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Let $k$ and $n$ be two positive integers such that $k<n$ and $k$ does not divide $n$. Show that one can fill a $n \times n$ square grid with $n^2$ real numbers such that sum of numbers in an arbitrary $k \times k$ square grid is negative, but sum of all $n^2$ numbers is positive.

Source: Homework

My idea: I presented $n = mk + r$ and write $-k^2$ into each square with coordinate of the form $(ik, jk)$ where $i, j$ are positive integers and 1 in another squares. Then the sum of each $k \times k$ square is $-1$ and sum of $n^2$ numbers is $n^2 - m^2(k^2+1)$. But this doesn't work for large $m$.

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    $\begingroup$ Note that the fact that the entries have to be real does not add any extra possibilities over just taking integer values: If we have a real solution we can slightly perturb the solution to make all the values rational (because it is a finite grid and all the sums of interest are elements of the open sets $\Bbb R^+$, $\Bbb R^-$), and then clear denominators to make it integral. $\endgroup$ – Mario Carneiro Oct 20 '15 at 15:23
  • $\begingroup$ So, for a $3\times3$ grid, would $\begin{array}&1&0&1\\0&-2&0\\1&0&1\end{array}$ work? $\endgroup$ – Akiva Weinberger Oct 20 '15 at 16:09
  • $\begingroup$ @AkivaWeinberger Yes, it does work. $\endgroup$ – primitiveroot Oct 20 '15 at 16:14
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This can be done so that all rows are the same. Here's the idea of my solution: let the first column all have a value of $a > 0$, to be chosen later. Let the next $k - 1$ columns all have $-1$ in them. Then repeat this pattern: one column of $a$'s, $k-1$ column's of $-1$'s. Then every $k \times k$ square has exactly one column of $a$'s and the rest are $-1$'s. Thus, the sum over this square is $ ka - k(k-1)$. If we want this value to be negative, then we need $a < k - 1$, i.e. $a = k-1 - \epsilon$ for some $\epsilon > 0$, to be decided later.

Now let's look at the sum over the whole matrix. If we write $n = mk + r$, then we have $m$ columns of $a$'s and $n - m$ columns of $-1$'s. The total sum is then $$ nma - (n - m)n$.

Note that $n = mk + r \implies n - m = m(k - 1) + r$. Thus, we have that the sum over the whole matrix is $$ nm((k - 1) - \epsilon) - ((k - 1)m - r)n = rn - nm\epsilon.$$

Thus, if we take $\epsilon = \frac{r}{2m}$, then we have that the sum is $$ rn - nm\epsilon = rn - nm \frac{r}{2m} = \frac{rn}{2} > 0$$ as desired.

EDIT: The final matrix looks like: $$ \left(\begin{array}{cccc|cccc|c|ccc} a& -1 & \cdots & -1 &a & -1 & \cdots & -1 &\cdots &a & \cdots & -1 \\ a& -1 & \cdots & -1 &a & -1 & \cdots & -1 &\cdots &a & \cdots & -1 \\ \vdots& \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots &\ddots &\vdots & \ddots & \vdots \\ a& -1 & \cdots & -1 &a & -1 & \cdots & -1 &\cdots &a & \cdots & -1 \\ \end{array} \right)$$

Where $a = k-1 -\frac{r}{2m}$ in each block but the last I have one column of $a$'s and $(k-1)$ columns of $-1$'s. In the last column, I have $r$ $-1$'s. note, if I multiply the whole matrix by $2m$, then it has integer entries.

