Introductory Number Theory. Homework 3
Due: Tuesday, June 7, 2011, 4:45PM
Note:
My words to work on your own seem to continue to be ignored. At least, don’t make it so obvious! At least,
UNDERSTAND WHAT YOU ARE COPYING!!!
I know you are copying without really understanding
when I see meaningless statements, or incomplete statements, that make sense when one sees the complete form in
someone else’s paper. I may not fail you for it, but rest assured that you I am aware of it. My apologies to the few
of you who work on their own and don’t allow others to copy from them.
1. Let
m,n
be integers, not both 0. Consider the set
A
=
{
mx
+
ny
:
x,y
∈
Z
}
. Show that this set coincides
with the set of all multiples of the greatest common divisor of
m,n
. That is, if
d
=
gcd
(
m,n
), show
A
=
{
kd
:
k
∈
Z
}
.
Solution.
By Bezout, there exist integers
a,b
such that
d
=
am
+
bn
. For every
k
∈
Z
we have
kd
=
kam
+
kbn
∈
A
, proving that the set
{
kd
:
k
∈
Z
} ⊂
A
.
But one also has to prove the converse
inclusion!
Assume now
x,y
∈
Z
. Since
d

m
and
d

n
, there exists integers
j,‘
such that
m
=
jd
,
n
=
‘d
,
hence
mx
+
ny
=
jdx
+
‘dy
= (
jx
+
‘y
)
d
=
kd
with
k
=
jx
+
‘y
. The converse inclusion follows.
2. If
a,m
∈
N
and
gcd
(
a,m
) = 1, then the order of an integer
a
modulo
m
is the
smallest
positive integer
k
such that
a
k
≡
1 (mod
m
). Prove: If
gcd
(
a,m
) = 1 and if
k
is the order of
a
modulo
m
, then
k

φ
(
m
).
Hint:
Divide
φ
(
m
) by
k
; what can you say about the remainder?
Solution.
Since
a
φ
(
m
)
≡
1 (mod
m
) by Euler’s theorem, we must have 1
≤
k
≤
φ
(
m
). By the division
algorithm, we can write
φ
(
m
) =
qk
+
r
for unique integers
q,r
, 0
≤
r < k
. Now, once again by Euler’s
theorem, and because
a
k
≡
1 (mod
m
),
1
≡
a
φ
(
m
)
=
a
qk
a
r
=
(
a
k
)
q
a
r
≡
1
q
a
r
=
a
r
(mod
m
);
i.e.,
a
r
≡
1 (mod
m
). If
r >
0 then 1
≤
r < k
, contradicting that
k
is the least positive integer such that