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

1




1










asked Dec 6 '18 at 22:32









HidawHidaw

505624




505624








  • 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

        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









            $endgroup$













            • $begingroup$
              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

            9,71741640




            9,71741640












            • $begingroup$
              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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            Berounka

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            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
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            Sphinx de Gizeh

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            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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