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  • $\begingroup$ As I suggested above, this can be turned into an integer solution: let every $k$-th column be $2m(k-1)-r$ and the other columns be $-2m$, in which case the sum over $k\times k$ is $-kr$ and the sum over $n\times n$ is $mnr$. $\endgroup$ – Mario Carneiro Oct 20 '15 at 15:37
  • $\begingroup$ I am afraid that you've made a little mistake in calculation. You wrote $nm((k - 1) - \epsilon) - ((k - 1)m - r)n = rn - nm\epsilon$ But I think it supposed to be $nm((k - 1) - \epsilon) - ((k - 1)m + r)n = -rn - nm\epsilon$ (since $n-m = (k-1)m + r$) and therefore the sum is $-rn - mn\epsilon$, which is clearly to be negative $\endgroup$ – primitiveroot Oct 20 '15 at 15:43
  • $\begingroup$ @MarioCarneiro, yes this is true. I added that to the solution. $\endgroup$ – Marcus M Oct 20 '15 at 15:43
  • $\begingroup$ @primitiveroot, where? $\endgroup$ – Marcus M Oct 20 '15 at 15:44
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    $\begingroup$ I also made some calculation. The sum of $n^2$ numbers are $nk - n(m+1)\epsilon$ therefore $\epsilon < \dfrac{k}{m+1}$ will work. And your idea was brilliant. Thank you! $\endgroup$ – primitiveroot Oct 20 '15 at 16:27
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Here's an answer to show that in fact the OP's original idea works, if the numbers are chosen correctly.

Let our matrix take values $b-a$ at indexes $(ik,jk)$ and $b$ elsewhere, with $n$ factored as $n=mk+r$. Then the sum across a $k\times k$ submatrix is $k^2b-a$, and the sum across the whole $n\times n$ matrix is $n^2b-m^2a$. The constraints to satisfy are thus $n^2b>m^2a$ and $k^2b<a$. This implies $m^2k^2b<m^2a<n^2b$, so $mk<n$ implies $b>0$. Dividing through by $m^2b$, we get

$$k^2<\frac ab<\frac{n^2}{m^2},$$

and we can feel free to choose $b=2m^2$ and $a=n^2+m^2k^2$ (the midpoint) or any other solution which satisfies the inequality. The OP's choice used $b=1$ and $a=k^2+1$, which satisfies the first inequality, but the second inequality in the worst case has $r=1$ so $\frac nm=k+\frac1m$ and $\frac{n^2}{m^2}=k^2+2\frac km+\frac1{m^2}$, so the $+1$ is a bit too generous and is undercut for large $m$. (This can also be taken as a proof that a sufficiently large integer solution cannot take $1$ for the "background" value $b$.)

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This is a slight variant on the construction in Marcus M's answer.

Having written $n=mk+r$ with positive integers $m$ and $0\lt r\lt k$, start with a matrix of the form

$$\pmatrix{ a&-1&\ldots&-1&\ldots&a&-1&\ldots&-1&b&\ldots&b\\ a&-1&\ldots&-1&\ldots&a&-1&\ldots&-1&b&\ldots&b\\ \vdots\\ a&-1&\ldots&-1&\ldots&a&-1&\ldots&-1&b&\ldots&b\\ }$$

where each row consists of an $a$ followed by $(k-1)$ $-1$'s, repeated $m$ times, with $r$ $b$'s at the end. Note that the sum of any $k$ consecutive numbers in any row is either $a-(k-1)$ or $bs-(k-s)$ with $1\le s\le r$, while the sum of the entire row is $m(a-(k-1))+br$.

Next, let $a=k-1$ and $b={k\over r}-1$. The assumption $0\lt r\lt k$ implies $a$ and $b$ are both positive. Our construction so far implies that the sum of any $k$ consecutive numbers in any row is either $$a-(k-1)=(k-1)-(k-1)=0$$ or $$bs-(k-s)=({k\over r}-1)s-(k-s)={s\over r}k-k\le0$$

while the sum of all the numbers in a row is

$$m(a-(k-1))+br=m((k-1)-(k-1))+({k\over r}-1)r=k-r\ge1$$

Thus the sum of the numbers in any $k\times k$ block is less than or equal to $0$, while the sum of the entire $n\times n$ matrix is greater than or equal to $n$.

Now this isn't quite what's required, but it's easy to fix it: Just subtract a very small number (anything less than $1/n$ will do) from every number in the matrix, including all the $-1$'s. Doing so ensures that every $k\times k$ block now has a negative sum while the sum for the entire matrix is still positive.

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