Can a sum of $n$ consecutive perfect squares be written as a sum of $n-1$ different perfect squares?












0














So, can $sum_{i=1}^n i^2$ be written as a sum of $n-1$ different perfect squares?
Surely if we are looking at this problem with small numbers, the answer is both yes and no.
If we take $n$ to be 3, there can not be two different perfect squares numbers which satisfy the statement.
If we take $n$ to be 4, we can see that $3^{2}+4^{2}=5^{2}$ which means that the statement is satisfied.



So, my truly question is: how can we find if a certain $n$ number is solution or not for this statement? (for example: Can $sum_{i=1}^{2015} i^2$ be written as a sum of $2014$ different perfect squares?)






square-numbers







share|cite|improve this question






















  • I was pointing out the main idea of that sum. $1^{2}+2^{2}+3^{2}+4^{2} = 1^{2}+2^{2}+5^{2}$
    – KroTeK
    Dec 11 '18 at 22:30


















0














So, can $sum_{i=1}^n i^2$ be written as a sum of $n-1$ different perfect squares?
Surely if we are looking at this problem with small numbers, the answer is both yes and no.
If we take $n$ to be 3, there can not be two different perfect squares numbers which satisfy the statement.
If we take $n$ to be 4, we can see that $3^{2}+4^{2}=5^{2}$ which means that the statement is satisfied.



So, my truly question is: how can we find if a certain $n$ number is solution or not for this statement? (for example: Can $sum_{i=1}^{2015} i^2$ be written as a sum of $2014$ different perfect squares?)






square-numbers







share|cite|improve this question






















  • I was pointing out the main idea of that sum. $1^{2}+2^{2}+3^{2}+4^{2} = 1^{2}+2^{2}+5^{2}$
    – KroTeK
    Dec 11 '18 at 22:30
















0












0








0







So, can $sum_{i=1}^n i^2$ be written as a sum of $n-1$ different perfect squares?
Surely if we are looking at this problem with small numbers, the answer is both yes and no.
If we take $n$ to be 3, there can not be two different perfect squares numbers which satisfy the statement.
If we take $n$ to be 4, we can see that $3^{2}+4^{2}=5^{2}$ which means that the statement is satisfied.



So, my truly question is: how can we find if a certain $n$ number is solution or not for this statement? (for example: Can $sum_{i=1}^{2015} i^2$ be written as a sum of $2014$ different perfect squares?)






square-numbers







share|cite|improve this question













So, can $sum_{i=1}^n i^2$ be written as a sum of $n-1$ different perfect squares?
Surely if we are looking at this problem with small numbers, the answer is both yes and no.
If we take $n$ to be 3, there can not be two different perfect squares numbers which satisfy the statement.
If we take $n$ to be 4, we can see that $3^{2}+4^{2}=5^{2}$ which means that the statement is satisfied.



So, my truly question is: how can we find if a certain $n$ number is solution or not for this statement? (for example: Can $sum_{i=1}^{2015} i^2$ be written as a sum of $2014$ different perfect squares?)






square-numbers




square-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 22:08









KroTeK

62




62












  • I was pointing out the main idea of that sum. $1^{2}+2^{2}+3^{2}+4^{2} = 1^{2}+2^{2}+5^{2}$
    – KroTeK
    Dec 11 '18 at 22:30




















  • I was pointing out the main idea of that sum. $1^{2}+2^{2}+3^{2}+4^{2} = 1^{2}+2^{2}+5^{2}$
    – KroTeK
    Dec 11 '18 at 22:30


















I was pointing out the main idea of that sum. $1^{2}+2^{2}+3^{2}+4^{2} = 1^{2}+2^{2}+5^{2}$
– KroTeK
Dec 11 '18 at 22:30






I was pointing out the main idea of that sum. $1^{2}+2^{2}+3^{2}+4^{2} = 1^{2}+2^{2}+5^{2}$
– KroTeK
Dec 11 '18 at 22:30












1 Answer
1






active

oldest

votes


















0














Let us replace 2015 by $N$.
What you found, is that: if there exists $1 leq u<v leq N$ such that $u^2+v^2$ is the square of some integer that is greater than $N$, then your statement holds.
Now, let $w=45^2+4^2=2041, u=2*45*4=360, v=45^2-4^2=2009.$
Then $w^2=u^2+v^2$.



To generalize, let $m$ be the smallest integer such that $m^2>N$. Let $k$ be the smallest integet such that $v=m^2-k^2 leq N$. Then, if $0<u=2*m*k<v$, $u^2+v^2=(m^2+k^2)^2$ and the statement holds.



Now, if $v=0$, then $N < m^2-(m-1)^2=2m-1 < 2sqrt{N}+1$, thus $N < 7$.



Thus, if $N > 6$, $k<m$ and $2mk+k^2 leq 3mk$.



Now, $k < 1+sqrt{m^2-N} leq 1+sqrt{2sqrt{N}+1} < 1+sqrt{2m+1} leq m/3$ if $m geq 24$ (corresponding to $N geq 23^2=529$) so $0<u<v$ holds.



