Solutions to Math 332 Homework 12
7. If a has order 4, then 4 = ord(a) | φ(p) = p − 1. So p = 4k + 1 for some k ∈ N. Conversely if p = 4k + 1
for some k ∈ N, let g be a primitive root of p (this exists by (6.7)), then by (6.3)(i),
ord(g)
p−1
4k
4k
=
=
=
= 4.
(k, ord(g))
(k, p − 1)
(k, 4k)
k
ord(g k ) =
Therefore p ≡ 1 (mod 4) iff elements of order 4 modulo p exists.
11. 6 is a primitive root of 41, so 6, 62 , . . . , 640 forms a reduced residue system modulo 41 by (6.11). Hence
we just need to find those 6k ’s such that ord(6k ) = 8. By (6.3)(i),
8 = ord(6k ) =
ord(6)
40
=
,
(k, ord(6))
(k, 40)
and we get (k, 40) = 5. So k ≡ 5, 15, 25, 35 (mod 41) and the required elements of order 8 are
65 , 615 , 625 , 635 . Now reduce modulo 41 to find the least positive residues:
65 ≡ (36)2 × 6 ≡ 25 × 6 ≡ 9 × 3 ≡ 27
1
5 3
3
6 5 ≡ (6 ) ≡ −14 ≡ 3
6
25
6
35
≡6
15
(mod 41)
5 2
× (6 ) ≡ 3 × 272 ≡ 14
15 2
5
(mod 41)
(mod 41)
2
≡ (6 ) × 6 ≡ 3 × 27 ≡ 38
(mod 41).
36. Let ind2 (−1) = n. Then 2n ≡ −1 (mod 211). So by 6-2, ord(2n ) = 2. Now use (6.3)(i) to get
2 = ord(2n ) =
ord(2)
210
=
,
(n, ord(2))
(n, 210)
ie. (n, 210) = 105, so n = 105.
37. (a) Taking ind2 on both sides of the congruence, we get
ind2 (9) + ind2 (x) = ind2 (14)
(mod φ(19)).
From the table, ind2 (9) = 8 and ind2 (14) = 7, so
ind2 (x) ≡ −1 ≡ 17
(mod 18).
From the table again, we get x ≡ 10 (mod 19).
(c) Taking ind2 on both sides of the congruence, we get
ind2 (5) + 6 ind2 (x) = ind2 (17)
(mod φ(19)).
From the table, ind2 (5) = 16 and ind2 (17) = 10, so
6 ind2 (x) ≡ −6
(mod 18),
or equivalently, using (2.2)(v),
ind2 (x) ≡ −1 ≡ 2
(mod 3).
Hence
ind2 (x) ≡ 2, 5, 8, 11, 14, 17
(mod 18).
From the table again, we get
x ≡ 4, 13, 9, 15, 6, 10
(mod 19).
46. Taking ind7 on both sides of the congruence1 to get
5 ind7 (x) ≡ ind7 (26) ≡ 45
(mod 70).
By (2.2)(v), this is equivalent to
ind7 (x) ≡ 9
(mod 14).
So
ind7 (x) ≡ 9, 23, 37, 51, 65
(mod 70)
and it follows that
x ≡ 79 , 723 , 737 , 751 , 765
≡ 47, 53, 22, 52, 39
(mod 71)
(mod 71).
47. (c) Multiply numbers on both sides of the congruence to reduce the coefficient of x20 to 1:
11 × 4x20 ≡ 11 × 23
20
x
≡ −5
(mod 43)
(mod 43).
Now apply (6.18). Note that we may simplify our calculations using Euler’s Criterion (5.9):
43−1
43−1
−5
−1
5
(−5) (20,42) = (−5) 2 =
≡
≡ (−1)(−1) ≡ 1 (mod 43)
43
43
43
by (5.11) and (5.13)(iv). So by (6.18), x20 ≡ −5 (mod 43) (and thus 4x20 ≡ 23 (mod 43)) has
exactly (20, 42) = 2 solutions.
