Find $p$ if $2p-1$ is a perfect square












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I need to find prime numbers $p$ such that $2p-1$ is a perfect square. I tried hard, but could not get a proper solution for this. I could guess $p=13$ works, but I need a proper rigorous solution for this. I cant think of one.






number-theory







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  • 1




    $begingroup$
    And $p = 5, 41, 61, 113, 181,$ etc. What exactly are you trying to prove about such $p$?
    $endgroup$
    – anomaly
    Jan 9 at 5:09










  • $begingroup$
    Note that $p = 5$ also works as $2p - 1 = 9 = 3^2$.
    $endgroup$
    – John Omielan
    Jan 9 at 5:09






  • 1




    $begingroup$
    This is oeis.org/A027862
    $endgroup$
    – Ross Millikan
    Jan 9 at 5:15
















0












$begingroup$


I need to find prime numbers $p$ such that $2p-1$ is a perfect square. I tried hard, but could not get a proper solution for this. I could guess $p=13$ works, but I need a proper rigorous solution for this. I cant think of one.






number-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    And $p = 5, 41, 61, 113, 181,$ etc. What exactly are you trying to prove about such $p$?
    $endgroup$
    – anomaly
    Jan 9 at 5:09










  • $begingroup$
    Note that $p = 5$ also works as $2p - 1 = 9 = 3^2$.
    $endgroup$
    – John Omielan
    Jan 9 at 5:09






  • 1




    $begingroup$
    This is oeis.org/A027862
    $endgroup$
    – Ross Millikan
    Jan 9 at 5:15














0












0








0





$begingroup$


I need to find prime numbers $p$ such that $2p-1$ is a perfect square. I tried hard, but could not get a proper solution for this. I could guess $p=13$ works, but I need a proper rigorous solution for this. I cant think of one.






number-theory







share|cite|improve this question









$endgroup$




I need to find prime numbers $p$ such that $2p-1$ is a perfect square. I tried hard, but could not get a proper solution for this. I could guess $p=13$ works, but I need a proper rigorous solution for this. I cant think of one.






number-theory




number-theory






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asked Jan 9 at 5:01









YellowYellow

16011




16011








  • 1




    $begingroup$
    And $p = 5, 41, 61, 113, 181,$ etc. What exactly are you trying to prove about such $p$?
    $endgroup$
    – anomaly
    Jan 9 at 5:09










  • $begingroup$
    Note that $p = 5$ also works as $2p - 1 = 9 = 3^2$.
    $endgroup$
    – John Omielan
    Jan 9 at 5:09






  • 1




    $begingroup$
    This is oeis.org/A027862
    $endgroup$
    – Ross Millikan
    Jan 9 at 5:15














  • 1




    $begingroup$
    And $p = 5, 41, 61, 113, 181,$ etc. What exactly are you trying to prove about such $p$?
    $endgroup$
    – anomaly
    Jan 9 at 5:09










  • $begingroup$
    Note that $p = 5$ also works as $2p - 1 = 9 = 3^2$.
    $endgroup$
    – John Omielan
    Jan 9 at 5:09






  • 1




    $begingroup$
    This is oeis.org/A027862
    $endgroup$
    – Ross Millikan
    Jan 9 at 5:15








1




1




$begingroup$
And $p = 5, 41, 61, 113, 181,$ etc. What exactly are you trying to prove about such $p$?
$endgroup$
– anomaly
Jan 9 at 5:09




$begingroup$
And $p = 5, 41, 61, 113, 181,$ etc. What exactly are you trying to prove about such $p$?
$endgroup$
– anomaly
Jan 9 at 5:09












$begingroup$
Note that $p = 5$ also works as $2p - 1 = 9 = 3^2$.
$endgroup$
– John Omielan
Jan 9 at 5:09




$begingroup$
Note that $p = 5$ also works as $2p - 1 = 9 = 3^2$.
$endgroup$
– John Omielan
Jan 9 at 5:09




1




1




$begingroup$
This is oeis.org/A027862
$endgroup$
– Ross Millikan
Jan 9 at 5:15




$begingroup$
This is oeis.org/A027862
$endgroup$
– Ross Millikan
Jan 9 at 5:15










2 Answers
2






active

oldest

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1












$begingroup$

OEIS A027862 lists the primes $p$ of the form $p=n^2+(n+1)^2$. It remarks that these are the ones where $2p-1$ is a square. We can show that if $p$ is of this form $2p-1$ is a square because $$2p-1=2(n^2+(n+1)^2)-1=4n^2+4n+1=(2n+1)^2$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    All primes excluding $2$ and $3$ are of the form $6kpm 1$. We have that:



    $$2(6k+1)-1=12k+1$$
    This leads to $$12k=m^2-1$$
    $$12k=(m+1)(m-1)$$
    $$to m+1=12a, m-1=frac ka$$
    Or vice versa, subject to $a,k in Bbb Z, a|k$



    Now use:
    $$2(6k-1)-1=12k-3$$
    $$to 12k-4=m^2-1$$
    $$to 4(3k-1)=(m+1)(m-1)$$
    Go from there.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Excluding 2. And 3.
      $endgroup$
      – Lucas Henrique
      Jan 9 at 5:36










    • $begingroup$
      I'm stupid. Thanks
      $endgroup$
      – Rhys Hughes
      Jan 9 at 5:42












