The set of the primitive roots modulo $p$, with $p$ a fermat prime












0












$begingroup$


"Let $p$ be a prime of the form $2^{2^{n}}+1$, with $n in mathbb{N} $ (This means $p$ is a Fermat prime)



Using Euler's Criterion, prove that the set of primitive roots mod $p$ is equal to the set of quadratic non-residues mod $p$.



Use this to show 7 is a primitive root mod $p$"



I'll work up to the point I am stuck:



Suppose $a$ is a primitive root of mod $p$



Then $ord_{p}(a)=p-1=2^{2^{k}}$



(This means $a^{2^{2^{k}}}equiv 1$ $text{mod}$ $p$)



Euler's Criterion tells us:



$left(frac{a}{p}right)equiv a^{frac{p-1}{2}}$ $text{mod}$ $p$



From our defintions:



$a^{frac{p-1}{2}}= a^{2^{2^{k}-1}} equiv a^{2^{-1}}$ $text{mod}$ $p$



Is correct so far? Where do i go from here? And is this a sufficient method to prove the sets are the same?






modular-arithmetic quadratic-residues primitive-roots







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    "Let $p$ be a prime of the form $2^{2^{n}}+1$, with $n in mathbb{N} $ (This means $p$ is a Fermat prime)



    Using Euler's Criterion, prove that the set of primitive roots mod $p$ is equal to the set of quadratic non-residues mod $p$.



    Use this to show 7 is a primitive root mod $p$"



    I'll work up to the point I am stuck:



    Suppose $a$ is a primitive root of mod $p$



    Then $ord_{p}(a)=p-1=2^{2^{k}}$



    (This means $a^{2^{2^{k}}}equiv 1$ $text{mod}$ $p$)



    Euler's Criterion tells us:



    $left(frac{a}{p}right)equiv a^{frac{p-1}{2}}$ $text{mod}$ $p$



    From our defintions:



    $a^{frac{p-1}{2}}= a^{2^{2^{k}-1}} equiv a^{2^{-1}}$ $text{mod}$ $p$



    Is correct so far? Where do i go from here? And is this a sufficient method to prove the sets are the same?






    modular-arithmetic quadratic-residues primitive-roots







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      "Let $p$ be a prime of the form $2^{2^{n}}+1$, with $n in mathbb{N} $ (This means $p$ is a Fermat prime)



      Using Euler's Criterion, prove that the set of primitive roots mod $p$ is equal to the set of quadratic non-residues mod $p$.



      Use this to show 7 is a primitive root mod $p$"



      I'll work up to the point I am stuck:



      Suppose $a$ is a primitive root of mod $p$



      Then $ord_{p}(a)=p-1=2^{2^{k}}$



      (This means $a^{2^{2^{k}}}equiv 1$ $text{mod}$ $p$)



      Euler's Criterion tells us:



      $left(frac{a}{p}right)equiv a^{frac{p-1}{2}}$ $text{mod}$ $p$



      From our defintions:



      $a^{frac{p-1}{2}}= a^{2^{2^{k}-1}} equiv a^{2^{-1}}$ $text{mod}$ $p$



      Is correct so far? Where do i go from here? And is this a sufficient method to prove the sets are the same?






      modular-arithmetic quadratic-residues primitive-roots







      share|cite|improve this question











      $endgroup$




      "Let $p$ be a prime of the form $2^{2^{n}}+1$, with $n in mathbb{N} $ (This means $p$ is a Fermat prime)



      Using Euler's Criterion, prove that the set of primitive roots mod $p$ is equal to the set of quadratic non-residues mod $p$.



      Use this to show 7 is a primitive root mod $p$"



      I'll work up to the point I am stuck:



      Suppose $a$ is a primitive root of mod $p$



      Then $ord_{p}(a)=p-1=2^{2^{k}}$



      (This means $a^{2^{2^{k}}}equiv 1$ $text{mod}$ $p$)



      Euler's Criterion tells us:



      $left(frac{a}{p}right)equiv a^{frac{p-1}{2}}$ $text{mod}$ $p$



      From our defintions:



      $a^{frac{p-1}{2}}= a^{2^{2^{k}-1}} equiv a^{2^{-1}}$ $text{mod}$ $p$



      Is correct so far? Where do i go from here? And is this a sufficient method to prove the sets are the same?






      modular-arithmetic quadratic-residues primitive-roots




      modular-arithmetic quadratic-residues primitive-roots






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 5:24









      user1101010

      7991730




      7991730










      asked Dec 12 '18 at 4:44









      DinoDino

      836




      836






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Use the criterion that $a$ is a primitive root modulo $p$ iff
          $a^{(p-1)/q}notequiv1pmod p$ for all prime factors $q$ of $(p-1)$.
          Here $p-1=2^{2^n}$ so the only relevant $q$ is $q=2$.



