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







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$endgroup$








  • 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

31811




31811








  • 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









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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

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    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

    6,52942244




    6,52942244










    answered Dec 30 '18 at 1:30









    Sandeep SilwalSandeep Silwal

    5,86811237




    5,86811237












    • $begingroup$
      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

        1,912415




        1,912415






























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