Math 230 E
Fall 2011
Homework 4
Drew Armstrong
Problem 1.
(a) Consider
a, b, d
∈
with gcd(
a, d
) = 1.
Prove:
If
d

ab
then
d

b
.
(b) Consider
a, b, m
∈
with gcd(
a, b
) = 1.
Prove:
If
a

m
and
b

m
then
ab

m
.
(c) Consider
a, b, c, n
∈
with gcd(
c, n
) = 1.
Prove:
If
ca
≡
cb
mod
n
then
a
≡
b
mod
n
.
Proof.
First we prove (a). Consider
a, b, d
∈
with
a, d
coprime, and suppose that
d

ab
. We
will show that
d

b
.
Since
d

ab
, there exists
k
∈
such that
dk
=
ab
.
Then since
a, d
are
coprime, B´
ezout’s Lemma implies that there exist
x, y
∈
such that
ax
+
dy
= 1. Finally, we
multiply both sides of this equality by
b
to get
b
(
ax
+
dy
) =
b
·
1
,
bax
+
bdy
=
b,
dkx
+
dby
=
b,
d
(
kx
+
by
) =
b.
We conclude that
d

b
. To show (b), consider
a, b, m
∈
with
a, b
coprime. Suppose further
that
a

m
and
b

m
. We will show that
ab

m
. There are two possible solutions.
Solution 1.
(From Scratch)
There exist
k, ‘
∈
such that
ak
=
m
and
b‘
=
m
.
Then since
a, b
are
coprime, B´
ezout’s Lemma gives us
x, y
∈
such that
ax
+
by
= 1. Multiply both sides by
m
to get
m
(
ax
+
by
) =
m
·
1
,
max
+
mby
=
m,
b‘ax
+
akby
=
m,
ab
(
‘x
+
ky
) =
m.
Hence
ab

m
.
Solution 2.
(Quote Part (a))
There exist
k, ‘
∈
such that
ak
=
m
=
b‘
,
hence
ak
=
b‘
. Since
b

ak
with
a, b
coprime, part (a) implies that
b

k
. That is, there exists
x
∈
such that
k
=
bx
.
Substitute this into
ak
=
m
to get
abx
=
m
.
We conclude that
ab

m
. Finally, we will prove (c). Consider
a, b, c, n
∈
with
c, n
coprime and
ca
≡
cb
mod
n
.
We will show that
a
≡
b
mod
n
. Indeed,
ca
≡
cb
mod
n
means that
n

(
ca

cb
), or
n

c
(
a

b
).
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 Fall '11
 Armstrong
 Math, Prime number, Euclidean algorithm, NK

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