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












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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







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


















0












$begingroup$


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









$endgroup$












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
















0












0








0





$begingroup$


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









$endgroup$




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









KroTeKKroTeK

32




32












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




















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


















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






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












1 Answer
1






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oldest

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0












$begingroup$

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











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    1 Answer
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    active

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    1 Answer
    1






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    active

    oldest

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    active

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    0












    $begingroup$

    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











    $endgroup$


















      0












      $begingroup$

      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











      $endgroup$
















        0












        0








        0





        $begingroup$

        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











        $endgroup$



        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









        MindlackMindlack

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