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
Tutorial 6
1.
Show that, if
n
is any positive integer,
n
! + 1 is not divisible by any number less
than
n
. Deduce that there are an infinite number of prime numbers.
Solution.
If
m < n
then
m
is one of the factors of
n
!; so
n
!
≡
0 (mod
m
).
Hence
n
!+1
≡
1
≡
0 (mod
m
), which means that
m
(
n
!+1). Now, by the Fundamental
Theorem of Arithmetic, there must be some prime number
p
such that
p

(
n
!+1),
and in view of what we have just shown it must be that
p > n
. This is true for
all positive integers
n
: we have shown that for every positive integer
n
there is
a prime
p > n
.
Choose
p
1
to be a prime greater than 1, and then (recursively) for each
i >
1
choose
p
i
to be a prime greater than
p
i

1
. This generates an infinite increasing
sequence of prime numbers, showing that the number of primes is infinite.
2.
Which positive integers have exactly two positive integer divisors? Which have
three? And which four?
Solution.
Every integer greater than 1 has at least two divisors: itself and 1. These are
the only divisors if and only if the number is prime.
If an integer
n
greater than 1 is not prime then it can be expressed as a product
rs
with 1
< r < n
and 1
< s < n
.
Then
n
is divisible by at least 1,
r
,
s
and
n
. This does not necessarily mean 4 divisors, since
r
may be equal to
s
. It
is certainly the case that
n
= 1 and
s
is not equal to 1 or
n
; so there are at least
three distinct divisors. And if there are only three then
r
must equal
s
, and
s
must be prime, for otherwise there would be a
t
with 1
< t < s
and
t

s
, and
t
would be another divisor of
n
. So if
n
has exactly three divisors then it must be
the square of a prime. Conversely, it is clear that if
n
=
p
2
is the square of a
prime, then
n
does have exactly three divisors, namely 1
, p
and
n
.
If
n
=
rs
, where
r
and
s
are distinct primes then
n
has exactly four divisors:
1,
r
,
s
and
n
.
If
n
=
p
3
, where
p
is prime, then
n
has exactly 4 divisors: 1,
p
,
p
2
and
p
3
. We can use arguments similar to those above to show that these
are the only circumstances under which
n
has exactly 4 divisors. Alternatively,
we can make use of the formula proved in lectures: if
n
=
p
k
1
1
p
k
2
2
· · ·
p
k
r
r
is the
prime factorization of
n
then
τ
(
n
) = (
k
1
+ 1)(
k
2
+ 1)
· · ·
(
k
r
+ 1).
Note that
2
by hypothesis the primes
p
1
,
p
2
,
. . .
,
p
r
are distinct from one another and are
divisors of
n
; so
k
i
≥
1 for all
i
.
Thus
k
i
+ 1
≥
2 for all
i
, and so if
r
≥
3
then
τ
(
n
)
≥
(
k
1
+ 1)(
k
2
+ 1)(
k
3
+ 1)
≥
2
3
= 8.
If
r
= 1 then
τ
(
n
) =
k
1
+ 1