$10$ 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?
 A: 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:
$$(113573, 111570, 109565, 107558, 105549, 103538, 101525, 117573, 121565, 123558). $$
A: 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.
