Mixing
$10$ distinct integers with sum of any $9$ a perfect square
$begingroup$
Do there exist $10$ distinct integers such that the sum of any $9$ of them is a perfect square?
number-theory
asked Jan 6 '13 at 20:10
foshofosho
74121122
$endgroup$
add a comment |
$begingroup$
Do there exist $10$ distinct integers such that the sum of any $9$ of them is a perfect square?
number-theory
asked Jan 6 '13 at 20:10
foshofosho
74121122
$endgroup$
1
$begingroup$
Can you find three distinct numbers so that the sum of any pair is a square?
$endgroup$
– Will Jagy
Jan 6 '13 at 20:18
$begingroup$
It says distinct integers, not consecutive ones.
$endgroup$
– Clive Newstead
Jan 6 '13 at 20:20
$begingroup$
oh right...didn't read correctly, but they are still distinct?
$endgroup$
– fosho
Jan 6 '13 at 20:23
$begingroup$
Misread question...
$endgroup$
– fosho
Jan 6 '13 at 20:37
$begingroup$
Could you tell us where this question comes from?
$endgroup$
– Julien
Jan 6 '13 at 20:45
add a comment |
$begingroup$
Do there exist $10$ distinct integers such that the sum of any $9$ of them is a perfect square?
number-theory
asked Jan 6 '13 at 20:10
foshofosho
74121122
$endgroup$
Do there exist $10$ distinct integers such that the sum of any $9$ of them is a perfect square?
number-theory
number-theory
asked Jan 6 '13 at 20:10
foshofosho
74121122
asked Jan 6 '13 at 20:10
foshofosho
74121122
edited Jan 6 '13 at 20:43
user4594
asked Jan 6 '13 at 20:10
foshofosho
74121122
asked Jan 6 '13 at 20:10
foshofosho
74121122
asked Jan 6 '13 at 20:10
foshofosho
74121122
74121122
1
$begingroup$
Can you find three distinct numbers so that the sum of any pair is a square?
$endgroup$
– Will Jagy
Jan 6 '13 at 20:18
$begingroup$
It says distinct integers, not consecutive ones.
$endgroup$
– Clive Newstead
Jan 6 '13 at 20:20
$begingroup$
oh right...didn't read correctly, but they are still distinct?
$endgroup$
– fosho
Jan 6 '13 at 20:23
$begingroup$
Misread question...
$endgroup$
– fosho
Jan 6 '13 at 20:37
$begingroup$
Could you tell us where this question comes from?
$endgroup$
– Julien
Jan 6 '13 at 20:45
add a comment |
1
$begingroup$
Can you find three distinct numbers so that the sum of any pair is a square?
$endgroup$
– Will Jagy
Jan 6 '13 at 20:18
$begingroup$
It says distinct integers, not consecutive ones.
$endgroup$
– Clive Newstead
Jan 6 '13 at 20:20
$begingroup$
oh right...didn't read correctly, but they are still distinct?
$endgroup$
– fosho
Jan 6 '13 at 20:23
$begingroup$
Misread question...
$endgroup$
– fosho
Jan 6 '13 at 20:37
$begingroup$
Could you tell us where this question comes from?
$endgroup$
– Julien
Jan 6 '13 at 20:45
1
1
$begingroup$
Can you find three distinct numbers so that the sum of any pair is a square?
$endgroup$
– Will Jagy
Jan 6 '13 at 20:18
$begingroup$
Can you find three distinct numbers so that the sum of any pair is a square?
$endgroup$
– Will Jagy
Jan 6 '13 at 20:18
$begingroup$
It says distinct integers, not consecutive ones.
$endgroup$
– Clive Newstead
Jan 6 '13 at 20:20
$begingroup$
It says distinct integers, not consecutive ones.
$endgroup$
– Clive Newstead
Jan 6 '13 at 20:20
$begingroup$
oh right...didn't read correctly, but they are still distinct?
$endgroup$
– fosho
Jan 6 '13 at 20:23
$begingroup$
oh right...didn't read correctly, but they are still distinct?
