Find the number of pairs of natural numbers with the difference between the squares of two natural numbers being equal to 63.

Option 1 : 2

**Given:**

Difference between the squares of two natural numbers is equal to 63.

**Concepts used:**

a^{2} - b^{2} = (a - b) × (a + b)

**Calculation:**

Let two natural numbers be x and y.

Difference between squares of x and y = 63

⇒ x^{2} – y^{2} = 63

⇒ (x – y) × (x + y) = 63

Possible values of (x - y) and (x + y) to satisfy the equation - (3,21) and (7,9)

On taking the pair (3,21)**.**

x - y = 3 ----(1)

x + y = 21 ----(2)

Adding equation (1) and equation (2),

⇒ 2x = 24

⇒ x = 24/2

⇒ x = 12

From equation (1),

12 - y = 3

⇒ -y = 3 - 12

⇒ y = 9

The value of x and y for the first pair is (12,9).

Similar to this we will get another pair (x,y) for the factors (7,9) and hence total number of pairs will be 2.

**∴ The total number of possible pairs is 2.**