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

Answer: (a) 3

x² − y² = (x − y)(x + y) = 63, so each way of writing 63 as a product of two factors gives one pair. 63 = 1 × 63 = 3 × 21 = 7 × 9, giving (32, 31), (12, 9) and (8, 1).

1. x² − y² = (x − y)(x + y) = 63, with x − y < x + y.
2. Factor pairs of 63: (1, 63), (3, 21), (7, 9) — all odd, so x and y come out whole.
3. (1, 63): x = 32, y = 31. (3, 21): x = 12, y = 9. (7, 9): x = 8, y = 1.
4. 3 pairs.
5. Check: 32² − 31² = 63, 144 − 81 = 63, 64 − 1 = 63.

Remember · Difference of squares: factor the number as (x − y)(x + y) with both factors of the same parity; each factor pair gives one solution.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). · Report an error