          In your question you wrote $a^{2^{-1}}pmod p$. I don't know what you mean
          by that.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Your idea makes complete sense! How would we then show that 7 is a primitive root mod p? By showing it's a quadratic non-residue? How do we do that?
            $endgroup$
            – Dino
            Dec 12 '18 at 6:01










          • $begingroup$
            @Dino By quadratic reciprocity?
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 6:46










          • $begingroup$
            So that tells us $left(frac{7}{p}right) = left(frac{p}{7}right)$, so I need to know the value of $p$ mod $7$. I have no idea how to work out that? I think its the multiple powers that are throwing me off
            $endgroup$
            – Dino
            Dec 12 '18 at 7:09










          • $begingroup$
            You need to work out $2^m$ modulo $7$ for $m=2^n$. The sequence $(2^m)$ is periodic modulo $7$. @Dino
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 7:18










          • $begingroup$
            I recognise it is 4,2,4,2..., but how do you show it without just brute forcing it? I can't pull out a multiple of 7 easily out of $2^{m}$
            $endgroup$
            – Dino
            Dec 12 '18 at 7:37













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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Use the criterion that $a$ is a primitive root modulo $p$ iff
          $a^{(p-1)/q}notequiv1pmod p$ for all prime factors $q$ of $(p-1)$.
          Here $p-1=2^{2^n}$ so the only relevant $q$ is $q=2$.



          In your question you wrote $a^{2^{-1}}pmod p$. I don't know what you mean
          by that.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Your idea makes complete sense! How would we then show that 7 is a primitive root mod p? By showing it's a quadratic non-residue? How do we do that?
            $endgroup$
            – Dino
            Dec 12 '18 at 6:01










          • $begingroup$
            @Dino By quadratic reciprocity?
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 6:46










          • $begingroup$
            So that tells us $left(frac{7}{p}right) = left(frac{p}{7}right)$, so I need to know the value of $p$ mod $7$. I have no idea how to work out that? I think its the multiple powers that are throwing me off
            $endgroup$
            – Dino
            Dec 12 '18 at 7:09










          • $begingroup$
            You need to work out $2^m$ modulo $7$ for $m=2^n$. The sequence $(2^m)$ is periodic modulo $7$. @Dino
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 7:18










          • $begingroup$
            I recognise it is 4,2,4,2..., but how do you show it without just brute forcing it? I can't pull out a multiple of 7 easily out of $2^{m}$
            $endgroup$
            – Dino
            Dec 12 '18 at 7:37


















          2












          $begingroup$

          Use the criterion that $a$ is a primitive root modulo $p$ iff
          $a^{(p-1)/q}notequiv1pmod p$ for all prime factors $q$ of $(p-1)$.
          Here $p-1=2^{2^n}$ so the only relevant $q$ is $q=2$.



          In your question you wrote $a^{2^{-1}}pmod p$. I don't know what you mean
          by that.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Your idea makes complete sense! How would we then show that 7 is a primitive root mod p? By showing it's a quadratic non-residue? How do we do that?
            $endgroup$
            – Dino
            Dec 12 '18 at 6:01










          • $begingroup$
            @Dino By quadratic reciprocity?
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 6:46










          • $begingroup$
            So that tells us $left(frac{7}{p}right) = left(frac{p}{7}right)$, so I need to know the value of $p$ mod $7$. I have no idea how to work out that? I think its the multiple powers that are throwing me off
            $endgroup$
            – Dino
            Dec 12 '18 at 7:09










          • $begingroup$
            You need to work out $2^m$ modulo $7$ for $m=2^n$. The sequence $(2^m)$ is periodic modulo $7$. @Dino
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 7:18










          • $begingroup$
            I recognise it is 4,2,4,2..., but how do you show it without just brute forcing it? I can't pull out a multiple of 7 easily out of $2^{m}$
            $endgroup$
            – Dino
            Dec 12 '18 at 7:37
















          2












          2








          2





          $begingroup$

          Use the criterion that $a$ is a primitive root modulo $p$ iff
          $a^{(p-1)/q}notequiv1pmod p$ for all prime factors $q$ of $(p-1)$.
          Here $p-1=2^{2^n}$ so the only relevant $q$ is $q=2$.



          In your question you wrote $a^{2^{-1}}pmod p$. I don't know what you mean
          by that.






          share|cite|improve this answer









          $endgroup$



          Use the criterion that $a$ is a primitive root modulo $p$ iff
          $a^{(p-1)/q}notequiv1pmod p$ for all prime factors $q$ of $(p-1)$.
          Here $p-1=2^{2^n}$ so the only relevant $q$ is $q=2$.



