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The University of Sydney
MATH2068
Number Theory and Cryptography
(http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/)
Semester 2, 2008
Lecturer: R. Howlett
Assignment 2
1.
(
i
)
Use the formula for
φ
(
n
) in terms of the factorization of
n
to compute
φ
(13),
φ
(21),
φ
(26),
φ
(28),
φ
(36) and
φ
(42).
(
ii
) Find all positive integers
n
such that
φ
(
n
) is odd.
(
iii
) Prove that if
n
is a positive integer and
p
a prime divisor of
n
then
p
≤
φ
(
n
) + 1, and classify all those cases for which equality holds.
(
iv
) Find all positive integers
n
such that
φ
(
n
) = 2 or
φ
(
n
) = 6. Then ﬁnd all
positive integers
n
such that
φ
(
n
) = 12. [Hint: Let
n
=
p
k
m
where
p
is
prime and
m
is not divisible by
p
, and use multiplicativity of
φ
.]
Solution.
(
i
)
If
n
=
p
k
1
1
p
k
2
2
···
p
k
r
r
where the
p
i
are pairwise distinct primes amd the
k
i
are
positive integers then (by the formulae on page 39 of the book of notes or Lecture
14)
φ
(
n
) =
p
k
1

1
1
(
p
1

1)
p
k
2

1
2
(
p
2

1)
···
p
k
r

1
r
(
p
r

1). Applying this with
r
= 1
and
p
k
1
1
= 13
1
gives
φ
(13) = 13
0
(13

1) = 12. To deal with 21 we need
r
= 2
and
p
k
1
1
p
k
2
2
= 3
1
×
7
1
, and we get
φ
(21) = 3
0
(3

1)7
0
(7

1) = 2
×
6 = 12. The
others are similar:
φ
(26) =
φ
(2
1
×
13
1
) = 2
0
(2

1)13
0
(13

1) = 12
,
φ
(28) =
φ
(2
2
×
7) = 2
1
(2

1)7
0
(7

1) = 12
,
φ
(36) =
φ
(2
2
×
3
2
) = 2
1
(2

1)3
1
(3

1) = 12
,
φ
(42) =
φ
(2
×
3
×
7) = 2
0
(2

1)3
0
(3

1)7
0
(7

1) = 12
.
(
ii
)
If
n
= 1 then by deﬁnition
φ
(
n
) = 1. (See the discussion of this point in
Lecture 18.) So
φ
(
n
) is odd when
n
= 1. If
n >
1 then by the Fundamental
Theorem of Arithmetic we can express
n
in the form
n
=
p
k
1
1
p
k
2
2
···
p
k
r
r
, where
the
p
i
are distinct primes, the
k
i
are positive integers, and
r
≥
1. As in Part (
i
)
this gives
φ
(
n
) =
p
k
1

1
1
(
p
1

1)
p
k
2

1
2
(
p
2

1)
···
p
k
r

1
r
(
p
r

1)
,
(
*
)
and if
φ
(
n
) is odd then all the factors in this expression must be odd. In partic
ular,
p
i

1 must be odd for each
i
. But there is only one prime
p
such that
p

1
is odd, namely the prime
p
= 2. So 2 is the only prime divisor of
n
. That is,
in the factorization
n
=
p
k
1
1
p
k
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 One '08
 Howlett
 Number Theory, Cryptography

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