a) (a,b) = 1 so the (unique) prime factorization of b contains NO primes in the prime factorization of a. Similarly (a,c) = 1 implies that the prime factorization of c contains no primes in the prime factorization of a. This means that the prime factorization of bc also contains no primes in the prime factorization of a. Thus (a,bc) = 1.

b) You really want to "turn this around" and start with (a_1,b) = 1 and (a_2,b) = 1 implies (a_1*a_2,b) = 1. (After all, the symbol is symmetric.)

Now, what about

(a_1,b) = 1, (a_2,b) = 1, and (a_3,b) = 1.

Let a_1*a_2 = d. Thus we need to prove

(a_1*a_2*a_3,b) = (d*a_3,b) = 1.

We know that (d,b) = (a_1*a_2,b) = 1 from a) and (a_3,b) = 1.

Thus by a) we know that (d*a_3,b) = 1.

Can you generalize this procedure?

-Dan