MATH 406: Homework 9.4 Solutions
1. (a) Show that 3 is a primitive root mod 14.
Solution: We have φ(14) = 6 so ord14 3|6. Note that 31 ≡ 3, 32 ≡ 9 and 33 ≡ −1 (all
mod 14) so ord14 3 = φ(14) = 6 meaning 3 is a primitive root.
(b) Construct a table of indices mod 14 using the primitive root 3.
Solution: We have
x
1 3 5 9 11 13
ind3 x 6 1 5 2 4
3
(c) Use this table to solve the congruence x3 ≡ 13 mod 14.
Solution: We have
x3 ≡ 13 mod 14
ind3 x3 ≡ ind3 13 mod φ(14) = 6
3ind3 x ≡ 3 mod 6
ind3 x ≡ 1 mod 2
ind3 x ≡ 1, 3, 5 mod 6
x ≡ 3, 13, 5 mod 14
(d) Use this table to solve the congruence 111−x ≡ 9 mod 14.
Solution: We have
111−x ≡ 9 mod 14
ind3 111−x ≡ ind3 9 mod φ(14) = 6
(1 − x)ind3 11 ≡ 2 mod 6
4(1 − x) ≡ 2 mod 6
4 − 4x ≡ 2 mod 6
2x ≡ 4 mod 6
x ≡ 2 mod 3
For problems 2-5:
Let m = 29 and let r = 8. It is a fact that r = 8 is a primitive root modulo 29, you need
not prove this. Consider the following table of indices. Note that three are missing.
a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
ind8 a
28
19
11
10
26
2
4
1
a
17
27
21
6
b
a
ind8 a
15
9
16
20
17
7
18
13
19
3
20
8
21
15
22
18
23
16
24
12
25
c
26
25
27
5
28
14
2. Solve for a, b and c. There are several ways to do this. Do it however you choose, just be clear
about your work.
Solution: This is just one of many ways:
• a ≡ ind8 9 ≡ ind8 32 ≡ 2ind8 3 ≡ 2(11) ≡ 22 mod 28.
• b ≡ ind8 14 ≡ ind8 2 · 7 ≡ ind8 2 + ind8 7 ≡ 19 + 4 ≡ 23 mod 28.
• c ≡ ind8 25 ≡ ind8 52 ≡ 2ind8 5 ≡ 2(26) ≡ 24 mod 28
3. Using the table of indices, find all solutions to the congruence 19x4 ≡ 21 mod 29. Solving this
does not require a, b or c.
Solution: We have
19x4 ≡ 21 mod 29
ind8 (19x4 ) ≡ ind8 21 mod φ(29) = 28
ind8 19 + 4ind8 x ≡ 15 mod 28
3 + 4ind8 x ≡ 15 mod 28
4ind8 x ≡ 12 mod 28
ind8 x ≡ 3 mod 7
ind8 x ≡ 3, 10, 17, 24 mod 28
x ≡ 19, 4, 10, 25 mod 29
4. Using the table of indices, find all solutions to the congruence 27x3 ≡ 24 mod 29. Solving this
does not require a, b or c.
Solution: We have
27x3 ≡ 24 mod 29
ind8 27x3 ≡ ind8 24 mod φ(29)
ind8 27 + 3ind8 x ≡ 12 mod 28
5 + 3ind8 x ≡ 12 mod 28
3ind8 x ≡ 7 mod 28
(19)3ind8 x ≡ (19)7 mod 28
ind8 x ≡ 21 mod 28
x ≡ 12 mod 29
5. Using the table of indices, solve the congruence 32x ≡ 23 mod 29. Solving this does not require
a, b or c.
Solution: We have
32x ≡ 23 mod 29
ind8 32x ≡ ind8 23 mod φ(29)
(2x)ind8 3 ≡ 16 mod 28
(2x)(11) ≡ 16 mod 28
22x ≡ 16 mod 28
11x ≡ 8 mod 14
−3x ≡ 8 mod 14
(−5)(−3x) ≡ (−5)(8) mod 14
x ≡ 2 mod 14
6. The change of base formula for logarithms says that
logb a =
logc a
logc b
which may be rewritten without division as
(logb a) (logc b) = logc a
State and prove an analogous formula for indices.
Solution: Suppose mod n we have a and b relatively prime to n and b and c both primitive
roots of n.
bindb a ≡ a mod n
indc bindb a ≡ indc a mod φ(n)
(indb a) (indc b) ≡ indc a mod φ(n)
7. Find two primitive roots for 17 and construct a table of indices for each. Use each of these
tables to solve the congruences x12 ≡ 13 mod 17 and 7x ≡ 7 mod 17.
Solution: A primitive root for 17 must be an r with ord17 r = φ(17) = 16. We check on
scratch paper and find:
r = 3: 31 ≡ 3, 32 ≡ 9, 34 ≡ 13, 38 ≡ 16, 316 ≡ 1 (all mod 17) so 3 is a primitive root.
r = 5: 51 ≡ 5, 52 ≡ 8, 54 ≡ 13, 58 ≡ 16, 516 ≡ 1 (all mod 17) so 5 is a primitive root.
Our two
x
ind3 x
and
x
ind5 x
tables are:
1
2 3
16 14 1
4
12
5
5
6
15
1
16
4
12
5
1
6
3
2
6
3
13
7
11
7
15
8
10
8
2
9
2
9
10
10
3
10
7
11
7
11
11
12
13
12
9
Then we solve x12 ≡ 13 mod 17 first using r = 3:
x12 ≡ 13 mod 17
ind3 x12 ≡ ind3 13 mod 16
12ind3 x ≡ 4 mod 16
3ind3 x ≡ 1 mod 4
ind3 x ≡ 3 mod 4
ind3 x ≡ 3, 7, 11, 15 mod 16
x ≡ 10, 11, 7, 6 mod 17
and then using r = 5:
x12 ≡ 13 mod 17
ind5 x12 ≡ ind5 13 mod 16
12ind5 x ≡ 4 mod 16
3ind5 x ≡ 1 mod 4
ind5 x ≡ 3 mod 4
ind5 x ≡ 3, 7, 11, 15 mod 16
x ≡ 6, 10, 11, 7 mod 17
13
4
13
4
14
9
14
5
15
6
15
14
16
8
16
8
And we solve 7x ≡ 7 mod 17 first using r = 3:
7x ≡ 7 mod 17
ind3 7x ≡ ind3 7 mod 16
xind3 7 ≡ 11 mod 16
11x ≡ 11 mod 16
x ≡ 1 mod 16
and then using r = 5:
7x ≡ 7 mod 17
ind5 7x ≡ ind5 7 mod 16
xind5 7 ≡ 10 mod 16
15x ≡ 15 mod 16
x ≡ 1 mod 16
8. Let r be a primitive root for an odd prime p. Show that indr (p − 1) = 21 (p − 1).
Solution: Since r is a primitive root we know that ordp r = φ(p) = p − 1. Now then observe
that
rp−1 ≡ 1 mod p
so that
p|rp−1 − 1
Since p is odd we know that p − 1 is even and we can factor to get
p| r(p−1)/2 + 1 r(p−1)/2 − 1
so that p divides one of the two. But if p|r(p−1)/2 − 1 then r(p−1)/2 ≡ 1 mod p contradicting
ordp r = p − 1 and so we must have p|r(p−1)/2 + 1 so that r(p−1)/2 ≡ −1 ≡ p − 1 mod p which
is the same as indr (p − 1) = 21 (p − 1).