SELECTED SOLUTIONS TO HOMEWORK 7
1. 6.4
Theorem 17.
(1) Suppose a, b ∈ Z>0 and p is prime. Then if p|ab
then p|a and p|b.
(2) Suppose a1 , . . . , an is a sequence of positive integers and p is
prime. If p|a1 . . . an , then p divides one of the ai .
(1) Suppose a, b ∈ Z>0 and that p is prime. We’ll show that
if p|ab and p 6 |a, then p|b, so we suppose that p|ab and p 6 |a.
Let d = (a, p). As p is prime d is either 1 or p, so as p 6 |a we
have d = 1.
So, there exist x, y ∈ Z with ax + py = 1. Multiplying this
by b gives us abx + bpy = b. But because p|ab there exists some
k ∈ Z so that kp = ab, giving us kpx + bpy = b, so p|b, as we
wanted to show.
(2) We use induction on n. The base case is n = 1, and if p|a1 ,
of course p|a1 . Now, suppose that for any list of n integers,
if p|a1 . . . an , p divides of of the n integers, and consider a list
of n + 1 integers a1 . . . an+1 . Suppose that p|(a1 . . . an+1 ), or
p|(a1 . . . an )(an+1 .
The first part of this theorem tells us that p|(a1 . . . an ) or
p|an+1 (here is where we are using that p is prime). If p|an+1
then we have that p divides one of the ai , as we wanted to show.
If p|(a1 . . . an ) then our inductive hypothesis applies to tell us
that p divides one of the ai , so we are done by induction.
Proof.
2. Extra Problems
Theorem 2. For a, b ∈ N>0 , let d = (a, b). We have that d|a and d|b,
so suppose that du = a and dv = b. Then we have the following: if
ax + by = c and ax0 + by 0 = c (where x, y, x0 , y 0 , c ∈ Z), then there is
some k ∈ Z so that x0 = x + kv, and y 0 = y − ku.
Proof. First, suppose that aα + bβ = 0, so that duα + dvβ = 0, where
d, u, v are as in the statement of the theorem. Then uα = −vβ, so
v|uα. Now, (u, v) = 1, because if (u, v) = k where k > 1, then dk|a
and dk|b, contradicting that d = (a, b). Therefore, since v 6 |u, we have
1
2
SELECTED SOLUTIONS TO HOMEWORK 7
that v|α, so there exists a k ∈ Z so that vk = α. Plugging this in to
uα = −vβ gives us −uk = β.
Now, suppose that ax + by = c, and ax0 + by 0 = c. Then a(x0 −
x) + b(y 0 − y) = 0, so by the first paragraph there exists a k ∈ Z with
x0 − x = vk, and y 0 − y = −uk, so x0 = x + kv and y 0 = y − ku, as we
wanted to show.
Theorem 4. Let n ∈ N>0 have prime factorization n = pa11 . . . pakk ,
where the pi and ai are as above. Then for all m ∈ Z, n|m if and only
if for all i ∈ {1, . . . , k}, pai i |m.
Proof. We use induction. The base base is k = 1, and of course if
1
pa11 |m, then pa1 |m. Now, suppose that for all m ∈ Z, and for all n ∈ Z
with k prime factors, if n = pa11 . . . pakk |m, then for all i ∈ {1, . . . , k}
pai i |m, and suppose that some integer with k + 1 prime factors divides
ak+1
m, so pa11 . . . pakk pk+1
|m.
ak+1
a1
We have that (p1 . . . pakk , pk+1
) = 1, so by the problem referenced
ak+1
a
a1
k
on homework 5, p1 . . . pk |m and pk+1
|m. Because pa11 . . . pakk |m our
inductive hypothesis tells us that for all i ∈ {1, . . . , k} pai i |m, so for all
i ∈ {1, . . . , k + 1} pai i |m, and we are done by induction.