Is $n^{(p-1)/2}equiv -1pmod p$ implies $n$ is a primitive root modulo $p$?Can the relation of $a^{p} equiv...

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Is $n^{(p-1)/2}equiv -1pmod p$ implies $n$ is a primitive root modulo $p$?


Can the relation of $a^{p} equiv apmod {2p}$ if (p prime and p>2) be added to Fermat's little theorem?Exhaustive (?) list of cases giving $sigma(n) equiv 0 pmod 4$ whenever $n$ is oddShowing that ${ nmid nu(n) equiv 0 pmod e}$ is independent of primitive rootWhy is $c+q$ still a primitive root modulo $q$?For which $n$ is $2^{2x+2} equiv 2pmod n quadtext{and}quad 2^{2x+2} equiv 4pmod {n-1} $Let $p$ be a prime. Then for every integer $a$ there is an integer $x$ such that $x^3 equiv a pmod p$What is the order of $3pmod{3215}$?Is $x^2 equiv 1 pmod{p^k} iff x equiv pm 1 pmod {p^k}$ for odd prime $p$?Show $p-a^2$ is a primitive root modulo $p$ with 1 < $a$ < $p-1$Show that $−g$ is also a primitive root of $p$ if $pequiv 1 pmod{4}$, but that $ord_p(−g) = frac{p−1}{2}$ if $p equiv 3 pmod{4}$.













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Let $p$ be an odd prime, $n$ be any integer, if $$n^{(p-1)/2}equiv -1pmod p,$$
is it always true that $k=p-1$ is the smallest positive integer satisfy $$n^kequiv 1pmod p?$$
This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$



I think this is not obvious, because $(p-1)/2$ does not have any relation with $(p-1)/3$, I have found an example, $7^6equiv 1pmod {19}$, but then I checked that $7^9equiv 1pmod {19}$ is also true.



Is there any counterexample? Or any proof?






number-theory multiplicative-function







share|cite|improve this question









$endgroup$

















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    $begingroup$


    Let $p$ be an odd prime, $n$ be any integer, if $$n^{(p-1)/2}equiv -1pmod p,$$
    is it always true that $k=p-1$ is the smallest positive integer satisfy $$n^kequiv 1pmod p?$$
    This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$



    I think this is not obvious, because $(p-1)/2$ does not have any relation with $(p-1)/3$, I have found an example, $7^6equiv 1pmod {19}$, but then I checked that $7^9equiv 1pmod {19}$ is also true.



    Is there any counterexample? Or any proof?






    number-theory multiplicative-function







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $p$ be an odd prime, $n$ be any integer, if $$n^{(p-1)/2}equiv -1pmod p,$$
      is it always true that $k=p-1$ is the smallest positive integer satisfy $$n^kequiv 1pmod p?$$
      This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$



      I think this is not obvious, because $(p-1)/2$ does not have any relation with $(p-1)/3$, I have found an example, $7^6equiv 1pmod {19}$, but then I checked that $7^9equiv 1pmod {19}$ is also true.



      Is there any counterexample? Or any proof?






      number-theory multiplicative-function







      share|cite|improve this question









      $endgroup$




      Let $p$ be an odd prime, $n$ be any integer, if $$n^{(p-1)/2}equiv -1pmod p,$$
      is it always true that $k=p-1$ is the smallest positive integer satisfy $$n^kequiv 1pmod p?$$
      This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$



      I think this is not obvious, because $(p-1)/2$ does not have any relation with $(p-1)/3$, I have found an example, $7^6equiv 1pmod {19}$, but then I checked that $7^9equiv 1pmod {19}$ is also true.



      Is there any counterexample? Or any proof?






      number-theory multiplicative-function




      number-theory multiplicative-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 15 hours ago









      kelvin hong 方kelvin hong 方

      78018




      78018






















          2 Answers
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          $begingroup$

          No. Take $n=5$, $p=13$. Then $frac{p-1}{2}=6$, $5^2=-1$ mod $p$, hence $5^6=-1$ mod $p$.
          However, $5^4=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I should find another way to solve my question, thanks for your counter-example!
            $endgroup$
            – kelvin hong 方
            15 hours ago



















          1












          $begingroup$


          This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$




          Yes, it is possible, e.g. $bmod p!=!7!:, nequiv -1,iff n^{large 3}equiv -1,$ and $ n^{large 2}equiv 1$



