What are the Legendre symbols $left(frac{10}{31}right)$ and $left(frac{-15}{43}right)$?












2












$begingroup$


I have the following two Legendre symbols that need calculated:



$left(frac{10}{31}right)$ $=$ $-left(frac{31}{10}right)$ $=$ $-left(frac{1}{10}right)$ $=$ $-(-1)$ $=$ $-1$



$left(frac{-15}{43}right)$ $=$ $left(frac{43}{15}right)$ $=$ $left(frac{13}{15}right)$ $=$ $-1$



is that correct?



I just want to make sure I am understanding this concept.






number-theory elementary-number-theory proof-verification legendre-symbol quadratic-reciprocity







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If you are using the Legendre symbol (and not the Jacobi symbol), then the denominators are supposed to be prime. Clearly, $10$ and $15$ are not prime, and the quadratic reciprocity law you are using is highly dubious since it works only with odd integers.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:35










  • $begingroup$
    @Batominovski ohh I did not know! Thank you so much for the comment
    $endgroup$
    – Hidaw
    Dec 6 '18 at 22:37










  • $begingroup$
    And please use a more descriptive title the next time.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:37






  • 2




    $begingroup$
    Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 22:45
















2












$begingroup$


I have the following two Legendre symbols that need calculated:



$left(frac{10}{31}right)$ $=$ $-left(frac{31}{10}right)$ $=$ $-left(frac{1}{10}right)$ $=$ $-(-1)$ $=$ $-1$



$left(frac{-15}{43}right)$ $=$ $left(frac{43}{15}right)$ $=$ $left(frac{13}{15}right)$ $=$ $-1$



is that correct?



I just want to make sure I am understanding this concept.






number-theory elementary-number-theory proof-verification legendre-symbol quadratic-reciprocity







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If you are using the Legendre symbol (and not the Jacobi symbol), then the denominators are supposed to be prime. Clearly, $10$ and $15$ are not prime, and the quadratic reciprocity law you are using is highly dubious since it works only with odd integers.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:35










  • $begingroup$
    @Batominovski ohh I did not know! Thank you so much for the comment
    $endgroup$
    – Hidaw
    Dec 6 '18 at 22:37










  • $begingroup$
    And please use a more descriptive title the next time.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:37






  • 2




    $begingroup$
    Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 22:45














2












2








2





$begingroup$


I have the following two Legendre symbols that need calculated:



$left(frac{10}{31}right)$ $=$ $-left(frac{31}{10}right)$ $=$ $-left(frac{1}{10}right)$ $=$ $-(-1)$ $=$ $-1$



$left(frac{-15}{43}right)$ $=$ $left(frac{43}{15}right)$ $=$ $left(frac{13}{15}right)$ $=$ $-1$



is that correct?



I just want to make sure I am understanding this concept.






number-theory elementary-number-theory proof-verification legendre-symbol quadratic-reciprocity







share|cite|improve this question











$endgroup$




I have the following two Legendre symbols that need calculated:



$left(frac{10}{31}right)$ $=$ $-left(frac{31}{10}right)$ $=$ $-left(frac{1}{10}right)$ $=$ $-(-1)$ $=$ $-1$



$left(frac{-15}{43}right)$ $=$ $left(frac{43}{15}right)$ $=$ $left(frac{13}{15}right)$ $=$ $-1$



is that correct?



I just want to make sure I am understanding this concept.






number-theory elementary-number-theory proof-verification legendre-symbol quadratic-reciprocity




number-theory elementary-number-theory proof-verification legendre-symbol quadratic-reciprocity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 7:56









Batominovski

33k33293




33k33293










asked Dec 6 '18 at 22:32









HidawHidaw

509624




509624








  • 3




    $begingroup$
    If you are using the Legendre symbol (and not the Jacobi symbol), then the denominators are supposed to be prime. Clearly, $10$ and $15$ are not prime, and the quadratic reciprocity law you are using is highly dubious since it works only with odd integers.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:35










  • $begingroup$
    @Batominovski ohh I did not know! Thank you so much for the comment
    $endgroup$
    – Hidaw
    Dec 6 '18 at 22:37










  • $begingroup$
    And please use a more descriptive title the next time.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:37






  • 2




    $begingroup$
    Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 22:45














  • 3




    $begingroup$
    If you are using the Legendre symbol (and not the Jacobi symbol), then the denominators are supposed to be prime. Clearly, $10$ and $15$ are not prime, and the quadratic reciprocity law you are using is highly dubious since it works only with odd integers.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:35










  • $begingroup$
    @Batominovski ohh I did not know! Thank you so much for the comment
    $endgroup$
    – Hidaw
    Dec 6 '18 at 22:37