$endgroup$
– fosho
Jan 6 '13 at 20:23
$begingroup$
Misread question...
$endgroup$
– fosho
Jan 6 '13 at 20:37
$begingroup$
Misread question...
$endgroup$
– fosho
Jan 6 '13 at 20:37
$begingroup$
Could you tell us where this question comes from?
$endgroup$
– Julien
Jan 6 '13 at 20:45
$begingroup$
Could you tell us where this question comes from?
$endgroup$
– Julien
Jan 6 '13 at 20:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Assume we have such integers $a_1, ldots, a_{10}$.
Let $a=sum a_i$.
Then we need that the numbers $b_i:=a-a_i$ are squares.
In other words, we need $10$ distinct squares $b_i$ such that their sum equals $$sum_{i=1}^{10} b_i=sum_{i=1}^{10} (a-a_i)=9a.$$
This is not more than requiring the sum of ten squares to be a multiple of $9$.
Note that for $minmathbb Z$ we have $m^2equiv 0, 1, 4text{ or }7pmod 9$.
The sum of three numbers $in{0,1,4,7}$ can be any residue class mod $9$. Therefore:
Select $7$ arbitrary distinct squares $b_1, ldots, b_7$.
Then select three further distinct squares $b_8,b_9,b_{10}$ such that
$b_8+b_9+b_{10}equiv -(b_1+ldots+b_7)pmod 9$.
Finally let $a_i=frac19sum_{j=1}^{10} b_j - b_i$.
Example: Let $b_i=i^2$ for $i=1, ldots 7$.
Then $-(b_1+ldots+b_7)equiv 4pmod 9$.
So we want to obtain $4pmod 9$ as sum of three numbers $in{0,1,4,7}$. This is possible (only) as $0+0+4$. So we may take $b_8=9^2$, $b_9=12^2$, $b_{10}=11^2$.
Then $a=frac19sum b_i=54$ and we arrive at
$$(a_1, ldots,a_{10})=(53,50,45,38,29,18,5,-27,-90,-67). $$
In case you don't like the appearence of negative numbers - they occur here only because the biggest square exceeds $frac{10}9$ of the average square. If one starts with bigger numbers, this need not be the case.
Here's a strictly very positive example:
$$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558). $$
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
This is how I did it (except I looked $mod 3$ not $mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$.
$endgroup$
– Michael Biro
Jan 6 '13 at 21:13
$begingroup$
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$.
$endgroup$
– Robert Israel
Jan 7 '13 at 2:16
$begingroup$
@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is.
$endgroup$
– Hagen von Eitzen
Jan 7 '13 at 15:13
$begingroup$
Actually I'm not sure it's the smallest.
$endgroup$
– Robert Israel
Jan 7 '13 at 15:18
add a comment |
$begingroup$
EEEEDDDDDDIIIIITTTTTTT: this is not an answer to the question as it stands. My hope was to convince the OP to put in some effort on the simplest cases of the question before jumping to ten numbers. This did not actually work, of course.
Here are some ways to do this for three distinct numbers, adding any pair (in the same row) gives a square:
=============================
30 19 6
44 20 5
47 34 2
48 33 16
60 21 4
66 34 15
69 52 12
70 51 30
78 22 3
86 35 14
90 54 10
92 52 29
94 75 6
95 74 26
96 73 48
98 23 2
===========================
Here are some ways to do this for four numbers, add any three in the same row and you get a square:
===========================
58 41 22 1
78 57 34 9
89 66 41 14
103 59 34 7
113 86 57 26
116 68 41 12
124 97 68 4
126 97 66 33
130 61 34 5
136 88 32 1
144 88 57 24
145 70 41 10
151 99 39 6
152 121 88 16
154 121 86 49
157 130 37 2
159 63 34 3
159 99 66 31
167 134 99 23
169 134 97 58
176 72 41 8
177 90 57 22
183 123 55 18
189 158 53 14
190 65 34 1
191 123 86 47
193 160 88 8
194 101 66 29
197 162 125 2
199 162 123 39
200 136 64 25
===========================
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
I think the answer is yes. Here is a simple idea:
Consider the system of equations
$$S-x_i= y_i^2, 1 leq i leq 10,,$$
where $S=x_1+..+x_n$.
