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Let $n \ge 3$ be a positive integer.

Inside a $n$ × $n$ array there are placed $n^2$ positive numbers with sum $n^3$.

Prove that we can find a square $2$ × $2$ of $4$ elements of the array, having the sides parallel with the sides of the array, and for which the sum of the elements in the square is greater than $3n$.

Here's my attempt:

Case 1: Assuming all elements as equal. Visualising the $n$ x $n$ array as a square matrix, we have, $$ \begin{pmatrix} n & n & n & \cdots & n \\ n & \color{red}{n} & \color{red}{n} & \cdots & n \\ n&\color{red}n&\color{red}n&\cdots&n\\ \vdots & \vdots& \vdots & \ddots & \vdots \\ n & n & n & \cdots & n \end{pmatrix} $$

Clearly the sum of all elements is $n^3$. Moreover the sum of elements in any $2$ x $2$ matrix is $4n \gt 3n$.

Case 2: Unequal numbers.

$$ \begin{pmatrix} n_{11} & n_{12} & \cdots& n_{1n}\\ n_{21} & n_{22} & \cdots &n_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ n_{n1} & n_{n2} & \cdots & n_{nn} \end{pmatrix} $$

Although the numbers are unequal, it is given that they will add up to $n^3$. Hence the total weight of the matrix remains constant. Now let's try to shift the individual weights.

  • Shift the weights of one $2$ x $2$ region such that the weight of that region adds up to less that $3n$.
  • This implies the weights of the rest of $2$ x $2$ regions either remain the same or is increased.
  • Since weights of the other regions can either be increased or remain the same, there has to be at least one region greater than $3n$.

Hence my proof. Is there a more mathematical approach to this?

Thanks!

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    I don't follow your proposed "shifting" strategy. Are you sure it works? How would your strategy work for a $3{\times}3$ matrix? – 2017-02-01
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    Are you able to adapt Test123's solution for the case where $n$ is even to handle the case where $n$ is odd? – 2017-02-01

1 Answers 1

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Let $S$ denote the sum of all elements of the matrix. By assumption $S=n^3$. By contradiction assume that each $2\times 2$ sub-matrix has sum of its 4 elements less than $3n$.

Assume that $n$ is even. Then we can divide the matrix into $\frac{n}{2}\times\frac{n}{2}$ disjoint submatrices of size $2\times 2$. Hence we have that: $$ S \leq \frac{n^2}{4} 3n=\frac{3n^3}{4}

Now assume that $n$ is odd. Can you proceed analogously?

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    In case $n=4$ there exist $9$ submatrices ... – 2017-02-01
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    @MichaelHoppe I mean disjoint submatrices of size $2\times 2$. I didn't say that the total number of $2\times 2$ submatrices is $n^2/4$. – 2017-02-01
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    @Test123 -- Very nice! – 2017-02-01