Well done to Maulik aged 11 who sent in some nice work on this problem. Neil's solution is given below.

For a 2 x 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 = 2 xy + 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 = 2 xy + 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 3x 3square 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:

x (y + 2) + (x + 1) (y + 1) + (x + 2) y = xy + 2 x + xy + x + y + 1 + xy + 2 y = 3 xy + 3 x + 3 y + 1

The diagonal from top left to bottom right is:

xy + (x + 1) (y + 1) + (x + 2) (y + 2) = xy + xy + x + y + 1 + xy + 2 x + 2 y + 4 = 3 xy + 3 x + 3 y + 5

Let's say that

3 xy + 3 x + 3 y = 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:
summary of results

Note that
1 = 1²
4 = 2²
10 = 1² + 3²
20 = 2² + 4²
35 =1² + 3² + 5²
56 = 2² + 4² + 6²
84 = 1² + 3² + 5² + 7²

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:

pascal's triangle