Let $A$ be the coefficients matrix of this system. Then all the entries of $I+A$
are $1$, thus $operatorname{rank}(I+A)=1$. This shows that $lambda=0$ is an eigenvalue of $I+A$ of multiplicity $n-1$, and hence the remaining eigenvalue is $lambda=tr(I+A)=n.$
Hence the eigenvalues of $A$ are $lambda_1=...=lambda_{n-1}=-1$ and $lambda_n=(n-1)$. This shows that $det(A)=(-1)^{n-1}(n-1)$.
Now pick distinct $y_1,..,y_n$ positive integers, each divisible by $n-1$. Then, by Cramer's rule, all the solutions to the system
$$S-x_i= y_i^2 1 leq i leq 10,,$$
are integers (since when you calculate the determinant of $A_i$, you can pull an $(n-1)^2$ from the i-th column, and you are left with a matrix with integer entries).
The only thing left to do is proving that $x_i$ are pairwise distinct. Let $i neq j$. Then
$$S-x_i =y_i^2 ,;, S-x_j=y_j^2 Rightarrow x_i-x_j=y_j^2-y_i^2 neq 0 ,.$$
Remark You can easily prove that $det(A)=(-1)^{n-1}(n-1)$ by row reduction: Add all the other rows to the last one, get an $(n-1)$ common factor from that one, and the n subtract the last row from each of the remaining ones.
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Assume we have such integers $a_1, ldots, a_{10}$.
Let $a=sum a_i$.
Then we need that the numbers $b_i:=a-a_i$ are squares.
In other words, we need $10$ distinct squares $b_i$ such that their sum equals $$sum_{i=1}^{10} b_i=sum_{i=1}^{10} (a-a_i)=9a.$$
This is not more than requiring the sum of ten squares to be a multiple of $9$.
Note that for $minmathbb Z$ we have $m^2equiv 0, 1, 4text{ or }7pmod 9$.
The sum of three numbers $in{0,1,4,7}$ can be any residue class mod $9$. Therefore:
Select $7$ arbitrary distinct squares $b_1, ldots, b_7$.
Then select three further distinct squares $b_8,b_9,b_{10}$ such that
$b_8+b_9+b_{10}equiv -(b_1+ldots+b_7)pmod 9$.
Finally let $a_i=frac19sum_{j=1}^{10} b_j - b_i$.
Example: Let $b_i=i^2$ for $i=1, ldots 7$.
Then $-(b_1+ldots+b_7)equiv 4pmod 9$.
So we want to obtain $4pmod 9$ as sum of three numbers $in{0,1,4,7}$. This is possible (only) as $0+0+4$. So we may take $b_8=9^2$, $b_9=12^2$, $b_{10}=11^2$.
Then $a=frac19sum b_i=54$ and we arrive at
$$(a_1, ldots,a_{10})=(53,50,45,38,29,18,5,-27,-90,-67). $$
In case you don't like the appearence of negative numbers - they occur here only because the biggest square exceeds $frac{10}9$ of the average square. If one starts with bigger numbers, this need not be the case.
Here's a strictly very positive example:
$$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558). $$
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
This is how I did it (except I looked $mod 3$ not $mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$.
$endgroup$
– Michael Biro
Jan 6 '13 at 21:13
$begingroup$
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$.
$endgroup$
– Robert Israel
Jan 7 '13 at 2:16
$begingroup$
@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is.
$endgroup$
– Hagen von Eitzen
Jan 7 '13 at 15:13
$begingroup$
Actually I'm not sure it's the smallest.
$endgroup$
– Robert Israel
Jan 7 '13 at 15:18
add a comment |
$begingroup$
Assume we have such integers $a_1, ldots, a_{10}$.
Let $a=sum a_i$.
Then we need that the numbers $b_i:=a-a_i$ are squares.
