Question

Given integers a, b, c,

g.c.d.(a, b, c) = 1 if and only if g.c.d.(a, b) = 1 and g.c.d.(a, c) = 1

Answer 1

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

g.c.d.(a, b, c) = 1 (proved previously)

(⇒) if g.c.d.(a, b, c) = 1 we will show g.c.d.(a, b) = g.c.d.(a, c) = 1.

let d = g.c.d(a, b) ⇒ d|a, d|b ⇒ d|bc and then

d|g.c.d(a, bc) = 1 ⇒ d|1 ⇒ d = 1

⇒ g.c.d(a, b) = 1, similarly g.c.d(a, c) = 1. and proof is complete.

(Another proof of the first side)

Let g.c.d(a, c) = 1 = g.c.d(a, c) and assume that g.c.d(a, bc) = d1 > 1.

Then d1 must have a prime divisor p. since d1|bc, it follows that p|bc; in

consequence, p|b or p|c.

If p|b, p|a ⇒ g.c.d(a, b) ≥ p (contradiction)

In the same way the condition p|c leads to the contradiction g.c.d(a, c) ≥ p.

Thus d1 = 1

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

g.c.d.(a, b, c) = 1 (proved previously)

(⇒) if g.c.d.(a, b, c) = 1 we will show g.c.d.(a, b) = g.c.d.(a, c) = 1.

let d = g.c.d(a, b) ⇒ d|a, d|b ⇒ d|bc and then

d|g.c.d(a, bc) = 1 ⇒ d|1 ⇒ d = 1

⇒ g.c.d(a, b) = 1, similarly g.c.d(a, c) = 1. and proof is complete.

(Another proof of the first side)

Let g.c.d(a, c) = 1 = g.c.d(a, c) and assume that g.c.d(a, bc) = d1 > 1.

Then d1 must have a prime divisor p. since d1|bc, it follows that p|bc; in

consequence, p|b or p|c.

If p|b, p|a ⇒ g.c.d(a, b) ≥ p (contradiction)

In the same way the condition p|c leads to the contradiction g.c.d(a, c) ≥ p.

Thus d1 = 1

                 ( proved)