49. Let p = 6k − 1. So (3, p − 1) = (3, 6k − 2) = 1. Hence
p−1
a (3,p−1) = ap−1 ≡ 1
(mod p)
by Fermat’s Theorem (3.6). So by (6.18), x3 ≡ a (mod p) has exactly (3, p − 1) = 1 solution (ie. a
unique solution).
E4.
(i) Let p = 6k + 1, q = 12k + 1 and s = 18k + 1. p, q, s are clearly distinct, and so if they are primes,
then m = pqs is square-free. Now note that
m − 1 = (6k + 1)(12k + 1)(18k + 1) − 1 = 1296k 3 + 396k 2 + 36k = 36k(36k 2 + 11k + 1)
and so is divisible by 6k = p − 1, 12k = q − 1 and 18k = s − 1. Hence m is Carmichael by Korselt’s
Criterion.
(ii) If k ≡ 2 (mod 5), then 12k + 1 ≡ 2k + 1 ≡ 5 ≡ 0 (mod 5), thus not prime. If k ≡ 3 (mod 5), then
18k+1 ≡ −2k+1 ≡ −5 ≡ 0 (mod 5), thus not prime. If k ≡ 4 (mod 5), then 6k+1 ≡ k+1 ≡ 5 ≡ 0
(mod 5), thus not prime.
(iii) By (ii), m cannot be a Carmichael number if k ≡ 2, 3, 4 (mod 5), so we must have k ≡ 0, 1
(mod 5), ie. k = 5r or k = 5r + 1 for some r ∈ N ∪ {0}. If k = 5r, then
m = (30r + 1)(60r + 1)(90r + 1).
(1)
m = (30r + 7)(60r + 13)(90r + 19).
(2)
If k = 5r + 1, then
1 If
the congruence does not have a solution, we will see it from the resulting linear congruence.
(iv) As in (ii): If r ≡ 1 (mod 7), then 90r + 1 ≡ −r + 1 ≡ 0 (mod 7), thus not prime. If r ≡ 3
(mod 7), then 30r + 1 ≡ 2r + 1 ≡ 7 ≡ 0 (mod 7), thus not prime. If r ≡ 5 (mod 7), then
60r + 1 ≡ −3r + 1 ≡ −14 ≡ 0 (mod 7), thus not prime. So if r ≡ 1, 3, 5 (mod 7), then m of the
form (1) is not Carmichael. If r ≡ 2 (mod 7), then 60r + 13 ≡ −3r − 1 ≡ −7 ≡ 0 (mod 7), thus
not prime. If r ≡ 5 (mod 7), then 90r + 19 ≡ −r − 2 ≡ −7 ≡ 0 (mod 7), thus not prime. If r ≡ 0
(mod 7) and r 6= 0, then 30r + 7 ≡ 0 (mod 7), thus not prime (note that 30r + 7 = 7 has been
excluded). So if r ≡ 0, 2, 5 (mod 7) and r 6= 0, then m of the form (2) is not Carmichael.
E9. Given L = 360 and P = {7, 11, 13, 19, 31, 37, 41, 61, 73, 181}. Following the counting argument in
Example 1, since
2|P| − 1
210 − 1
=
≈ 11,
φ(360)
96
we should expect2 about 11 different products of primes in P to be congruent to 1 (mod 360). Some
examples (not exhaustive) of these are:
7 × 31 × 73 ≡ 1
(mod 360)
37 × 73 × 181 ≡ 1
7 × 11 × 13 × 41 ≡ 1
(mod 360)
(mod 360)
13 × 37 × 61 × 181 ≡ 1
7 × 11 × 13 × 37 × 41 × 73 × 181 ≡ 1
(mod 360)
(mod 360).
By Erdös’ Criterion, these are Carmichael numbers.
2 Note that the counting argument is only a heuristic and does not constitute a proof. So we may get more or less than what
is predicted by the argument.