In other words, we need $10$ distinct squares $b_i$ such that their sum equals $$sum_{i=1}^{10} b_i=sum_{i=1}^{10} (a-a_i)=9a.$$
This is not more than requiring the sum of ten squares to be a multiple of $9$.
Note that for $minmathbb Z$ we have $m^2equiv 0, 1, 4text{ or }7pmod 9$.
The sum of three numbers $in{0,1,4,7}$ can be any residue class mod $9$. Therefore:
Select $7$ arbitrary distinct squares $b_1, ldots, b_7$.
Then select three further distinct squares $b_8,b_9,b_{10}$ such that
$b_8+b_9+b_{10}equiv -(b_1+ldots+b_7)pmod 9$.
Finally let $a_i=frac19sum_{j=1}^{10} b_j - b_i$.
Example: Let $b_i=i^2$ for $i=1, ldots 7$.
Then $-(b_1+ldots+b_7)equiv 4pmod 9$.
So we want to obtain $4pmod 9$ as sum of three numbers $in{0,1,4,7}$. This is possible (only) as $0+0+4$. So we may take $b_8=9^2$, $b_9=12^2$, $b_{10}=11^2$.
Then $a=frac19sum b_i=54$ and we arrive at
$$(a_1, ldots,a_{10})=(53,50,45,38,29,18,5,-27,-90,-67). $$
In case you don't like the appearence of negative numbers - they occur here only because the biggest square exceeds $frac{10}9$ of the average square. If one starts with bigger numbers, this need not be the case.
Here's a strictly very positive example:
$$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558). $$
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
This is how I did it (except I looked $mod 3$ not $mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$.
$endgroup$
– Michael Biro
Jan 6 '13 at 21:13
$begingroup$
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$.
$endgroup$
– Robert Israel
Jan 7 '13 at 2:16
$begingroup$
@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is.
$endgroup$
– Hagen von Eitzen
Jan 7 '13 at 15:13
$begingroup$
Actually I'm not sure it's the smallest.
$endgroup$
– Robert Israel
Jan 7 '13 at 15:18
add a comment |
$begingroup$
Assume we have such integers $a_1, ldots, a_{10}$.
Let $a=sum a_i$.
Then we need that the numbers $b_i:=a-a_i$ are squares.
In other words, we need $10$ distinct squares $b_i$ such that their sum equals $$sum_{i=1}^{10} b_i=sum_{i=1}^{10} (a-a_i)=9a.$$
This is not more than requiring the sum of ten squares to be a multiple of $9$.
Note that for $minmathbb Z$ we have $m^2equiv 0, 1, 4text{ or }7pmod 9$.
The sum of three numbers $in{0,1,4,7}$ can be any residue class mod $9$. Therefore:
Select $7$ arbitrary distinct squares $b_1, ldots, b_7$.
Then select three further distinct squares $b_8,b_9,b_{10}$ such that
$b_8+b_9+b_{10}equiv -(b_1+ldots+b_7)pmod 9$.
Finally let $a_i=frac19sum_{j=1}^{10} b_j - b_i$.
Example: Let $b_i=i^2$ for $i=1, ldots 7$.
Then $-(b_1+ldots+b_7)equiv 4pmod 9$.
So we want to obtain $4pmod 9$ as sum of three numbers $in{0,1,4,7}$. This is possible (only) as $0+0+4$. So we may take $b_8=9^2$, $b_9=12^2$, $b_{10}=11^2$.
Then $a=frac19sum b_i=54$ and we arrive at
$$(a_1, ldots,a_{10})=(53,50,45,38,29,18,5,-27,-90,-67). $$
In case you don't like the appearence of negative numbers - they occur here only because the biggest square exceeds $frac{10}9$ of the average square. If one starts with bigger numbers, this need not be the case.
Here's a strictly very positive example:
$$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558). $$
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
Assume we have such integers $a_1, ldots, a_{10}$.
Let $a=sum a_i$.
Then we need that the numbers $b_i:=a-a_i$ are squares.
