IMC 2012 Q24

The key to the solution is to note that the squares of the odd numbers occur on the diagonal leading downwards and to the right from the cell which contains the number 1, and the squares of the even numbers occur on the diagonal which leads upwards and to the left of the cell which contains the number 4.
The squares of the even numbers have the form \((2n)^2\), that is, \(4n^2\).We see that the number \(4n^2 + 1\) occurs to the left of the cell containing \(4n^2\). Below \(4n^2 + 1\) there occur the numbers \(4n^2 + 2, 4n^2 + 3, \ldots, 4n^2 + 2n + 1\), and then in the cells to the right of the cell containing \(4n^2 + 2n + 1\), there occur the numbers \(4n^2 + 2n + 2, 4n^2 + 2n + 3, \ldots, 4n^2 + 4n + 1 = (2n + 1)^2\).
Now \(44^2 = 1936\) and \(45^2 = 2025\). Thus 2011 is in the same row as 2025 and to the left of it, in the sequence 1981,….,2012,…,2025, and below these occur the numbers 2163,…., \(2209 = 47^2\), with 2208 below 2025 as shown below. It follows that 2195 is the number below 2012. In the diagram below the square numbers are shown in bold.