If g. c. d (a, b) = 1, then g. c. d (a — b, a b) = ………

Given g.c.d. (a,b) = 1

From Euclid’s algorithm we can say that if b|a, then g.c.d(a,b) = b

Let g. c. d (a—b, a + b) = k

Therefore we can say that k is a factor of both (a–b) and (a + b).

We can write a–b = rk for some r N

And also a + b = sk for some sN

Now, (a + b) + (a–b) = rk + sk

a + a + b–b = k(r + s)

2a = k(r + s)………..eq(1)

(a + b) – (a – b) = rk – sk

a + b–a + b = k(r–s)

2b = k(r–s)………..eq(2)

Also, g. c. d (a, b) = 1

Therefore, 2 × g. c. d (a, b) = 2 × 1

g. c. d (2a, 2b) = 2

g. c. d [(r + s)k, (r – s)k] = 2 (from eq (1) and eq (2))

k × g. c. d(r + s, r – s) = 2

= 2 × 1

So, k × g. c. d(r + s, r – s) = 2

k × g. c. d(r + s, r – s) = 2 × 1 = 2 × g.c.d(a,b)

By comparing we get k = 2

We know that 1 and 2 are co–prime numbers.

Similarly, we get k = 1

So, g. c. d (a—b, a + b) = k = 2

or g. c. d (a—b, a + b) = k = 1

g. c. d (a—b, a + b) = 1 or 2