In other words, we need $10$ distinct squares $b_i$ such that their sum equals $$sum_{i=1}^{10} b_i=sum_{i=1}^{10} (a-a_i)=9a.$$
This is not more than requiring the sum of ten squares to be a multiple of $9$.
Note that for $minmathbb Z$ we have $m^2equiv 0, 1, 4text{ or }7pmod 9$.
The sum of three numbers $in{0,1,4,7}$ can be any residue class mod $9$. Therefore:
Select $7$ arbitrary distinct squares $b_1, ldots, b_7$.
Then select three further distinct squares $b_8,b_9,b_{10}$ such that
$b_8+b_9+b_{10}equiv -(b_1+ldots+b_7)pmod 9$.
Finally let $a_i=frac19sum_{j=1}^{10} b_j - b_i$.
Example: Let $b_i=i^2$ for $i=1, ldots 7$.
Then $-(b_1+ldots+b_7)equiv 4pmod 9$.
So we want to obtain $4pmod 9$ as sum of three numbers $in{0,1,4,7}$. This is possible (only) as $0+0+4$. So we may take $b_8=9^2$, $b_9=12^2$, $b_{10}=11^2$.
Then $a=frac19sum b_i=54$ and we arrive at
$$(a_1, ldots,a_{10})=(53,50,45,38,29,18,5,-27,-90,-67). $$
In case you don't like the appearence of negative numbers - they occur here only because the biggest square exceeds $frac{10}9$ of the average square. If one starts with bigger numbers, this need not be the case.
Here's a strictly very positive example:
$$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558). $$
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
edited Jan 6 '13 at 21:29
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 6 '13 at 21:10
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
$begingroup$
This is how I did it (except I looked $mod 3$ not $mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$.
$endgroup$
– Michael Biro
Jan 6 '13 at 21:13
$begingroup$
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$.
$endgroup$
– Robert Israel
Jan 7 '13 at 2:16
$begingroup$
@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is.
$endgroup$
– Hagen von Eitzen
Jan 7 '13 at 15:13
$begingroup$
Actually I'm not sure it's the smallest.
$endgroup$
– Robert Israel
Jan 7 '13 at 15:18
add a comment |
$begingroup$
This is how I did it (except I looked $mod 3$ not $mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$.
$endgroup$
– Michael Biro
Jan 6 '13 at 21:13
$begingroup$
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$.
$endgroup$
– Robert Israel
Jan 7 '13 at 2:16
$begingroup$
@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is.
$endgroup$
– Hagen von Eitzen
Jan 7 '13 at 15:13
$begingroup$
Actually I'm not sure it's the smallest.
$endgroup$
– Robert Israel
Jan 7 '13 at 15:18
$begingroup$
This is how I did it (except I looked $mod 3$ not $mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$.
$endgroup$
– Michael Biro
Jan 6 '13 at 21:13
$begingroup$
This is how I did it (except I looked $mod 3$ not $mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$.
$endgroup$
– Michael Biro
Jan 6 '13 at 21:13
$begingroup$
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$.
$endgroup$
– Robert Israel
Jan 7 '13 at 2:16
$begingroup$
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$.
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– Robert Israel
Jan 7 '13 at 2:16
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@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is.
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– Hagen von Eitzen
Jan 7 '13 at 15:13
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@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is.
$endgroup$
– Hagen von Eitzen
Jan 7 '13 at 15:13
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Actually I'm not sure it's the smallest.
$endgroup$
– Robert Israel
Jan 7 '13 at 15:18
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Actually I'm not sure it's the smallest.
$endgroup$
– Robert Israel
Jan 7 '13 at 15:18
add a comment |
$begingroup$
EEEEDDDDDDIIIIITTTTTTT: this is not an answer to the question as it stands. My hope was to convince the OP to put in some effort on the simplest cases of the question before jumping to ten numbers. This did not actually work, of course.
