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April 6–10, 2015
Week #11 Practice Quiz
Math 430
SOLUTIONS
Name:
Problem 1. (6 points) Let n = 11 and consider the primitive root r = 2. We have the
following table of indices:
x
ind2 (x)
1
10
2
1
3
8
4
2
5
4
6
9
7
7
8
3
9
6
10
5
Solve
3x+2 ≡ 42x−1
(mod 11)
for x.
SOLUTION: We take ind2 on both sides of the given congruence to get
(x + 2)ind2 (3) ≡ (2x − 1)ind2 (4)
(mod 10)
Plugging in the values from the table, we have
8(x + 2) ≡ 2(2x − 1)
(mod 10)
or
8x + 6 ≡ 4x − 2
(mod 10)
Thus, 4x ≡ 2 (mod 10), or 2x ≡ 1 (mod 5). Multiplying through by 3, we obtain x ≡ 3 (mod 5).
Thus, the solutions for x are x = 3, 8, 13, 18, . . ..
Problem 2. (6 points) Show that if p is an odd prime and x4 ≡ −1 (mod p) has a
solution, then p ≡ 1 (mod 8).
SOLUTION: From x4 ≡ −1 (mod p), we take the index of both sides (with respect to any primitive
root r of p) to get
4indr (x) ≡ indr (−1) (mod p − 1).
Note that indr (−1) = (p − 1)/2. Therefore, we have
4indr (x) ≡
p−1
2
(mod p − 1).
Hence, there exists an integer k such that
4indr (x) −
p−1
= k(p − 1).
2
That is,
8indr (x) − (p − 1) = 2k(p − 1),
so that
8indr (x) = (2k + 1)(p − 1).
This shows that
8|p − 1.
Hence, p ≡ 1 (mod 8).
Problem 3. (4 points each) Given that 3 is a primitive root of 43, find
(a): all positive integers less than 43 having order 6 modulo 43.
SOLUTION: Since 3 is a primitive root of 43, we know that order(3) = 42. Any other integer
< 43 which is relatively prime to 43 can be written in the form 3h for some integer h, by Theorem
8.4. Moreover, by Theorem 8.3
42
.
order(3h ) =
gcd(h, 42)
In order for this to equal 6, we must have gcd(h, 42) = 7, which implies that h = 7 or h = 35. Thus,
the two numbers 37 and 335 have order 6, modulo 43. We quickly compute directly that
37 ≡ 37
(mod 43)
335 ≡ 7
(mod 43).
and
So the answer is 7 and 37.
(b): all positive integers less than 43 having order 21 modulo 43.
SOLUTION: Using the same analysis as in part (a), we find that we must have gcd(h, 42) = 2,
which implies that h = 2, 4, 8, 10, 16, 20, 22, 26, 32, 34, 38, 40. Modulo 43, we compute 3 to each of
these powers:
32 ≡ 9
34 ≡ 38
38 ≡ 25
310 ≡ 10
316 ≡ 23
320 ≡ 14
322 ≡ 40
326 ≡ 15
332 ≡ 13
334 ≡ 31
338 ≡ 17
340 ≡ 24