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

33711




33711








  • 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.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%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

    6,63342245




    6,63342245










    answered Dec 30 '18 at 1:30









    Sandeep SilwalSandeep Silwal

    5,88811237




    5,88811237












    • $begingroup$
      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

        1,922415




        1,922415






























            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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more