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












2














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




















  • 3




    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.
    – Batominovski
    Dec 6 at 22:35










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










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






  • 2




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
















2














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




















  • 3




    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.
    – Batominovski
    Dec 6 at 22:35










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










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






  • 2




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














2












2








2







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















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 at 7:56









Batominovski

33.7k33292




33.7k33292










asked Dec 6 at 22:32









Hidaw

500624




500624








  • 3




    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.
    – Batominovski
    Dec 6 at 22:35










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










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






  • 2




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














  • 3




    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.
    – Batominovski
    Dec 6 at 22:35










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










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






  • 2




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








3




3




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.
– Batominovski
Dec 6 at 22:35




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.
– Batominovski
Dec 6 at 22:35












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




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












And please use a more descriptive title the next time.
– Batominovski
Dec 6 at 22:37




And please use a more descriptive title the next time.
– Batominovski
Dec 6 at 22:37




2




2




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




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










2 Answers
2






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oldest

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2














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





























    1














    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





















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










    • 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
      – Hidaw
      Dec 7 at 20:28











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    2 Answers
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    2 Answers
    2






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    oldest

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    active

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    active

    oldest

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    2














    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


























      2














      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
























        2












        2








        2






        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












        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 at 22:53









        Batominovski

        33.7k33292




        33.7k33292























            1














            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





















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










            • 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
              – Hidaw
              Dec 7 at 20:28
















            1














            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





















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










            • 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
              – Hidaw
              Dec 7 at 20:28














            1












            1








            1






            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












            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 at 22:53









            Antonios-Alexandros Robotis

            9,12541640




            9,12541640












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










            • 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
              – Hidaw
              Dec 7 at 20:28


















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










            • 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
              – Hidaw
              Dec 7 at 20:28
















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




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












            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
            – Hidaw
            Dec 7 at 20:28




            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
            – Hidaw
            Dec 7 at 20:28


















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            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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