  • $begingroup$
    And please use a more descriptive title the next time.
    $endgroup$
    – Batominovski
    Dec 6 '18 at 22:37






  • 2




    $begingroup$
    Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 22:45








3




3




$begingroup$
If you are using the Legendre symbol (and not the Jacobi symbol), then the denominators are supposed to be prime. Clearly, $10$ and $15$ are not prime, and the quadratic reciprocity law you are using is highly dubious since it works only with odd integers.
$endgroup$
– Batominovski
Dec 6 '18 at 22:35




$begingroup$
If you are using the Legendre symbol (and not the Jacobi symbol), then the denominators are supposed to be prime. Clearly, $10$ and $15$ are not prime, and the quadratic reciprocity law you are using is highly dubious since it works only with odd integers.
$endgroup$
– Batominovski
Dec 6 '18 at 22:35












$begingroup$
@Batominovski ohh I did not know! Thank you so much for the comment
$endgroup$
– Hidaw
Dec 6 '18 at 22:37




$begingroup$
@Batominovski ohh I did not know! Thank you so much for the comment
$endgroup$
– Hidaw
Dec 6 '18 at 22:37












$begingroup$
And please use a more descriptive title the next time.
$endgroup$
– Batominovski
Dec 6 '18 at 22:37




$begingroup$
And please use a more descriptive title the next time.
$endgroup$
– Batominovski
Dec 6 '18 at 22:37




2




2




$begingroup$
Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:45




$begingroup$
Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:45










2 Answers
2






active

oldest

votes


















2












$begingroup$

Here is an approach I would take. Note that
$$left(frac{10}{31}right)=left(frac{-21}{31}right)=left(frac{-1}{31}right)left(frac{3}{31}right)left(frac{7}{31}right)=(-1)Biggl(-left(frac{31}{3}right)Biggr)Biggl(-left(frac{31}{7}right)Biggr),.$$




That is, $$left(frac{10}{31}right)=-left(frac{1}{3}right)left(frac{3}{7}right)=-(+1)Biggl(-left(frac{7}{3}right)Biggr)=left(frac{1}{3}right)=+1,.$$ This can be verified by noting that $6^2equiv 5pmod{31}$ and $8^2equiv 2pmod{31}$, so $$17^2equiv (6cdot 8)^2equiv 5cdot2=10pmod{31},.$$





For the second part, note that
$$left(frac{-15}{43}right)=left(frac{-1}{43}right)left(frac{3}{43}right)left(frac{5}{43}right)=(-1)Biggl(-left(frac{43}{3}right)Biggr)left(frac{43}{5}right),.$$




Therefore, $$left(frac{-15}{43}right)=left(frac{1}{3}right)left(frac{3}{5}right)=left(frac{3}{5}right),.$$ It is easy to verify that $left(dfrac{3}{5}right)=-1$, whence $$left(frac{-15}{43}right)=-1,.$$ You can check that $12^2equiv 15pmod{43}$, so $left(dfrac{15}{43}right)=+1$, whereas $left(dfrac{-1}{43}right)=-1$, confirming the calculations.







share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    As remarked in comments, to use quadratic reciprocity we need to work with Legendre Symbols
    $$ left(frac{p}{q}right)$$
    for $p,q$ prime.
    You should repeatedly use the property
    $$ left(frac{ab}{p}right)=left(frac{a}{p}right)left(frac{b}{p}right)$$
    to make sure that you are calculating with both parts of the symbol being prime. That is, write
    $$ left(frac{10}{31}right)=left(frac{2}{31}right)left(frac{5}{31}right)$$
    then iteratively apply quadratic reciprocity as you intended to.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR.
      $endgroup$
      – Oscar Lanzi
      Dec 6 '18 at 23:23










    • $begingroup$
      If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what
      $endgroup$
      – Hidaw
      Dec 7 '18 at 20:28











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Here is an approach I would take. Note that
    $$left(frac{10}{31}right)=left(frac{-21}{31}right)=left(frac{-1}{31}right)left(frac{3}{31}right)left(frac{7}{31}right)=(-1)Biggl(-left(frac{31}{3}right)Biggr)Biggl(-left(frac{31}{7}right)Biggr),.$$




    That is, $$left(frac{10}{31}right)=-left(frac{1}{3}right)left(frac{3}{7}right)=-(+1)Biggl(-left(frac{7}{3}right)Biggr)=left(frac{1}{3}right)=+1,.$$ This can be verified by noting that $6^2equiv 5pmod{31}$ and $8^2equiv 2pmod{31}$, so $$17^2equiv (6cdot 8)^2equiv 5cdot2=10pmod{31},.$$





