106
Solutions to Review Exercises
(c)
I
{x
E Z
I
X
2

12 is divisible by 3}
{x
E Z
I
x
2

1
=
3k
for some integer
k}
{x
E
Z
I
xI
=
3Uor
some
t'
E
Z or
x
+
1
=
3Uor
some
t'
E
Z}
(3Z
+
1) U (3Z 
1)
(Since
x
2

1
=
(x
+
l)(x

1), if 31
(x
2

1), then 3
I
(x

1) or 3
I
(x
+
1).)
16. (a) Suppose there are just a finite number of primes,
PI, P2,
... ,
Pn.
Then look at the number
PIP2
...
Pn
+
1. This can't be prime since it is bigger than each
Pi.
So it is divisible by some
prime, hence by some
Pi.
But this implies
Pi
11, which is not true.
(b) Every integer
n
2: 2 can be written in the form
n
=
PIP2
...
Pr
for unique primes PI.
P2,···, Pr;
equivalently, every integer
n
2: 2 can be written
n
=
qfl
q~2
...
q':B
as the product of powers of
distinct prime numbers
ql, q2,
... ,
qs.
These primes and the exponents
aI, a2,
... ,
as
are unique.
17. 2
119
1
=
2(17)(7)
1
=
(217 _1)(2
102
+
2
85
+
2
68
+
2
51
+
2
34
+
217
+
1) is not prime; 3
109
1 is
not prime because it is even; 4
109
1
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 any
 Graph Theory, Prime number, unique solution

Click to edit the document details