How many ordered pairs of positive integers (a, b) satisfy a² − b² = 45?

Factor: (a − b)(a + b) = 45. Since a, b are positive integers with a > b, both factors are positive, and a − b and a + b have the same parity — since 45 is odd, both factors must be odd, which is automatic here as 45 has only odd divisors.

List factor pairs (p, q) with p < q, pq = 45, p = a − b, q = a + b: (1, 45), (3, 15), (5, 9).

For each pair, a = (p + q)/2, b = (q − p)/2: (1,45) → a=23, b=22; (3,15) → a=9, b=6; (5,9) → a=7, b=2.

All three give positive integers, so there are 3 ordered pairs.

Tip: For a² − b² = N with N odd, every factor pair of N (both odd automatically) yields exactly one valid (a, b).