Here are some ways to do this for three distinct numbers, adding any pair (in the same row) gives a square:
=============================
30 19 6
44 20 5
47 34 2
48 33 16
60 21 4
66 34 15
69 52 12
70 51 30
78 22 3
86 35 14
90 54 10
92 52 29
94 75 6
95 74 26
96 73 48
98 23 2
===========================
Here are some ways to do this for four numbers, add any three in the same row and you get a square:
===========================
58 41 22 1
78 57 34 9
89 66 41 14
103 59 34 7
113 86 57 26
116 68 41 12
124 97 68 4
126 97 66 33
130 61 34 5
136 88 32 1
144 88 57 24
145 70 41 10
151 99 39 6
152 121 88 16
154 121 86 49
157 130 37 2
159 63 34 3
159 99 66 31
167 134 99 23
169 134 97 58
176 72 41 8
177 90 57 22
183 123 55 18
189 158 53 14
190 65 34 1
191 123 86 47
193 160 88 8
194 101 66 29
197 162 125 2
199 162 123 39
200 136 64 25
===========================
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
EEEEDDDDDDIIIIITTTTTTT: this is not an answer to the question as it stands. My hope was to convince the OP to put in some effort on the simplest cases of the question before jumping to ten numbers. This did not actually work, of course.
Here are some ways to do this for three distinct numbers, adding any pair (in the same row) gives a square:
=============================
30 19 6
44 20 5
47 34 2
48 33 16
60 21 4
66 34 15
69 52 12
70 51 30
78 22 3
86 35 14
90 54 10
92 52 29
94 75 6
95 74 26
96 73 48
98 23 2
===========================
Here are some ways to do this for four numbers, add any three in the same row and you get a square:
===========================
58 41 22 1
78 57 34 9
89 66 41 14
103 59 34 7
113 86 57 26
116 68 41 12
124 97 68 4
126 97 66 33
130 61 34 5
136 88 32 1
144 88 57 24
145 70 41 10
151 99 39 6
152 121 88 16
154 121 86 49
157 130 37 2
159 63 34 3
159 99 66 31
167 134 99 23
169 134 97 58
176 72 41 8
177 90 57 22
183 123 55 18
189 158 53 14
190 65 34 1
191 123 86 47
193 160 88 8
194 101 66 29
197 162 125 2
199 162 123 39
200 136 64 25
===========================
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
EEEEDDDDDDIIIIITTTTTTT: this is not an answer to the question as it stands. My hope was to convince the OP to put in some effort on the simplest cases of the question before jumping to ten numbers. This did not actually work, of course.
Here are some ways to do this for three distinct numbers, adding any pair (in the same row) gives a square:
=============================
30 19 6
44 20 5
47 34 2
48 33 16
60 21 4
66 34 15
69 52 12
70 51 30
78 22 3
86 35 14
90 54 10
92 52 29
94 75 6
95 74 26
96 73 48
98 23 2
===========================
Here are some ways to do this for four numbers, add any three in the same row and you get a square:
===========================
58 41 22 1
78 57 34 9
89 66 41 14
103 59 34 7
113 86 57 26
116 68 41 12
124 97 68 4
126 97 66 33
130 61 34 5
136 88 32 1
144 88 57 24
145 70 41 10
151 99 39 6
152 121 88 16
154 121 86 49
157 130 37 2
159 63 34 3
159 99 66 31
167 134 99 23
169 134 97 58
176 72 41 8
177 90 57 22
183 123 55 18
189 158 53 14
190 65 34 1
191 123 86 47
193 160 88 8
194 101 66 29
197 162 125 2
199 162 123 39
200 136 64 25
===========================
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
EEEEDDDDDDIIIIITTTTTTT: this is not an answer to the question as it stands. My hope was to convince the OP to put in some effort on the simplest cases of the question before jumping to ten numbers. This did not actually work, of course.