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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    OEIS A027862 lists the primes $p$ of the form $p=n^2+(n+1)^2$. It remarks that these are the ones where $2p-1$ is a square. We can show that if $p$ is of this form $2p-1$ is a square because $$2p-1=2(n^2+(n+1)^2)-1=4n^2+4n+1=(2n+1)^2$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      OEIS A027862 lists the primes $p$ of the form $p=n^2+(n+1)^2$. It remarks that these are the ones where $2p-1$ is a square. We can show that if $p$ is of this form $2p-1$ is a square because $$2p-1=2(n^2+(n+1)^2)-1=4n^2+4n+1=(2n+1)^2$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        OEIS A027862 lists the primes $p$ of the form $p=n^2+(n+1)^2$. It remarks that these are the ones where $2p-1$ is a square. We can show that if $p$ is of this form $2p-1$ is a square because $$2p-1=2(n^2+(n+1)^2)-1=4n^2+4n+1=(2n+1)^2$$






        share|cite|improve this answer









        $endgroup$



        OEIS A027862 lists the primes $p$ of the form $p=n^2+(n+1)^2$. It remarks that these are the ones where $2p-1$ is a square. We can show that if $p$ is of this form $2p-1$ is a square because $$2p-1=2(n^2+(n+1)^2)-1=4n^2+4n+1=(2n+1)^2$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 5:36









        Ross MillikanRoss Millikan

        301k24200375




        301k24200375























            1












            $begingroup$

            All primes excluding $2$ and $3$ are of the form $6kpm 1$. We have that:



            $$2(6k+1)-1=12k+1$$
            This leads to $$12k=m^2-1$$
            $$12k=(m+1)(m-1)$$
            $$to m+1=12a, m-1=frac ka$$
            Or vice versa, subject to $a,k in Bbb Z, a|k$



            Now use:
            $$2(6k-1)-1=12k-3$$
            $$to 12k-4=m^2-1$$
            $$to 4(3k-1)=(m+1)(m-1)$$
            Go from there.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Excluding 2. And 3.
              $endgroup$
              – Lucas Henrique
              Jan 9 at 5:36










            • $begingroup$
              I'm stupid. Thanks
              $endgroup$
              – Rhys Hughes
              Jan 9 at 5:42
















            1












            $begingroup$

            All primes excluding $2$ and $3$ are of the form $6kpm 1$. We have that:



            $$2(6k+1)-1=12k+1$$
            This leads to $$12k=m^2-1$$
            $$12k=(m+1)(m-1)$$
            $$to m+1=12a, m-1=frac ka$$
            Or vice versa, subject to $a,k in Bbb Z, a|k$



            Now use:
            $$2(6k-1)-1=12k-3$$
            $$to 12k-4=m^2-1$$
            $$to 4(3k-1)=(m+1)(m-1)$$
            Go from there.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Excluding 2. And 3.
              $endgroup$
              – Lucas Henrique
              Jan 9 at 5:36










            • $begingroup$
              I'm stupid. Thanks
              $endgroup$
              – Rhys Hughes
              Jan 9 at 5:42














            1












            1








            1





            $begingroup$

            All primes excluding $2$ and $3$ are of the form $6kpm 1$. We have that:



            $$2(6k+1)-1=12k+1$$
            This leads to $$12k=m^2-1$$
            $$12k=(m+1)(m-1)$$
            $$to m+1=12a, m-1=frac ka$$
            Or vice versa, subject to $a,k in Bbb Z, a|k$



            Now use:
            $$2(6k-1)-1=12k-3$$
            $$to 12k-4=m^2-1$$
            $$to 4(3k-1)=(m+1)(m-1)$$
            Go from there.






            share|cite|improve this answer











            $endgroup$



            All primes excluding $2$ and $3$ are of the form $6kpm 1$. We have that:



            $$2(6k+1)-1=12k+1$$
            This leads to $$12k=m^2-1$$
            $$12k=(m+1)(m-1)$$
            $$to m+1=12a, m-1=frac ka$$
            Or vice versa, subject to $a,k in Bbb Z, a|k$



            Now use:
            $$2(6k-1)-1=12k-3$$
            $$to 12k-4=m^2-1$$
            $$to 4(3k-1)=(m+1)(m-1)$$
            Go from there.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 9 at 5:42

























            answered Jan 9 at 5:21









            Rhys HughesRhys Hughes

            7,0791630




            7,0791630












            • $begingroup$
              Excluding 2. And 3.
              $endgroup$
              – Lucas Henrique
              Jan 9 at 5:36










            • $begingroup$
              I'm stupid. Thanks
              $endgroup$
              – Rhys Hughes
              Jan 9 at 5:42


















            • $begingroup$
              Excluding 2. And 3.
              $endgroup$
              – Lucas Henrique
              Jan 9 at 5:36










            • $begingroup$
              I'm stupid. Thanks
              $endgroup$
              – Rhys Hughes
              Jan 9 at 5:42
















            $begingroup$
            Excluding 2. And 3.
            $endgroup$
            – Lucas Henrique
            Jan 9 at 5:36




            $begingroup$
            Excluding 2. And 3.
            $endgroup$
            – Lucas Henrique
            Jan 9 at 5:36












            $begingroup$
            I'm stupid. Thanks
            $endgroup$
            – Rhys Hughes
            Jan 9 at 5:42




            $begingroup$
            I'm stupid. Thanks
            $endgroup$
            – Rhys Hughes
            Jan 9 at 5:42


















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