Question

Write a mathematical proof that shows that a square at position (i, j) is diagonal to a square at (x, y) if and only if i+j == x+y or i-j == x-y. You can use the following definition of a diagonal square. Two squares (i, j) and (x, y) are diagonal if one of the following cases is true:

i-m = x and j-m = y

i-m = x and j+m = y

i+m = x and j-m = y

i+m = x and j + m = y

Hint: Go through each of the four definitions and show that each of them resolves to either i+j == x+y or i-j == x-y.

Answer #1

**Consider case
1:-**

Given i-m = x and j-m = y. So subtracting these two equations,
we get (i-m) - (j-m) = x-y i.e. **i-j = x-y**
---------- (1)

**Consider case
2:-**

Given i-m = x and j+m = y. So adding these two equations, we get
(i-m) + (j+m) = x+y i.e. **i+j = x+y** ----------
(2)

**Consider case
3:-**

Given i+m = x and j-m = y. So adding these two equations, we get
(i+m) + (j-m) = x+y i.e. **i+j = x+y** ----------
(3)

**Consider case
4:-**

Given i+m = x and j+m = y. So subtracting these two equations,
we get (i+m) - (j+m) = x-y i.e. **i-j = x-y**
---------- (4)

We can observe that (1) , (4) are same equations i.e.
**i-j = x-y**, similarly (2) , (3) are also same
equations i.e.**i+j = x+y**

** Since it is given that one of the cases must be true, we can conclude that a square at position (i, j) is diagonal to a square at (x, y) if and only if i+j == x+y or i-j == x-y.

Hence proved.

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