# Q. How many pairs of natural numbers are there such that the difference of whose squares is 63?

(a) 3

(b) 4

(c) 5

(d) 2

Correct Answer: (a) 3

#### Question from UPSC Prelims 2020 CSAT Paper

**Explanation :**

## Pair of numbers whose difference of squares is 63

Let’s say the two natural numbers are x and y, where x > y. The problem statement can be rewritten as: x^2 – y^2 = 63.

Factoring the left side of this equation, we get: (x + y)(x – y) = 63.

Since x and y are natural numbers and x > y, both (x + y) and (x – y) must also be natural numbers. So we need to find pairs of natural numbers whose product is 63.

The only such pairs are (1, 63), (3, 21), and (7, 9). Solving for x and y in each case gives us the following pairs of natural numbers whose squares differ by 63:

(32, 31)

(12, 9)

(8, 1)

So there are 3 such pairs of natural numbers.