If x and y are positive integers such that x 2 -y 2 =13, then the possible value of x 2 +y 2 will be

Write x² − y² as (x − y)(x + y) = 13. Since 13 is prime and x, y are positive integers, the only factorisation with x + y larger than x − y is 1 × 13. So x − y = 1 and x + y = 13, giving x = 7, y = 6. Then x² + y² = 49 + 36 = 85, which is option B.

Why the others fail: each is a sum of two squares, but none comes from a pair whose squares differ by 13.

• 90 = 81 + 9 (from 9 and 3), with difference 72.
• 72 = 36 + 36, with difference 0.
• 65 = 64 + 1 or 49 + 16, with differences 63 or 33.

The concept is factorising a difference of squares and using the prime factorisation of the constant. Identity-based factorisation is a basic tool for quantitative reasoning in Prelims Paper II, so build the habit of factorising before guessing.

Trap: Aspirants try random integer pairs, get stuck, and pick 65 because it looks like a tidy sum of squares.