As a conclusion, if $N$ is large enough (at least $529$) your result holds.






share|cite|improve this answer























    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%2f3035899%2fcan-a-sum-of-n-consecutive-perfect-squares-be-written-as-a-sum-of-n-1-differ%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Let us replace 2015 by $N$.
    What you found, is that: if there exists $1 leq u<v leq N$ such that $u^2+v^2$ is the square of some integer that is greater than $N$, then your statement holds.
    Now, let $w=45^2+4^2=2041, u=2*45*4=360, v=45^2-4^2=2009.$
    Then $w^2=u^2+v^2$.



    To generalize, let $m$ be the smallest integer such that $m^2>N$. Let $k$ be the smallest integet such that $v=m^2-k^2 leq N$. Then, if $0<u=2*m*k<v$, $u^2+v^2=(m^2+k^2)^2$ and the statement holds.



    Now, if $v=0$, then $N < m^2-(m-1)^2=2m-1 < 2sqrt{N}+1$, thus $N < 7$.



    Thus, if $N > 6$, $k<m$ and $2mk+k^2 leq 3mk$.



    Now, $k < 1+sqrt{m^2-N} leq 1+sqrt{2sqrt{N}+1} < 1+sqrt{2m+1} leq m/3$ if $m geq 24$ (corresponding to $N geq 23^2=529$) so $0<u<v$ holds.



    As a conclusion, if $N$ is large enough (at least $529$) your result holds.






    share|cite|improve this answer




























      0














      Let us replace 2015 by $N$.
      What you found, is that: if there exists $1 leq u<v leq N$ such that $u^2+v^2$ is the square of some integer that is greater than $N$, then your statement holds.
      Now, let $w=45^2+4^2=2041, u=2*45*4=360, v=45^2-4^2=2009.$
      Then $w^2=u^2+v^2$.



      To generalize, let $m$ be the smallest integer such that $m^2>N$. Let $k$ be the smallest integet such that $v=m^2-k^2 leq N$. Then, if $0<u=2*m*k<v$, $u^2+v^2=(m^2+k^2)^2$ and the statement holds.



      Now, if $v=0$, then $N < m^2-(m-1)^2=2m-1 < 2sqrt{N}+1$, thus $N < 7$.



      Thus, if $N > 6$, $k<m$ and $2mk+k^2 leq 3mk$.



      Now, $k < 1+sqrt{m^2-N} leq 1+sqrt{2sqrt{N}+1} < 1+sqrt{2m+1} leq m/3$ if $m geq 24$ (corresponding to $N geq 23^2=529$) so $0<u<v$ holds.



      As a conclusion, if $N$ is large enough (at least $529$) your result holds.






      share|cite|improve this answer


























        0












        0








        0






        Let us replace 2015 by $N$.
        What you found, is that: if there exists $1 leq u<v leq N$ such that $u^2+v^2$ is the square of some integer that is greater than $N$, then your statement holds.
        Now, let $w=45^2+4^2=2041, u=2*45*4=360, v=45^2-4^2=2009.$
        Then $w^2=u^2+v^2$.



        To generalize, let $m$ be the smallest integer such that $m^2>N$. Let $k$ be the smallest integet such that $v=m^2-k^2 leq N$. Then, if $0<u=2*m*k<v$, $u^2+v^2=(m^2+k^2)^2$ and the statement holds.



        Now, if $v=0$, then $N < m^2-(m-1)^2=2m-1 < 2sqrt{N}+1$, thus $N < 7$.



        Thus, if $N > 6$, $k<m$ and $2mk+k^2 leq 3mk$.



        Now, $k < 1+sqrt{m^2-N} leq 1+sqrt{2sqrt{N}+1} < 1+sqrt{2m+1} leq m/3$ if $m geq 24$ (corresponding to $N geq 23^2=529$) so $0<u<v$ holds.



        As a conclusion, if $N$ is large enough (at least $529$) your result holds.






        share|cite|improve this answer














        Let us replace 2015 by $N$.
        What you found, is that: if there exists $1 leq u<v leq N$ such that $u^2+v^2$ is the square of some integer that is greater than $N$, then your statement holds.
        Now, let $w=45^2+4^2=2041, u=2*45*4=360, v=45^2-4^2=2009.$
        Then $w^2=u^2+v^2$.



        To generalize, let $m$ be the smallest integer such that $m^2>N$. Let $k$ be the smallest integet such that $v=m^2-k^2 leq N$. Then, if $0<u=2*m*k<v$, $u^2+v^2=(m^2+k^2)^2$ and the statement holds.



        Now, if $v=0$, then $N < m^2-(m-1)^2=2m-1 < 2sqrt{N}+1$, thus $N < 7$.



        Thus, if $N > 6$, $k<m$ and $2mk+k^2 leq 3mk$.



        Now, $k < 1+sqrt{m^2-N} leq 1+sqrt{2sqrt{N}+1} < 1+sqrt{2m+1} leq m/3$ if $m geq 24$ (corresponding to $N geq 23^2=529$) so $0<u<v$ holds.



        As a conclusion, if $N$ is large enough (at least $529$) your result holds.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 22:50

























        answered Dec 11 '18 at 22:38









        Mindlack

        1,70717




        1,70717






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3035899%2fcan-a-sum-of-n-consecutive-perfect-squares-be-written-as-a-sum-of-n-1-differ%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

            Bressuire

            Image
            .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 ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .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 ...
            Read more