Here are some ways to do this for three distinct numbers, adding any pair (in the same row) gives a square:
=============================
30 19 6
44 20 5
47 34 2
48 33 16
60 21 4
66 34 15
69 52 12
70 51 30
78 22 3
86 35 14
90 54 10
92 52 29
94 75 6
95 74 26
96 73 48
98 23 2
===========================
Here are some ways to do this for four numbers, add any three in the same row and you get a square:
===========================
58 41 22 1
78 57 34 9
89 66 41 14
103 59 34 7
113 86 57 26
116 68 41 12
124 97 68 4
126 97 66 33
130 61 34 5
136 88 32 1
144 88 57 24
145 70 41 10
151 99 39 6
152 121 88 16
154 121 86 49
157 130 37 2
159 63 34 3
159 99 66 31
167 134 99 23
169 134 97 58
176 72 41 8
177 90 57 22
183 123 55 18
189 158 53 14
190 65 34 1
191 123 86 47
193 160 88 8
194 101 66 29
197 162 125 2
199 162 123 39
200 136 64 25
===========================
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
edited Jan 6 '13 at 22:22
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
answered Jan 6 '13 at 20:47
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
$begingroup$
I think the answer is yes. Here is a simple idea:
Consider the system of equations
$$S-x_i= y_i^2, 1 leq i leq 10,,$$
where $S=x_1+..+x_n$.
Let $A$ be the coefficients matrix of this system. Then all the entries of $I+A$
are $1$, thus $operatorname{rank}(I+A)=1$. This shows that $lambda=0$ is an eigenvalue of $I+A$ of multiplicity $n-1$, and hence the remaining eigenvalue is $lambda=tr(I+A)=n.$
Hence the eigenvalues of $A$ are $lambda_1=...=lambda_{n-1}=-1$ and $lambda_n=(n-1)$. This shows that $det(A)=(-1)^{n-1}(n-1)$.
Now pick distinct $y_1,..,y_n$ positive integers, each divisible by $n-1$. Then, by Cramer's rule, all the solutions to the system
$$S-x_i= y_i^2 1 leq i leq 10,,$$
are integers (since when you calculate the determinant of $A_i$, you can pull an $(n-1)^2$ from the i-th column, and you are left with a matrix with integer entries).
The only thing left to do is proving that $x_i$ are pairwise distinct. Let $i neq j$. Then
$$S-x_i =y_i^2 ,;, S-x_j=y_j^2 Rightarrow x_i-x_j=y_j^2-y_i^2 neq 0 ,.$$
Remark You can easily prove that $det(A)=(-1)^{n-1}(n-1)$ by row reduction: Add all the other rows to the last one, get an $(n-1)$ common factor from that one, and the n subtract the last row from each of the remaining ones.
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
$endgroup$
add a comment |
$begingroup$
I think the answer is yes. Here is a simple idea:
Consider the system of equations
$$S-x_i= y_i^2, 1 leq i leq 10,,$$
where $S=x_1+..+x_n$.
Let $A$ be the coefficients matrix of this system. Then all the entries of $I+A$
are $1$, thus $operatorname{rank}(I+A)=1$. This shows that $lambda=0$ is an eigenvalue of $I+A$ of multiplicity $n-1$, and hence the remaining eigenvalue is $lambda=tr(I+A)=n.$
Hence the eigenvalues of $A$ are $lambda_1=...=lambda_{n-1}=-1$ and $lambda_n=(n-1)$. This shows that $det(A)=(-1)^{n-1}(n-1)$.
Now pick distinct $y_1,..,y_n$ positive integers, each divisible by $n-1$. Then, by Cramer's rule, all the solutions to the system
$$S-x_i= y_i^2 1 leq i leq 10,,$$
are integers (since when you calculate the determinant of $A_i$, you can pull an $(n-1)^2$ from the i-th column, and you are left with a matrix with integer entries).
The only thing left to do is proving that $x_i$ are pairwise distinct. Let $i neq j$. Then
$$S-x_i =y_i^2 ,;, S-x_j=y_j^2 Rightarrow x_i-x_j=y_j^2-y_i^2 neq 0 ,.$$
Remark You can easily prove that $det(A)=(-1)^{n-1}(n-1)$ by row reduction: Add all the other rows to the last one, get an $(n-1)$ common factor from that one, and the n subtract the last row from each of the remaining ones.
