Can a sum of $n$ consecutive perfect squares be written as a sum of $n-1$ different perfect squares?
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
asked Dec 11 '18 at 22:08
KroTeK
62
add a comment |
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
asked Dec 11 '18 at 22:08
KroTeK
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
add a comment |
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
asked Dec 11 '18 at 22:08
KroTeK
62
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
asked Dec 11 '18 at 22:08
KroTeK
62
asked Dec 11 '18 at 22:08
KroTeK
62
asked Dec 11 '18 at 22:08
KroTeK
62
asked Dec 11 '18 at 22:08
KroTeK
62
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 11 '18 at 22:38
Mindlack
1,70717
add a comment |
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1 Answer
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1 Answer
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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.
answered Dec 11 '18 at 22:38
Mindlack
1,70717
add a comment |
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.
answered Dec 11 '18 at 22:38
Mindlack
1,70717
add a comment |
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.
answered Dec 11 '18 at 22:38
Mindlack
1,70717
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.
answered Dec 11 '18 at 22:38
Mindlack
1,70717
edited Dec 11 '18 at 22:50
answered Dec 11 '18 at 22:38
Mindlack
1,70717
answered Dec 11 '18 at 22:38
Mindlack
1,70717
answered Dec 11 '18 at 22:38
Mindlack
1,70717
1,70717
add a comment |
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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