Question

8. (Section 1.3 #10) Let p be an integer other than 0, ±1. Prove p is prime if and only if for all a 2 Z
either (a, p) = 1 OR p | a.
9. (Section 1.3 #16) Prove that (a, b) = 1 if and only if there is no prime p such that p | a AND p | b.

Answer 1

8.

“a” be an integer which is not divisible by a. and p is prime. The only common divisors of a are −1 and 1and the only divisors of p are −1, 1, −p, p. Therefore (a, p) = 1. Since this argument works for every integer which does not divisible by p, we have for every integer a that either p|a or (a, p) = 1.

If suppose that p ∈ Z and for every a ∈ Z, either (a, p) = 1 or p|a. Suppose if p is not prime then p has a divisor a such that a is not equal to −1, 1, p, −p. If a is a divisor then −a is a divisor, we may choose a to be positive; thus we have a divisor a such that 1 < a < p. According to our hypothesis, either (a, p) = 1 or p|a. Since 0 < a < p, p cannot divide a. Since a is a common divisor of a and p, (a, p) > a > 1. This is a contradiction. Therefore, p must be prime.

9.

The only common divisors of a and b are +1 and -1 if (a, b) = 1, neither of which is prime. If there are no common divisors other than ±1; for if c is not equal to ±1 were a common divisor of a and b, then c would be a product of primes, say c = p1 . . . pn, and since p1|c, we would have that p1|a and p1|b, which contradicts (a, b) = 1.