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Well done to Maulik aged 11 who sent in some nice work on this problem. Neil's solution is given below.
For a 2 by 2 square with column headings of x and x+1, and row headings of y and y+1, Neil says that:
For the two by two square you can always express it algebraically like this:
So the diagonal from top right to bottom left is:
$(x+1)y+x(y+1) = xy+y+xy+x = 2xy+x+y$
Lets call that Z.
The diagonal from top left to bottom right is:
$xy+(x+1)(y+1) = xy+xy+x+y+1 = 2xy+x+y+1$
So the first diagonal is Z and the second Z+1 so the diagonal from top left to bottom right is always 1 more than the diagonal from top right to bottom left.
For a 3 by 3 square with column headings of x, x+1 and x+2, and row headings of y, y+1 and y+2, Neil says that:
The 3 by 3 square looks like this:
The diagonal from top right to bottom left is:
$\begin{split}x(y+2)+(x+1)(y+1)+(x+2)y &= xy+2x+xy+x+y+1+xy+2y \\&= 3xy+3x+3y+1\end{split}$
The diagonal from top left to bottom right is:
$ \begin{split}xy+(x+1)(y+1)+(x+2)(y+2) &= xy+xy+x+y+1+xy+2x+2y+4 \\&= 3xy+3x+3y+5 \end{split}$
Let's say that $3xy+3x+3y = W$
The diagonal from top right to bottom left is W+1.
The diagonal from top left to bottom right is W+5. So the difference between the diagonals is 4.
More generally
Continuing with the same method:
Note that
1 = 1 ²
4 = 2 ²
10 = 1 ² + 3 ²
20 = 2 ² + 4 ²
35 =1 ² + 3 ² + 5 ²
56 = 2 ² + 4 ² + 6 ²
84 = 1 ² + 3 ² + 5 ² + 7 ²
Neil goes on to point out that the differences are the tetrahedral numbers.
Tetrahedral numbers are the sum of consecutive triangular numbers.
The formula is 1/6n(n+1)(n+2).
The first few tetrahedral numbers are 1, 4, 10, 20, 35, 56, 84, 120, ...
The tetrahedral numbers are found in the fourth diagonal of Pascal's triangle: