Least quadratic residue under GRH: an explicit bound












10












$begingroup$


Let $m$ be a positive integer and $chi$ a primitive character mod $m$. Let $x$ be such that $chi(p)ne 1$ for all primes $p<x$. Assume GRH. How can one bound $x$ in terms of $m$ ? I do not need the best possible bound, but I need a good quality bound which is totally explicit in all parameters.



A related question: what is an explicit lower bound for $L(1,chi)$ under GRH?










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Yuri Bilu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    10












    $begingroup$


    Let $m$ be a positive integer and $chi$ a primitive character mod $m$. Let $x$ be such that $chi(p)ne 1$ for all primes $p<x$. Assume GRH. How can one bound $x$ in terms of $m$ ? I do not need the best possible bound, but I need a good quality bound which is totally explicit in all parameters.



    A related question: what is an explicit lower bound for $L(1,chi)$ under GRH?










    share|cite|improve this question









    New contributor




    Yuri Bilu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      10












      10








      10





      $begingroup$


      Let $m$ be a positive integer and $chi$ a primitive character mod $m$. Let $x$ be such that $chi(p)ne 1$ for all primes $p<x$. Assume GRH. How can one bound $x$ in terms of $m$ ? I do not need the best possible bound, but I need a good quality bound which is totally explicit in all parameters.



      A related question: what is an explicit lower bound for $L(1,chi)$ under GRH?










      share|cite|improve this question









      New contributor




      Yuri Bilu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $m$ be a positive integer and $chi$ a primitive character mod $m$. Let $x$ be such that $chi(p)ne 1$ for all primes $p<x$. Assume GRH. How can one bound $x$ in terms of $m$ ? I do not need the best possible bound, but I need a good quality bound which is totally explicit in all parameters.



      A related question: what is an explicit lower bound for $L(1,chi)$ under GRH?







      nt.number-theory analytic-number-theory l-functions






      share|cite|improve this question









      New contributor




      Yuri Bilu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Yuri Bilu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      YCor

      29.1k486140




      29.1k486140






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      Yuri Bilu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked Apr 8 at 1:21









      Yuri BiluYuri Bilu

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          1 Answer
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          20












          $begingroup$

          See the work of Lamzouri, Li, and Soundararajan (I link the arXiv version; the paper appeared in Math. Comp.). Assuming that $chi$ is a primitive quadratic character (as the title suggests) then Theorem 1.4 of that paper gives an explicit bound on the least prime quadratic residue on GRH. (Indeed that theorem gives an explicit bound on the least prime in any coset of a subgroup of $({Bbb Z}/q{Bbb Z})^times$.) Theorem 1.5 there gives explicit upper and lower bounds for $|L(1,chi)|$ for any primitive character $chi pmod q$ (not necessarily quadratic).






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Lucia, many thanks! This is exactly what I am looking for!
            $endgroup$
            – Yuri Bilu
            Apr 8 at 2:37






          • 2




            $begingroup$
            @YuriBilu: If you like Lucia's answer, please accept it officially (so that it turns green). Thanks! (And welcome to MO!)
            $endgroup$
            – GH from MO
            2 days ago














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          1 Answer
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          active

          oldest

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          20












          $begingroup$

          See the work of Lamzouri, Li, and Soundararajan (I link the arXiv version; the paper appeared in Math. Comp.). Assuming that $chi$ is a primitive quadratic character (as the title suggests) then Theorem 1.4 of that paper gives an explicit bound on the least prime quadratic residue on GRH. (Indeed that theorem gives an explicit bound on the least prime in any coset of a subgroup of $({Bbb Z}/q{Bbb Z})^times$.) Theorem 1.5 there gives explicit upper and lower bounds for $|L(1,chi)|$ for any primitive character $chi pmod q$ (not necessarily quadratic).






