•
When
p
= 2,
x
= 1 is a solution since
1
1
+ 1
≡
0
mod 2
.
•
Suppose that
p
≡
1
mod 4. Then 4

(
p

1), so by the Primitive Root Theorem,
there is an integer
a
such that
ord
p
(
a
) = 4
.
Thus,
a
4
≡
1
mod
p,
so
a
4

1
≡
(
a
2

1)(
a
2
+ 1)
≡
0
mod
p.
This implies that either
a
2

1
≡
0
mod
p
or
a
2
+ 1
≡
0
mod
p.
Note that
a
2

1
≡
0
mod
p
since ord
p
(
a
) = 4. Thus,
a
2
+ 1
≡
0
mod
p,
so
a
is a solution of
x
2
+ 1
≡
0
mod
p.
•
Suppose that
p
≡
3
mod 4. Then 4

(
p

3), so 4
(
p

1). Since
p
is odd,
p

1 is even, so 2

(
p

1). Thus, (
p

1
,
4) = 2. Now, suppose (by way of
contradiction) that there is an integer
a
such that
a
2
+ 1
≡
0
mod
p.
Then
a
4
≡
(
a
2
)
2
≡
(

1)
2
≡
1
mod
p.
By Fermat’s Little Theorem, we also know that
a
p

1
≡
1
mod
p.
Since 2 = (
p

1
,
4), there are integers
u
and
v
such that 2 = (
p

1)
s
+ 4
t
.
Thus, we have
a
2
≡
a
(
p

1)
s
+4
t
≡
(
a
p

1
)
s
(
a
4
)
t
≡
1
mod
p.
Then
a
2
≡
1
≡ 
1
mod
p,
87
Stanford University EPGY
Number Theory
so
2
≡
0
mod
p.
This implies that
p

2
,
which is a contradiction since
p
is an odd prime. Thus, there is no integer
a
such that
a
2
+ 1
≡
0
mod
p
, so the congruence
x
2
+ 1
≡
0
mod
p
does not have any solutions in this case.
Finally, we note that the ideas of
order
and
primitive roots
can be extended to all
(i.e. not necessarily prime) positive integers.
Definition 12.3
Suppose that (
a, n
) = 1. The
order
of
a
modulo
n
is the smallest
positive integer
b
such that
a
b
≡
1
mod
n.
By the EulerFermat Theorem, we know that
a
φ
(
n
)
≡
1
mod
n.
Thus, ord
n
(
a
)
≤
φ
(
n
).
Using an argument similar to that given in the proof of
Theorem 12.1, we can prove the following result.
Theorem 12.5
Suppose that (
a, n
) = 1 and that
a
b
≡
1
mod
n.
Then
ord
n
(
a
)

b,
and in particular,
ord
n
(
a
)

φ
(
n
)
.
Definition 12.4
If (
a, n
) = 1 and ord
n
(
a
) =
φ
(
n
), then we say that
a
is a
primitive
root
modulo
p
.
Example 12.6
3 is a primitive root modulo 10 since
φ
(10) = 4 and 3
1
≡
3
mod 10,
3
2
≡
9
mod 10, 3
3
≡
7
mod 10, 3
4
≡
1
mod 10.
88
Stanford University EPGY
Number Theory
Problem Set
1. Compute each of the following.
(a) ord
5
(3)
(b) ord
5
(4)
(c) ord
7
(3)
(d) ord
9
(2)
(e) ord
15
(2)
(f) ord
16
(3)
(g) ord
10
(3)
2. Find all primitive roots modulo 5.
3. Find all primitive roots modulo 7.
4. (a) Find all primitive roots modulo 13.
(b) For each number
d
dividing 12, list the
a
’s such that 1
≤
a <
13 and
ord
13
(1) =
d
.
5. Find all primes less than 20 for which 3 is a primitive root.
6. In this exercise, you will investigate the value of ordn(2) for odd integersn.
(a) Compute the value of ord
n
(2) for each odd number 3
≤
n
≤
19.
(b) In the table below, the value of ord
n
(2) is given for all odd numbers between
21 and 115.
ord
21
(2) = 6
ord
23
(2) = 11
ord
25
(2) = 20
ord
27
(2) = 18
ord
29
(2) = 28
ord
31
(2) = 5
ord
33
(2) = 10
ord
35
(2) = 12
ord
37
(2) = 36
ord
39
(2) = 12
ord
41
(2) = 20
ord
43
(2) = 14
ord
45
(2) = 12
ord
47
(2) = 23
ord
49
(2) = 21
ord
51
(2) = 8
ord
53
(2) = 52
ord
55
(2) = 20
ord
57
(2) = 18
ord
59
(2) = 58
ord
61
(2) = 60
ord
63
(2) = 6
ord
65
(2) = 12
ord
67
(2) = 66
ord
69
(2) = 22
ord
71
(2) = 35
ord
73
(2) = 9
ord
75
(2) = 20
ord
77
(2) = 30
ord
79
(2) = 39
ord
81
(2) = 54
ord
83
(2) = 82
ord
85
(2) = 8
ord
87
(2) = 28
ord
89
(2) = 11
ord
91
(2) = 12
ord
93
(2) = 10
ord
95
(2) = 36
ord
97
(2) = 48
ord
99
(2) = 30
ord
101
(2) = 100
ord
103
(2) = 51
ord
105
(2) = 12
ord
107
(2) = 106
ord
109
(2) = 36
ord
111
(2) = 36
ord
113
(2) = 28
ord
115
(2) = 44
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 Fall '16
 jena smith
 Number Theory, Prime number, Stanford University EPGY