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
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 Summer '11
 STAFF
 Prime number, DI, Greatest common divisor, Divisor, gcd, di divides

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