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Lucia, many thanks! This is exactly what I am looking for!
            $endgroup$
            – Yuri Bilu
            Apr 8 at 2:37






          • 2




            $begingroup$
            @YuriBilu: If you like Lucia's answer, please accept it officially (so that it turns green). Thanks! (And welcome to MO!)
            $endgroup$
            – GH from MO
            2 days ago


















          20












          $begingroup$

          See the work of Lamzouri, Li, and Soundararajan (I link the arXiv version; the paper appeared in Math. Comp.). Assuming that $chi$ is a primitive quadratic character (as the title suggests) then Theorem 1.4 of that paper gives an explicit bound on the least prime quadratic residue on GRH. (Indeed that theorem gives an explicit bound on the least prime in any coset of a subgroup of $({Bbb Z}/q{Bbb Z})^times$.) Theorem 1.5 there gives explicit upper and lower bounds for $|L(1,chi)|$ for any primitive character $chi pmod q$ (not necessarily quadratic).






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Lucia, many thanks! This is exactly what I am looking for!
            $endgroup$
            – Yuri Bilu
            Apr 8 at 2:37






          • 2




            $begingroup$
            @YuriBilu: If you like Lucia's answer, please accept it officially (so that it turns green). Thanks! (And welcome to MO!)
            $endgroup$
            – GH from MO
            2 days ago
















          20












          20








          20





          $begingroup$

          See the work of Lamzouri, Li, and Soundararajan (I link the arXiv version; the paper appeared in Math. Comp.). Assuming that $chi$ is a primitive quadratic character (as the title suggests) then Theorem 1.4 of that paper gives an explicit bound on the least prime quadratic residue on GRH. (Indeed that theorem gives an explicit bound on the least prime in any coset of a subgroup of $({Bbb Z}/q{Bbb Z})^times$.) Theorem 1.5 there gives explicit upper and lower bounds for $|L(1,chi)|$ for any primitive character $chi pmod q$ (not necessarily quadratic).






          share|cite|improve this answer









          $endgroup$



          See the work of Lamzouri, Li, and Soundararajan (I link the arXiv version; the paper appeared in Math. Comp.). Assuming that $chi$ is a primitive quadratic character (as the title suggests) then Theorem 1.4 of that paper gives an explicit bound on the least prime quadratic residue on GRH. (Indeed that theorem gives an explicit bound on the least prime in any coset of a subgroup of $({Bbb Z}/q{Bbb Z})^times$.) Theorem 1.5 there gives explicit upper and lower bounds for $|L(1,chi)|$ for any primitive character $chi pmod q$ (not necessarily quadratic).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 8 at 2:05









          LuciaLucia

          34.9k5151177




          34.9k5151177








          • 1




            $begingroup$
            Lucia, many thanks! This is exactly what I am looking for!
            $endgroup$
            – Yuri Bilu
            Apr 8 at 2:37






          • 2




            $begingroup$
            @YuriBilu: If you like Lucia's answer, please accept it officially (so that it turns green). Thanks! (And welcome to MO!)
            $endgroup$
            – GH from MO
            2 days ago
















          • 1




            $begingroup$
            Lucia, many thanks! This is exactly what I am looking for!
            $endgroup$
            – Yuri Bilu
            Apr 8 at 2:37






          • 2




            $begingroup$
            @YuriBilu: If you like Lucia's answer, please accept it officially (so that it turns green). Thanks! (And welcome to MO!)
            $endgroup$
            – GH from MO
            2 days ago










          1




          1




          $begingroup$
          Lucia, many thanks! This is exactly what I am looking for!
          $endgroup$
          – Yuri Bilu
          Apr 8 at 2:37




          $begingroup$
          Lucia, many thanks! This is exactly what I am looking for!
          $endgroup$
          – Yuri Bilu
          Apr 8 at 2:37




          2




          2




          $begingroup$
          @YuriBilu: If you like Lucia's answer, please accept it officially (so that it turns green). Thanks! (And welcome to MO!)
          $endgroup$
          – GH from MO
          2 days ago






          $begingroup$
          @YuriBilu: If you like Lucia's answer, please accept it officially (so that it turns green). Thanks! (And welcome to MO!)
          $endgroup$
          – GH from MO
          2 days ago












          Yuri Bilu is a new contributor. Be nice, and check out our Code of Conduct.










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