Can a number be a quadratic residue modulo all prime that do not divide it












2












$begingroup$


Is there a proof that for any number $a$, there must be at least one prime $p$ such that $(a/p)=-1$, where $(a/p)$ is the Legendre symbol?



In other words, for all $a$, is there at least one prime $p$ such that $a$ is a quadratic nonresidue modulo $p$?



EDIT: Due to the comments pointing out that there is no such $p$ for $a=x^2$, my question remains the same, except for only all $aneq x^2$.






number-theory elementary-number-theory legendre-symbol







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    What about $a=1$?
    $endgroup$
    – Somos
    Dec 30 '18 at 0:39






  • 5




    $begingroup$
    Or any other square, for that matter.
    $endgroup$
    – Robert Israel
    Dec 30 '18 at 0:42










  • $begingroup$
    See here and its linked questions for related results.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 23:00
















2












$begingroup$


Is there a proof that for any number $a$, there must be at least one prime $p$ such that $(a/p)=-1$, where $(a/p)$ is the Legendre symbol?



In other words, for all $a$, is there at least one prime $p$ such that $a$ is a quadratic nonresidue modulo $p$?



EDIT: Due to the comments pointing out that there is no such $p$ for $a=x^2$, my question remains the same, except for only all $aneq x^2$.






number-theory elementary-number-theory legendre-symbol







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    What about $a=1$?
    $endgroup$
    – Somos
    Dec 30 '18 at 0:39






  • 5




    $begingroup$
    Or any other square, for that matter.
    $endgroup$
    – Robert Israel
    Dec 30 '18 at 0:42










  • $begingroup$
    See here and its linked questions for related results.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 23:00














2












2








2


0



$begingroup$


Is there a proof that for any number $a$, there must be at least one prime $p$ such that $(a/p)=-1$, where $(a/p)$ is the Legendre symbol?



In other words, for all $a$, is there at least one prime $p$ such that $a$ is a quadratic nonresidue modulo $p$?



EDIT: Due to the comments pointing out that there is no such $p$ for $a=x^2$, my question remains the same, except for only all $aneq x^2$.






number-theory elementary-number-theory legendre-symbol







share|cite|improve this question











$endgroup$




Is there a proof that for any number $a$, there must be at least one prime $p$ such that $(a/p)=-1$, where $(a/p)$ is the Legendre symbol?



In other words, for all $a$, is there at least one prime $p$ such that $a$ is a quadratic nonresidue modulo $p$?



EDIT: Due to the comments pointing out that there is no such $p$ for $a=x^2$, my question remains the same, except for only all $aneq x^2$.






number-theory elementary-number-theory legendre-symbol




number-theory elementary-number-theory legendre-symbol






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 0:45







Tejas Rao

















asked Dec 30 '18 at 0:35









Tejas RaoTejas Rao

34211




34211








  • 5




    $begingroup$
    What about $a=1$?
    $endgroup$
    – Somos
    Dec 30 '18 at 0:39






  • 5




    $begingroup$
    Or any other square, for that matter.
    $endgroup$
    – Robert Israel
    Dec 30 '18 at 0:42










  • $begingroup$
    See here and its linked questions for related results.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 23:00














  • 5




    $begingroup$
    What about $a=1$?
    $endgroup$
    – Somos
    Dec 30 '18 at 0:39






  • 5




    $begingroup$
    Or any other square, for that matter.
    $endgroup$
    – Robert Israel
    Dec 30 '18 at 0:42










  • $begingroup$
    See here and its linked questions for related results.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 23:00








5




5




$begingroup$
What about $a=1$?
$endgroup$
– Somos
Dec 30 '18 at 0:39




$begingroup$
What about $a=1$?
$endgroup$
– Somos
Dec 30 '18 at 0:39




5




5




$begingroup$
Or any other square, for that matter.
$endgroup$
– Robert Israel
Dec 30 '18 at 0:42




$begingroup$
Or any other square, for that matter.
$endgroup$
– Robert Israel
Dec 30 '18 at 0:42












$begingroup$
See here and its linked questions for related results.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 23:00




$begingroup$
See here and its linked questions for related results.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 23:00










2 Answers
2






active

oldest

votes


















2












$begingroup$

Here is an interesting exercise that appears in A Classical Introduction to Modern Number Theory.




Let $a$ be a non square integer. Then there are infinitely many primes $p$ for which $a$ is a quadratic non residue.




As a hint, use the Chinese Remainder Theorem in a clever way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the reference. May you provide a proof of the exercise as the book is behind a pay barrier?
    $endgroup$
    – Tejas Rao
    Dec 30 '18 at 2:08










  • $begingroup$
    Try your best and after a few days I can give you another hint if needed.
    $endgroup$
    – Sandeep Silwal
    Dec 30 '18 at 2:11



















0












$begingroup$

Here is a solution to the above post.



