Exercises for Chapter 7
1. Prove : If gcd(a,b) = c, then gcd(a2,b2)=c2 .
Proof. We use the Fundamental Theorem of Arithmetic to factor a,b and c into primes.
Let c = p1 p2 ! pt , a = q1 q2 ! qs p1 p2 ! pt , b = r1 r2 ! ru p1 p2 ! pt , where
pi , q j and rk are primes with qi ≠ rk for any i and k.
Then c 2 = p12 p2 2 ! pt 2 , a 2 = q12 q2 2 ! qs 2 p12 p2 2 ! pt 2 , b = r12 r2 2 ! ru 2 p12 p2 2 ! pt 2 .
It is evident that c2|a2b2. Any larger divisor of a2 and b2 would contain common factors
of qk and rj which is not possible.
2. Prove: If a prime p has the form 3k + 1, then it has the form 6k + 1.
Proof. The integer k in the form 3k + 1 is either even or odd, i.e., k = 2m, or k = 2m + 1
for some integer m.
Suppose k = 2m. Then p = 3k + 1 = 6m + 1.
Suppose k = 2m + 1. Then p = 3(2m + 1) + 1 = 6m + 4 = 2(3m + 2) which is not
prime. Therefore, a prime p of the form 3k + 1 occurs only when k is even and in this
case p takes the form 6m + 1.
3. Prove: If p is a prime and a is an integer such that p|a2, then p|a.
This follows directly from the prime divisibility property, page 44.
4. Prove: If c|ab, the c | gcd(a,c) gcd(b,c).
Proof. Let k = gcd(a,b,c). c is made up of factors that divide both a and b, that divide a
and not b and that divide b and not a. Then c = j k l where gcd(a,c) = j k, and gcd(b,c) =
k l. Then c = j k l|j k k l = gcd(a,c) gcd(b,c).