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MATH 3360
Sketch of solutions for Assignment 5
9C.49
Since
ϕ
(18) = 6, we have 5
6
= 5
·
5
5
≡
1(mod 18). Answer: 5
5
.
9C.50
[graded out of 2 points]
f
≡
1(mod
ϕ
(
m
))
⇒
f
=
kϕ
(
m
) + 1 for some
k
∈
Z
. Thus,
a
f
=
a
kϕ
(
m
)+1
=
(
a
ϕ
(
m
)
)
k
·
a
≡
1
·
a
(mod
m
) by Euler.
9C.62
[graded out of 5 points]
We are given that the order of 2 is
q

1, so 2
q

1
≡
1(mod
q
). We also have
(
q,
2) = 1, so 2 is a unit and by Euler 2
ϕ
(
q
)
≡
1(mod
q
). Since
q

1 is the
order of 2, we must have
q

1
≤
ϕ
(
q
). By deﬁnition of
ϕ
, we must also have
ϕ
(
q
)
≤
q

1, so
ϕ
(
q
) =
q

1
⇒
q
is prime.
Note that 2
10
≡
1(mod 341).
10A.7
[graded out of 2 points]
m
=
pq
,
ϕ
(
m
) = (
p

1)(
q

1) =
pq

p

q
+ 1 =
m

p

m/p
+ 1
⇒
pϕ
(
m
) =
pm

p
2

m
+
p
⇒
p
2
+ (
ϕ
(
m
)

m

1)
p
+
m
= 0. Also note
that
m
and
ϕ
(
m
) are both symmetric in
p
and
q
. Thus, a quadratic equation
x
2
+ (
ϕ
(
m
)

m

1)
x
+
m
= 0, has both
p
and
q
as its solution.
11B.11
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 BILLERA
 Math, Algebra

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