Set $a=p_1^{alpha_1}cdots p_k^{alpha_k}$ be prime decomposition of $a$. Clearly, there is a $j$ such that, $alpha_j$ is odd. Without loss of generality, assume $j=1$. Also, suppose $p_1neq 2$, which is a case that can be handled separately.



We will construct a prime $p$, such that $p$ is a quadratic residue modulo $p_i$, for every $i geq 2$, and is not a quadratic residue in modulo $p_1$. Take $a_1$ to be a quadratic non-residue in modulo $p_1$. Take $q_2,dots,q_k$ to be quadratic residues in modulo $p_k$. Now, there is a prime $p$ such that $pequiv a_1pmod{p_1}$, and $pequiv q_ipmod{p_i}$, and $pequiv 1pmod{4}$, for $i geq 2$. Existence of this prime is due to Chinese remainder theorem + Dirichlet's theorem.



Finally, by quadratic reciprocity, it is not hard to check, $p$ is a quadratic residue in mod $p_j$ for $jgeq 2$, and a non-residue in mod $p_1$. Construction is complete.






share|cite|improve this answer









$endgroup$














    Your Answer








    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%2f3056403%2fcan-a-number-be-a-quadratic-residue-modulo-all-prime-that-do-not-divide-it%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Here is an interesting exercise that appears in A Classical Introduction to Modern Number Theory.




    Let $a$ be a non square integer. Then there are infinitely many primes $p$ for which $a$ is a quadratic non residue.




    As a hint, use the Chinese Remainder Theorem in a clever way.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for the reference. May you provide a proof of the exercise as the book is behind a pay barrier?
      $endgroup$
      – Tejas Rao
      Dec 30 '18 at 2:08










    • $begingroup$
      Try your best and after a few days I can give you another hint if needed.
      $endgroup$
      – Sandeep Silwal
      Dec 30 '18 at 2:11
















    2












    $begingroup$

    Here is an interesting exercise that appears in A Classical Introduction to Modern Number Theory.




    Let $a$ be a non square integer. Then there are infinitely many primes $p$ for which $a$ is a quadratic non residue.




    As a hint, use the Chinese Remainder Theorem in a clever way.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for the reference. May you provide a proof of the exercise as the book is behind a pay barrier?
      $endgroup$
      – Tejas Rao
      Dec 30 '18 at 2:08










    • $begingroup$
      Try your best and after a few days I can give you another hint if needed.
      $endgroup$
      – Sandeep Silwal
      Dec 30 '18 at 2:11














    2












    2








    2





    $begingroup$

    Here is an interesting exercise that appears in A Classical Introduction to Modern Number Theory.




    Let $a$ be a non square integer. Then there are infinitely many primes $p$ for which $a$ is a quadratic non residue.




    As a hint, use the Chinese Remainder Theorem in a clever way.






    share|cite|improve this answer











    $endgroup$



    Here is an interesting exercise that appears in A Classical Introduction to Modern Number Theory.




    Let $a$ be a non square integer. Then there are infinitely many primes $p$ for which $a$ is a quadratic non residue.




    As a hint, use the Chinese Remainder Theorem in a clever way.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 30 '18 at 3:26









    dantopa

    6,69442245




    6,69442245










    answered Dec 30 '18 at 1:30









    Sandeep SilwalSandeep Silwal

    5,90811337




    5,90811337












    • $begingroup$
      Thank you for the reference. May you provide a proof of the exercise as the book is behind a pay barrier?
      $endgroup$
      – Tejas Rao
      Dec 30 '18 at 2:08










    • $begingroup$
      Try your best and after a few days I can give you another hint if needed.
      $endgroup$
      – Sandeep Silwal
      Dec 30 '18 at 2:11


















    • $begingroup$
      Thank you for the reference. May you provide a proof of the exercise as the book is behind a pay barrier?
      $endgroup$
      – Tejas Rao
      Dec 30 '18 at 2:08










    • $begingroup$
      Try your best and after a few days I can give you another hint if needed.
      $endgroup$
      – Sandeep Silwal
      Dec 30 '18 at 2:11
















    $begingroup$
    Thank you for the reference. May you provide a proof of the exercise as the book is behind a pay barrier?
    $endgroup$
    – Tejas Rao
    Dec 30 '18 at 2:08




    $begingroup$
    Thank you for the reference. May you provide a proof of the exercise as the book is behind a pay barrier?
    $endgroup$
    – Tejas Rao
    Dec 30 '18 at 2:08












    $begingroup$
    Try your best and after a few days I can give you another hint if needed.
    $endgroup$
    – Sandeep Silwal
    Dec 30 '18 at 2:11




    $begingroup$
    Try your best and after a few days I can give you another hint if needed.
    $endgroup$
    – Sandeep Silwal
    Dec 30 '18 at 2:11











    0












    $begingroup$

    Here is a solution to the above post.