    For the second part, note that
    $$left(frac{-15}{43}right)=left(frac{-1}{43}right)left(frac{3}{43}right)left(frac{5}{43}right)=(-1)Biggl(-left(frac{43}{3}right)Biggr)left(frac{43}{5}right),.$$




    Therefore, $$left(frac{-15}{43}right)=left(frac{1}{3}right)left(frac{3}{5}right)=left(frac{3}{5}right),.$$ It is easy to verify that $left(dfrac{3}{5}right)=-1$, whence $$left(frac{-15}{43}right)=-1,.$$ You can check that $12^2equiv 15pmod{43}$, so $left(dfrac{15}{43}right)=+1$, whereas $left(dfrac{-1}{43}right)=-1$, confirming the calculations.







    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Here is an approach I would take. Note that
      $$left(frac{10}{31}right)=left(frac{-21}{31}right)=left(frac{-1}{31}right)left(frac{3}{31}right)left(frac{7}{31}right)=(-1)Biggl(-left(frac{31}{3}right)Biggr)Biggl(-left(frac{31}{7}right)Biggr),.$$




      That is, $$left(frac{10}{31}right)=-left(frac{1}{3}right)left(frac{3}{7}right)=-(+1)Biggl(-left(frac{7}{3}right)Biggr)=left(frac{1}{3}right)=+1,.$$ This can be verified by noting that $6^2equiv 5pmod{31}$ and $8^2equiv 2pmod{31}$, so $$17^2equiv (6cdot 8)^2equiv 5cdot2=10pmod{31},.$$





      For the second part, note that
      $$left(frac{-15}{43}right)=left(frac{-1}{43}right)left(frac{3}{43}right)left(frac{5}{43}right)=(-1)Biggl(-left(frac{43}{3}right)Biggr)left(frac{43}{5}right),.$$




      Therefore, $$left(frac{-15}{43}right)=left(frac{1}{3}right)left(frac{3}{5}right)=left(frac{3}{5}right),.$$ It is easy to verify that $left(dfrac{3}{5}right)=-1$, whence $$left(frac{-15}{43}right)=-1,.$$ You can check that $12^2equiv 15pmod{43}$, so $left(dfrac{15}{43}right)=+1$, whereas $left(dfrac{-1}{43}right)=-1$, confirming the calculations.







      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Here is an approach I would take. Note that
        $$left(frac{10}{31}right)=left(frac{-21}{31}right)=left(frac{-1}{31}right)left(frac{3}{31}right)left(frac{7}{31}right)=(-1)Biggl(-left(frac{31}{3}right)Biggr)Biggl(-left(frac{31}{7}right)Biggr),.$$




        That is, $$left(frac{10}{31}right)=-left(frac{1}{3}right)left(frac{3}{7}right)=-(+1)Biggl(-left(frac{7}{3}right)Biggr)=left(frac{1}{3}right)=+1,.$$ This can be verified by noting that $6^2equiv 5pmod{31}$ and $8^2equiv 2pmod{31}$, so $$17^2equiv (6cdot 8)^2equiv 5cdot2=10pmod{31},.$$





        For the second part, note that
        $$left(frac{-15}{43}right)=left(frac{-1}{43}right)left(frac{3}{43}right)left(frac{5}{43}right)=(-1)Biggl(-left(frac{43}{3}right)Biggr)left(frac{43}{5}right),.$$




        Therefore, $$left(frac{-15}{43}right)=left(frac{1}{3}right)left(frac{3}{5}right)=left(frac{3}{5}right),.$$ It is easy to verify that $left(dfrac{3}{5}right)=-1$, whence $$left(frac{-15}{43}right)=-1,.$$ You can check that $12^2equiv 15pmod{43}$, so $left(dfrac{15}{43}right)=+1$, whereas $left(dfrac{-1}{43}right)=-1$, confirming the calculations.







        share|cite|improve this answer









        $endgroup$



        Here is an approach I would take. Note that
        $$left(frac{10}{31}right)=left(frac{-21}{31}right)=left(frac{-1}{31}right)left(frac{3}{31}right)left(frac{7}{31}right)=(-1)Biggl(-left(frac{31}{3}right)Biggr)Biggl(-left(frac{31}{7}right)Biggr),.$$




        That is, $$left(frac{10}{31}right)=-left(frac{1}{3}right)left(frac{3}{7}right)=-(+1)Biggl(-left(frac{7}{3}right)Biggr)=left(frac{1}{3}right)=+1,.$$ This can be verified by noting that $6^2equiv 5pmod{31}$ and $8^2equiv 2pmod{31}$, so $$17^2equiv (6cdot 8)^2equiv 5cdot2=10pmod{31},.$$





