




Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol




elementary-number-theory modular-arithmetic legendre-symbol






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 15:17







Alessar

















asked Nov 29 '18 at 14:41









AlessarAlessar

298115




298115








  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28














  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28








1




1




$begingroup$
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
$endgroup$
– B. Goddard
Nov 30 '18 at 12:42




$begingroup$
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
$endgroup$
– B. Goddard
Nov 30 '18 at 12:42












$begingroup$
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
$endgroup$
– Alessar
Nov 30 '18 at 13:08




$begingroup$
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
$endgroup$
– Alessar
Nov 30 '18 at 13:08




1




1




$begingroup$
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
$endgroup$
– B. Goddard
Nov 30 '18 at 13:31




$begingroup$
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
$endgroup$
– B. Goddard
Nov 30 '18 at 13:31












$begingroup$
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
$endgroup$
– Alessar
Dec 1 '18 at 11:28




$begingroup$
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
$endgroup$
– Alessar
Dec 1 '18 at 11:28










1 Answer
1






active

oldest

votes


















1












$begingroup$

See my other post use Gauss lemma to find $(frac{n}{p})$:



$(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






share|cite|improve this answer









$endgroup$













    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%2f3018706%2fproof-that-5-is-a-quadratic-residue-mod-p-with-p-odd-prime-iif-p-equi%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









    1












    $begingroup$

    See my other post use Gauss lemma to find $(frac{n}{p})$:



    $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      See my other post use Gauss lemma to find $(frac{n}{p})$:



      $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        See my other post use Gauss lemma to find $(frac{n}{p})$:



        $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






        share|cite|improve this answer









        $endgroup$



        See my other post use Gauss lemma to find $(frac{n}{p})$:



        $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 11:18









        Maestro13Maestro13

        1,081724




        1,081724






























            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%2f3018706%2fproof-that-5-is-a-quadratic-residue-mod-p-with-p-odd-prime-iif-p-equi%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more