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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029147%2fwhat-are-the-legendre-symbols-left-frac1031-right-and-left-frac-15%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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

            10.2k41641




            10.2k41641












            • $begingroup$
              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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029147%2fwhat-are-the-legendre-symbols-left-frac1031-right-and-left-frac-15%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            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...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more