        For the second part, note that
        $$left(frac{-15}{43}right)=left(frac{-1}{43}right)left(frac{3}{43}right)left(frac{5}{43}right)=(-1)Biggl(-left(frac{43}{3}right)Biggr)left(frac{43}{5}right),.$$




        Therefore, $$left(frac{-15}{43}right)=left(frac{1}{3}right)left(frac{3}{5}right)=left(frac{3}{5}right),.$$ It is easy to verify that $left(dfrac{3}{5}right)=-1$, whence $$left(frac{-15}{43}right)=-1,.$$ You can check that $12^2equiv 15pmod{43}$, so $left(dfrac{15}{43}right)=+1$, whereas $left(dfrac{-1}{43}right)=-1$, confirming the calculations.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 22:53









        BatominovskiBatominovski

        33k33293




        33k33293























            1












            $begingroup$

            As remarked in comments, to use quadratic reciprocity we need to work with Legendre Symbols
            $$ left(frac{p}{q}right)$$
            for $p,q$ prime.
            You should repeatedly use the property
            $$ left(frac{ab}{p}right)=left(frac{a}{p}right)left(frac{b}{p}right)$$
            to make sure that you are calculating with both parts of the symbol being prime. That is, write
            $$ left(frac{10}{31}right)=left(frac{2}{31}right)left(frac{5}{31}right)$$
            then iteratively apply quadratic reciprocity as you intended to.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR.
              $endgroup$
              – Oscar Lanzi
              Dec 6 '18 at 23:23










            • $begingroup$
              If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what
              $endgroup$
              – Hidaw
              Dec 7 '18 at 20:28
















            1












            $begingroup$

            As remarked in comments, to use quadratic reciprocity we need to work with Legendre Symbols
            $$ left(frac{p}{q}right)$$
            for $p,q$ prime.
            You should repeatedly use the property
            $$ left(frac{ab}{p}right)=left(frac{a}{p}right)left(frac{b}{p}right)$$
            to make sure that you are calculating with both parts of the symbol being prime. That is, write
            $$ left(frac{10}{31}right)=left(frac{2}{31}right)left(frac{5}{31}right)$$
            then iteratively apply quadratic reciprocity as you intended to.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR.
              $endgroup$
              – Oscar Lanzi
              Dec 6 '18 at 23:23










            • $begingroup$
              If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what
              $endgroup$
              – Hidaw
              Dec 7 '18 at 20:28














            1












            1








            1





            $begingroup$

            As remarked in comments, to use quadratic reciprocity we need to work with Legendre Symbols
            $$ left(frac{p}{q}right)$$
            for $p,q$ prime.
            You should repeatedly use the property
            $$ left(frac{ab}{p}right)=left(frac{a}{p}right)left(frac{b}{p}right)$$
            to make sure that you are calculating with both parts of the symbol being prime. That is, write
            $$ left(frac{10}{31}right)=left(frac{2}{31}right)left(frac{5}{31}right)$$
            then iteratively apply quadratic reciprocity as you intended to.






            share|cite|improve this answer









            $endgroup$



            As remarked in comments, to use quadratic reciprocity we need to work with Legendre Symbols
            $$ left(frac{p}{q}right)$$
            for $p,q$ prime.
            You should repeatedly use the property
            $$ left(frac{ab}{p}right)=left(frac{a}{p}right)left(frac{b}{p}right)$$
            to make sure that you are calculating with both parts of the symbol being prime. That is, write
            $$ left(frac{10}{31}right)=left(frac{2}{31}right)left(frac{5}{31}right)$$
            then iteratively apply quadratic reciprocity as you intended to.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 22:53









            Antonios-Alexandros RobotisAntonios-Alexandros Robotis

            10.3k41641




            10.3k41641












            • $begingroup$
              But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR.
              $endgroup$
              – Oscar Lanzi
              Dec 6 '18 at 23:23










            • $begingroup$
              If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what
              $endgroup$
              – Hidaw
              Dec 7 '18 at 20:28


















            • $begingroup$
              But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR.
              $endgroup$
              – Oscar Lanzi
              Dec 6 '18 at 23:23










            • $begingroup$
              If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what
              $endgroup$
              – Hidaw
              Dec 7 '18 at 20:28
















            $begingroup$
            But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR.
            $endgroup$
            – Oscar Lanzi
            Dec 6 '18 at 23:23




            $begingroup$
            But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR.
            $endgroup$
            – Oscar Lanzi
            Dec 6 '18 at 23:23












            $begingroup$
            If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what
            $endgroup$
            – Hidaw
            Dec 7 '18 at 20:28




            $begingroup$
            If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what
            $endgroup$
            – Hidaw
            Dec 7 '18 at 20:28


















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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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