Q. 34.2( 5 Votes )

Prove g. c. d (a - b, a + b) = 1 or 2, if g. c. d. (a, b) = 1

Answer :

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

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