Math 110A HW §2.4 – Solutions
4. Find the smallest integer in the given set.
(a)
{x ∈ Z|x > 0 and x = 4s + 6t for some s, t ∈ Z}
(b)
{x ∈ Z|x > 0 and x = 6s + 15t for some s, t ∈ Z}
From the proof of theorem 2.12, we know the smallest positive integer in each of these sets
is just (4, 6) for part a and (6, 15) for part b. In other words the smallest positive integer in
the set in part a is 2 and the smallest positive integer in the set in part b is 3.
10. Let a|c and b|c, and (a, b) = 1, prove that ab divides c.
Proof. Assume a|c, b|c and (a, b) = 1. So there exists integers m and n such that am+bn = 1.
Multiplying by c we get amc + bnc = c. Now, since a|c and b|c there are integers k and l such
that c = ak and c = bl. Therefore we have the following:
amc + bnc = c
am(bl) + bn(ak) = c
ab(ml + nk) = c
Since m, l, n, k ∈ Z, then ml + nk ∈ Z. Thus ab|c.
11. Prove that if d = (a, b), a|c and b|c, then ab|cd.
Proof. Since a|c and b|c there exist integers k and l such that c = ak and c = bl. Also since
d = (a, b), there exist integers m and n such that d = am + bn. Therefore we have the
following.
cd = c(am + bn) = cam + cbn = (bl)am + (ak)bn = ab(lm + kn)
Thus ab|cd.
15. Let a and b be integers, at least one of them not 0. Prove that an integer c can be expressed
as a linear combination of a and b if and only if (a, b)|c.
Proof.
(⇒) Assume c = ax + by for some x, y ∈ Z and let d = (a, b). Since d|a and d|b, then
d|(ax + by). In other words d|c.
(⇐) Assume (a, b)|c. Let d = (a, b), so there exists k ∈ Z such that c = dk. Moreover, since
d is the gcd of a and b there exist integers m and n such that d = am + bn. Multiplying
by k gives, dk = akm + bkn. Thus c = a(km) + b(kn). That is, c can be expressed as a
linear combination of a and b.
18. Prove that (ab, c) = 1 if and only if (a, c) = 1 and (b, c) = 1.
Proof.
(⇒) Assume (ab, c) = 1. Then there are integers m and n such that abm + cn = 1. Since
b, m ∈ Z, then bm ∈ Z. So 1 has been expressed as a linear combination of a and c.
Thus (a, c) = 1. Similarly, since a, m ∈ Z, then am ∈ Z. So 1 has been expressed as a
linear combination of b and c. Thus (b, c) = 1.
(⇐) Assume (a, c) = 1 and (b, c) = 1. So there are integers m, n, x and y such that 1 = am+cn
and 1 = bx+cy. Multiplying these two equations gives 1 = ab(mx)+c(amy +bnx+cny).
Thus we have written 1 as a linear combination of ab and c. Hence (ab, c) = 1.
Thus (ab, c) = 1 if and only if (a, c) = 1 = (b, c).
22. Let (a, b) = 1. Prove that (a, bn) = 1 for all positive integers n.
Proof. Let d = (a, bn). Suppose d 6= 1. Since d 6= 1, then there exists a prime p such that
p|d. Since p|d and d|a, we know p|a. Similarly, since p|d and d|bn, we know p|bn. By Euclid’s
lemma, since p|bn , we know p|b. Therefore we have p|a and p|b, thus p|(a, b). So p|1, which
would imply p = 1. This is a contradiction since p is prime. Therefore d = 1.
24. Prove that if d = (a, b), a = a0 d and b = b0d, then (a0 , b0) = 1.
Proof. Assume d = (a, b), a = a0 d and b = b0d. So there are integers m, n ∈ Z such that
d = am + bn. Therefore we have the following:
d = am + bn
d = a0dm + b0 dn
d = d(a0m + b0n)
1 = a 0 m + b0 n
Thus we have expressed 1 as a linear combination of a0 and b0, hence (a0, b0) = 1.