    Set $a=p_1^{alpha_1}cdots p_k^{alpha_k}$ be prime decomposition of $a$. Clearly, there is a $j$ such that, $alpha_j$ is odd. Without loss of generality, assume $j=1$. Also, suppose $p_1neq 2$, which is a case that can be handled separately.



    We will construct a prime $p$, such that $p$ is a quadratic residue modulo $p_i$, for every $i geq 2$, and is not a quadratic residue in modulo $p_1$. Take $a_1$ to be a quadratic non-residue in modulo $p_1$. Take $q_2,dots,q_k$ to be quadratic residues in modulo $p_k$. Now, there is a prime $p$ such that $pequiv a_1pmod{p_1}$, and $pequiv q_ipmod{p_i}$, and $pequiv 1pmod{4}$, for $i geq 2$. Existence of this prime is due to Chinese remainder theorem + Dirichlet's theorem.



    Finally, by quadratic reciprocity, it is not hard to check, $p$ is a quadratic residue in mod $p_j$ for $jgeq 2$, and a non-residue in mod $p_1$. Construction is complete.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is a solution to the above post.



      Set $a=p_1^{alpha_1}cdots p_k^{alpha_k}$ be prime decomposition of $a$. Clearly, there is a $j$ such that, $alpha_j$ is odd. Without loss of generality, assume $j=1$. Also, suppose $p_1neq 2$, which is a case that can be handled separately.



      We will construct a prime $p$, such that $p$ is a quadratic residue modulo $p_i$, for every $i geq 2$, and is not a quadratic residue in modulo $p_1$. Take $a_1$ to be a quadratic non-residue in modulo $p_1$. Take $q_2,dots,q_k$ to be quadratic residues in modulo $p_k$. Now, there is a prime $p$ such that $pequiv a_1pmod{p_1}$, and $pequiv q_ipmod{p_i}$, and $pequiv 1pmod{4}$, for $i geq 2$. Existence of this prime is due to Chinese remainder theorem + Dirichlet's theorem.



      Finally, by quadratic reciprocity, it is not hard to check, $p$ is a quadratic residue in mod $p_j$ for $jgeq 2$, and a non-residue in mod $p_1$. Construction is complete.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is a solution to the above post.



        Set $a=p_1^{alpha_1}cdots p_k^{alpha_k}$ be prime decomposition of $a$. Clearly, there is a $j$ such that, $alpha_j$ is odd. Without loss of generality, assume $j=1$. Also, suppose $p_1neq 2$, which is a case that can be handled separately.



        We will construct a prime $p$, such that $p$ is a quadratic residue modulo $p_i$, for every $i geq 2$, and is not a quadratic residue in modulo $p_1$. Take $a_1$ to be a quadratic non-residue in modulo $p_1$. Take $q_2,dots,q_k$ to be quadratic residues in modulo $p_k$. Now, there is a prime $p$ such that $pequiv a_1pmod{p_1}$, and $pequiv q_ipmod{p_i}$, and $pequiv 1pmod{4}$, for $i geq 2$. Existence of this prime is due to Chinese remainder theorem + Dirichlet's theorem.



        Finally, by quadratic reciprocity, it is not hard to check, $p$ is a quadratic residue in mod $p_j$ for $jgeq 2$, and a non-residue in mod $p_1$. Construction is complete.






        share|cite|improve this answer









        $endgroup$



        Here is a solution to the above post.



        Set $a=p_1^{alpha_1}cdots p_k^{alpha_k}$ be prime decomposition of $a$. Clearly, there is a $j$ such that, $alpha_j$ is odd. Without loss of generality, assume $j=1$. Also, suppose $p_1neq 2$, which is a case that can be handled separately.



        We will construct a prime $p$, such that $p$ is a quadratic residue modulo $p_i$, for every $i geq 2$, and is not a quadratic residue in modulo $p_1$. Take $a_1$ to be a quadratic non-residue in modulo $p_1$. Take $q_2,dots,q_k$ to be quadratic residues in modulo $p_k$. Now, there is a prime $p$ such that $pequiv a_1pmod{p_1}$, and $pequiv q_ipmod{p_i}$, and $pequiv 1pmod{4}$, for $i geq 2$. Existence of this prime is due to Chinese remainder theorem + Dirichlet's theorem.



        Finally, by quadratic reciprocity, it is not hard to check, $p$ is a quadratic residue in mod $p_j$ for $jgeq 2$, and a non-residue in mod $p_1$. Construction is complete.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 18:02









        AaronAaron

        2,035415




        2,035415






























            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%2f3056403%2fcan-a-number-be-a-quadratic-residue-modulo-all-prime-that-do-not-divide-it%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more