          In your question you wrote $a^{2^{-1}}pmod p$. I don't know what you mean
          by that.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 4:53









          Lord Shark the UnknownLord Shark the Unknown

          105k1160132




          105k1160132












          • $begingroup$
            Your idea makes complete sense! How would we then show that 7 is a primitive root mod p? By showing it's a quadratic non-residue? How do we do that?
            $endgroup$
            – Dino
            Dec 12 '18 at 6:01










          • $begingroup$
            @Dino By quadratic reciprocity?
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 6:46










          • $begingroup$
            So that tells us $left(frac{7}{p}right) = left(frac{p}{7}right)$, so I need to know the value of $p$ mod $7$. I have no idea how to work out that? I think its the multiple powers that are throwing me off
            $endgroup$
            – Dino
            Dec 12 '18 at 7:09










          • $begingroup$
            You need to work out $2^m$ modulo $7$ for $m=2^n$. The sequence $(2^m)$ is periodic modulo $7$. @Dino
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 7:18










          • $begingroup$
            I recognise it is 4,2,4,2..., but how do you show it without just brute forcing it? I can't pull out a multiple of 7 easily out of $2^{m}$
            $endgroup$
            – Dino
            Dec 12 '18 at 7:37




















          • $begingroup$
            Your idea makes complete sense! How would we then show that 7 is a primitive root mod p? By showing it's a quadratic non-residue? How do we do that?
            $endgroup$
            – Dino
            Dec 12 '18 at 6:01










          • $begingroup$
            @Dino By quadratic reciprocity?
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 6:46










          • $begingroup$
            So that tells us $left(frac{7}{p}right) = left(frac{p}{7}right)$, so I need to know the value of $p$ mod $7$. I have no idea how to work out that? I think its the multiple powers that are throwing me off
            $endgroup$
            – Dino
            Dec 12 '18 at 7:09










          • $begingroup$
            You need to work out $2^m$ modulo $7$ for $m=2^n$. The sequence $(2^m)$ is periodic modulo $7$. @Dino
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 7:18










          • $begingroup$
            I recognise it is 4,2,4,2..., but how do you show it without just brute forcing it? I can't pull out a multiple of 7 easily out of $2^{m}$
            $endgroup$
            – Dino
            Dec 12 '18 at 7:37


















          $begingroup$
          Your idea makes complete sense! How would we then show that 7 is a primitive root mod p? By showing it's a quadratic non-residue? How do we do that?
          $endgroup$
          – Dino
          Dec 12 '18 at 6:01




          $begingroup$
          Your idea makes complete sense! How would we then show that 7 is a primitive root mod p? By showing it's a quadratic non-residue? How do we do that?
          $endgroup$
          – Dino
          Dec 12 '18 at 6:01












          $begingroup$
          @Dino By quadratic reciprocity?
          $endgroup$
          – Lord Shark the Unknown
          Dec 12 '18 at 6:46




          $begingroup$
          @Dino By quadratic reciprocity?
          $endgroup$
          – Lord Shark the Unknown
          Dec 12 '18 at 6:46












          $begingroup$
          So that tells us $left(frac{7}{p}right) = left(frac{p}{7}right)$, so I need to know the value of $p$ mod $7$. I have no idea how to work out that? I think its the multiple powers that are throwing me off
          $endgroup$
          – Dino
          Dec 12 '18 at 7:09




          $begingroup$
          So that tells us $left(frac{7}{p}right) = left(frac{p}{7}right)$, so I need to know the value of $p$ mod $7$. I have no idea how to work out that? I think its the multiple powers that are throwing me off
          $endgroup$
          – Dino
          Dec 12 '18 at 7:09












          $begingroup$
          You need to work out $2^m$ modulo $7$ for $m=2^n$. The sequence $(2^m)$ is periodic modulo $7$. @Dino
          $endgroup$
          – Lord Shark the Unknown
          Dec 12 '18 at 7:18




          $begingroup$
          You need to work out $2^m$ modulo $7$ for $m=2^n$. The sequence $(2^m)$ is periodic modulo $7$. @Dino
          $endgroup$
          – Lord Shark the Unknown
          Dec 12 '18 at 7:18












          $begingroup$
          I recognise it is 4,2,4,2..., but how do you show it without just brute forcing it? I can't pull out a multiple of 7 easily out of $2^{m}$
          $endgroup$
          – Dino
          Dec 12 '18 at 7:37






          $begingroup$
          I recognise it is 4,2,4,2..., but how do you show it without just brute forcing it? I can't pull out a multiple of 7 easily out of $2^{m}$
          $endgroup$
          – Dino
          Dec 12 '18 at 7:37




















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