Coprime property transfer If HCF(m, n) = 1 with m > n, then what are HCF(m + n, m) and HCF(m - n, n), respectively?

Introduction / Context:This problem uses standard gcd identities. If gcd(m, n) = 1 (m, n are coprime), then adding or subtracting multiples does not introduce a common factor. In particular, gcd(m, m ± n) = gcd(m, n) = 1, and similarly for gcd(n, m ± n).

Given Data / Assumptions:

• gcd(m, n) = 1 and m > n.
• We need gcd(m + n, m) and gcd(m - n, n).

Concept / Approach:Use the identity gcd(a, b) = gcd(a, b ± a). Then gcd(m + n, m) = gcd(m + n - m, m) = gcd(n, m) = 1. Likewise, gcd(m - n, n) = gcd(m, n) when adjusting with multiples.

Step-by-Step Solution:gcd(m + n, m) = gcd(n, m) = 1.gcd(m - n, n) = gcd(m, n) = 1.

Verification / Alternative check:Try small coprime examples, e.g., m = 5, n = 2. Then gcd(7, 5) = 1 and gcd(3, 2) = 1. The property holds in general.

Why Other Options Are Wrong:Options involving 2 assume parity constraints; however, even if m and n share parity conditions, overall gcd remains 1 by the identities.

Common Pitfalls:Assuming that adding or subtracting numbers may introduce a factor like 2; this does not happen in gcd algebra unless parity forces both to be even, which cannot occur for coprime m and n simultaneously.

Final Answer:1 and 1