          Generally: $ , aequiv n^{large (p-1)/6}equiv -1iff ,a^{large 3}!equiv n^{large (p-1)/2}!equiv -1,$ and $ a^{large 2}!equiv n^{large (p-1)/3}!equiv 1$






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
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            2 Answers
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            $begingroup$

            No. Take $n=5$, $p=13$. Then $frac{p-1}{2}=6$, $5^2=-1$ mod $p$, hence $5^6=-1$ mod $p$.
            However, $5^4=1$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I should find another way to solve my question, thanks for your counter-example!
              $endgroup$
              – kelvin hong 方
              15 hours ago
















            2












            $begingroup$

            No. Take $n=5$, $p=13$. Then $frac{p-1}{2}=6$, $5^2=-1$ mod $p$, hence $5^6=-1$ mod $p$.
            However, $5^4=1$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I should find another way to solve my question, thanks for your counter-example!
              $endgroup$
              – kelvin hong 方
              15 hours ago














            2












            2








            2





            $begingroup$

            No. Take $n=5$, $p=13$. Then $frac{p-1}{2}=6$, $5^2=-1$ mod $p$, hence $5^6=-1$ mod $p$.
            However, $5^4=1$.






            share|cite|improve this answer









            $endgroup$



            No. Take $n=5$, $p=13$. Then $frac{p-1}{2}=6$, $5^2=-1$ mod $p$, hence $5^6=-1$ mod $p$.
            However, $5^4=1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 15 hours ago









            MindlackMindlack

            4,885211




            4,885211












            • $begingroup$
              I should find another way to solve my question, thanks for your counter-example!
              $endgroup$
              – kelvin hong 方
              15 hours ago


















            • $begingroup$
              I should find another way to solve my question, thanks for your counter-example!
              $endgroup$
              – kelvin hong 方
              15 hours ago
















            $begingroup$
            I should find another way to solve my question, thanks for your counter-example!
            $endgroup$
            – kelvin hong 方
            15 hours ago




            $begingroup$
            I should find another way to solve my question, thanks for your counter-example!
            $endgroup$
            – kelvin hong 方
            15 hours ago











            1












            $begingroup$


            This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$




            Yes, it is possible, e.g. $bmod p!=!7!:, nequiv -1,iff n^{large 3}equiv -1,$ and $ n^{large 2}equiv 1$



            Generally: $ , aequiv n^{large (p-1)/6}equiv -1iff ,a^{large 3}!equiv n^{large (p-1)/2}!equiv -1,$ and $ a^{large 2}!equiv n^{large (p-1)/3}!equiv 1$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$


              This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$




              Yes, it is possible, e.g. $bmod p!=!7!:, nequiv -1,iff n^{large 3}equiv -1,$ and $ n^{large 2}equiv 1$



              Generally: $ , aequiv n^{large (p-1)/6}equiv -1iff ,a^{large 3}!equiv n^{large (p-1)/2}!equiv -1,$ and $ a^{large 2}!equiv n^{large (p-1)/3}!equiv 1$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$


                This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$




                Yes, it is possible, e.g. $bmod p!=!7!:, nequiv -1,iff n^{large 3}equiv -1,$ and $ n^{large 2}equiv 1$



                Generally: $ , aequiv n^{large (p-1)/6}equiv -1iff ,a^{large 3}!equiv n^{large (p-1)/2}!equiv -1,$ and $ a^{large 2}!equiv n^{large (p-1)/3}!equiv 1$






                share|cite|improve this answer









                $endgroup$




                This is the little lemma I need in my solution to a bigger problem. I first hold a skeptical mind that this is false, so I want to find is it possible $$n^{(p-1)/2}equiv -1pmod pquad text{and} quad n^{(p-1)/3}equiv1pmod p?$$




                Yes, it is possible, e.g. $bmod p!=!7!:, nequiv -1,iff n^{large 3}equiv -1,$ and $ n^{large 2}equiv 1$



                Generally: $ , aequiv n^{large (p-1)/6}equiv -1iff ,a^{large 3}!equiv n^{large (p-1)/2}!equiv -1,$ and $ a^{large 2}!equiv n^{large (p-1)/3}!equiv 1$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 10 hours ago









                Bill DubuqueBill Dubuque

                212k29195650




                212k29195650






























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