# $10$ distinct integers with sum of any $9$ a perfect square

Do there exist $10$ distinct integers such that the sum of any $9$ of them is a perfect square?

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Can you find three distinct numbers so that the sum of any pair is a square? – Will Jagy Jan 6 '13 at 20:18
It says distinct integers, not consecutive ones. – Clive Newstead Jan 6 '13 at 20:20
oh right...didn't read correctly, but they are still distinct? – fosho Jan 6 '13 at 20:23
Misread question... – fosho Jan 6 '13 at 20:37
Could you tell us where this question comes from? – 1015 Jan 6 '13 at 20:45

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 $m\in\mathbb Z$ we have $m^2\equiv 0, 1, 4\text{ or }7\pmod 9$. 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 $a_i=\frac19\sum_{j=1}^{10} b_j - b_i$.

Example: Let $b_i=i^2$ for $i=1, \ldots 7$. Then $-(b_1+\ldots+b_7)\equiv 4\pmod 9$. So we want to obtain $4\pmod 9$ 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 $a=\frac19\sum b_i=54$ 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: $$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558).$$

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This is how I did it (except I looked $\mod 3$ not $\mod 9$). Here's my list of numbers: $61,58,53,46,37,13,-2,-38,-59,-107$. – Michael Biro Jan 6 '13 at 21:13
A smaller positive example: $(2623, 2404, 2183, 1960, 1508, 1279, 1048, 815, 580, 104)$ where the squares are $109^2, 110^2, 111^2, 112^2, 114^2, 115^2, 116^2, 117^2, 118^2, 120^2$. – Robert Israel Jan 7 '13 at 2:16
@RobertIsrael OK, I did not want to claim to have found anything like a smallest positive example - which I assume yours is. – Hagen von Eitzen Jan 7 '13 at 15:13
Actually I'm not sure it's the smallest. – Robert Israel Jan 7 '13 at 15:18

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:

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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


===========================

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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 $\rank(I+A)=1$. 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.

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