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

Given:

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

Concepts used:
a2 - b2 = (a - b) × (a + b)

Calculation:

Let two natural numbers be x and y. 

Difference between squares of x and y = 63

⇒ x2 – y2 = 63

⇒ (x – y) × (x + y) = 63

Possible values of (x - y) and (x + y) to satisfy the equation - (3,21), (7,9) and (1, 63)

• 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)

​x - y = 7   ----(1)

x + y = 9    ----(2)

Adding equation (1) and equation (2),

⇒ 2x = 16

⇒ x = 8

⇒ y = 1

The value of x and y for the second pair is (8,1).

Similar to this we will get another pair (x,y) for the factors (1,63)

​x - y = 1   ----(1)

x + y = 63    ----(2)

Adding equation (1) and equation (2),

⇒ 2x = 64

⇒ x = 32

⇒ y = 31
The value of x and y for the Third pair is (32,31).

and hence the total number of pairs will be 3 viz. (12,9), (8,1), (32,31)

∴ The total number of possible pairs is 3.

Alternate MethodFactor of 63 = 1, 3, 7, 9, 21, 63

(x + y)(x - y) = x² - y²

We took the number whose square is greater than 63

So, 9, 21, and 63

According to the formula 

Take the 9

⇒ (8 + 1)(8 - 1) = 8² - 1² = 63

Take 21

⇒ (12 + 9)(12 - 9) = 12² - 9² 

⇒ 144 - 81 = 63

Take 63

⇒ (32 + 31)(32 - 1) = 32² - 31² 

⇒ 1024 - 961 = 63

So, The Three possible pair is (8, 1), (12, 9), and (32, 31).

∴ The total number of possible pairs is 3.