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
$endgroup$
add a comment |
$begingroup$
I think the answer is yes. Here is a simple idea:
Consider the system of equations
$$S-x_i= y_i^2, 1 leq i leq 10,,$$
where $S=x_1+..+x_n$.
Let $A$ be the coefficients matrix of this system. Then all the entries of $I+A$
are $1$, thus $operatorname{rank}(I+A)=1$. This shows that $lambda=0$ is an eigenvalue of $I+A$ of multiplicity $n-1$, and hence the remaining eigenvalue is $lambda=tr(I+A)=n.$
Hence the eigenvalues of $A$ are $lambda_1=...=lambda_{n-1}=-1$ and $lambda_n=(n-1)$. This shows that $det(A)=(-1)^{n-1}(n-1)$.
Now pick distinct $y_1,..,y_n$ positive integers, each divisible by $n-1$. Then, by Cramer's rule, all the solutions to the system
$$S-x_i= y_i^2 1 leq i leq 10,,$$
are integers (since when you calculate the determinant of $A_i$, you can pull an $(n-1)^2$ from the i-th column, and you are left with a matrix with integer entries).
The only thing left to do is proving that $x_i$ are pairwise distinct. Let $i neq j$. Then
$$S-x_i =y_i^2 ,;, S-x_j=y_j^2 Rightarrow x_i-x_j=y_j^2-y_i^2 neq 0 ,.$$
Remark You can easily prove that $det(A)=(-1)^{n-1}(n-1)$ by row reduction: Add all the other rows to the last one, get an $(n-1)$ common factor from that one, and the n subtract the last row from each of the remaining ones.
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
$endgroup$
I think the answer is yes. Here is a simple idea:
Consider the system of equations
$$S-x_i= y_i^2, 1 leq i leq 10,,$$
where $S=x_1+..+x_n$.
Let $A$ be the coefficients matrix of this system. Then all the entries of $I+A$
are $1$, thus $operatorname{rank}(I+A)=1$. This shows that $lambda=0$ is an eigenvalue of $I+A$ of multiplicity $n-1$, and hence the remaining eigenvalue is $lambda=tr(I+A)=n.$
Hence the eigenvalues of $A$ are $lambda_1=...=lambda_{n-1}=-1$ and $lambda_n=(n-1)$. This shows that $det(A)=(-1)^{n-1}(n-1)$.
Now pick distinct $y_1,..,y_n$ positive integers, each divisible by $n-1$. Then, by Cramer's rule, all the solutions to the system
$$S-x_i= y_i^2 1 leq i leq 10,,$$
are integers (since when you calculate the determinant of $A_i$, you can pull an $(n-1)^2$ from the i-th column, and you are left with a matrix with integer entries).
The only thing left to do is proving that $x_i$ are pairwise distinct. Let $i neq j$. Then
$$S-x_i =y_i^2 ,;, S-x_j=y_j^2 Rightarrow x_i-x_j=y_j^2-y_i^2 neq 0 ,.$$
Remark You can easily prove that $det(A)=(-1)^{n-1}(n-1)$ by row reduction: Add all the other rows to the last one, get an $(n-1)$ common factor from that one, and the n subtract the last row from each of the remaining ones.
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
edited Jan 30 at 19:30
Martin Sleziak
44.9k10122277
44.9k10122277
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
answered Jan 6 '13 at 21:02
N. S.N. S.
105k7114210
105k7114210
add a comment |
add a comment |
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$begingroup$
Can you find three distinct numbers so that the sum of any pair is a square?
$endgroup$
– Will Jagy
Jan 6 '13 at 20:18
$begingroup$
It says distinct integers, not consecutive ones.
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– Clive Newstead
Jan 6 '13 at 20:20
$begingroup$
oh right...didn't read correctly, but they are still distinct?
$endgroup$
– fosho
Jan 6 '13 at 20:23
$begingroup$
Misread question...
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– fosho
Jan 6 '13 at 20:37
$begingroup$
Could you tell us where this question comes from?
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– Julien
Jan 6 '13 at 20:45