MTH 417
Topics in Algebra
S08
2/20/07
Mid-Term Practice Exam/Solutions
#1. Find all solutions of x2 ≡ 2 (mod 49).
Since 32 = 9 ≡ 2 (mod 7), x1 = 3 is a solution of x2 ≡ 2 (mod 7). Let x2 = x1 + 7k =
3 + 7k. Then
x22
≡
2
(mod 49)
⇐⇒ (9 + 42k + 49k 2 ) ≡
2
(mod 49)
⇐⇒
42k
≡ −7
(mod 49)
⇐⇒
6k
≡ −1
(mod 7)
⇐⇒
−k
≡ −1
(mod 7)
⇐⇒
k
≡
(mod 7)
1
Thus x2 = 3 + 7 · 1 = 10 is a solution of x2 ≡ 2 (mod 49). Note that 49 is odd and has
a unique prime divisor. Thus by Lemma 7.1, x2 ≡ 2 (mod 49) has exactly two solutions
modulo 49. It follows that the solutions of x2 ≡ 2 (mod 49) are ±10 (mod 49).
#2. Is [89]131 in Q131 ?
We compute:
89
131
=
131
89
=
42
=
Law of Quadratic Reciprocity
since 42 ≡ 131
89
2
89
3
89
7
89
(mod 89)
Theorem 7.5
Since 89 ≡ 1 (mod 8), Corollary 7.10 gives:
2
=1
89
With similar arguments:
3
89
2 + 87
2
=
=
=
= −1
89
3
3
3
and
7
89
=
89
7
=
5 + 84
7
5
7
2
=
=
=
= −1
7
5
5
1
Thus
89
131
=
2
89
3
89
7
89
= 1 · (−1) · (−1) = 1
So the definition of the Legendre symbol shows that
[89]131 ∈ Q131
#3. Find all solutions of the system of linear congruences
x≡2
(mod 6) and x ≡ 11
(mod 15)
We have 6 = 2 · 3 and 15 = 3 · 5. In particular, gcd(6, 15) = 3. Since 2 ≡ 11 (mod 3),
the General Chinese Remainder Theorem 3.12 shows that our system of linear congruence
has a solution and is equivalent to
x≡2
(mod 2),
x≡2
(mod 3) and x ≡ 11
(mod 5)
Note that the general solution of the first two equation is x ≡ 2 (mod 6). So the system
is equivalent to
x≡2
(mod 6) and x ≡ 1
(mod 5)
Since −1 · 5 = −5 ≡ 1 (mod 6) and 1 · 6 ≡ 1 (mod 5) the proof of the Chinese remainder
theorem shows that
2 · (−1) · 5 + 1 · 1 · 6 = −10 + 6 = −4
is a solution of the system of linear congruences. Moreover, 6 · 5 = 30 and so the general
solution is
x ≡ −4 (mod 30)
#4. Find all positive integers n with φ(n) = 256.
By Corollary 5.7
256 = φ(n) =
Y
pe−1 (p − 1)
pe kn
Since 256 is a power of 2, each pe−1 and each p − 1 must a power of two. So if p is a
prime divisor of n, then p = 2k + 1 for some k and if p 6= 2 then e = 1. Thus
n = 2e q1 q2 . . . qk
where e ∈ N and q1 < q2 < . . . < qk are odd Fermat primes. Note that qi − 1 ≤ 256 and so
l
qi ≤ 257. Since Fermat primes are of the form 22 + 1 each qi is one of 3, 5, 17 or 257.
2
Suppose first that qk = 257. From (q1 − 1)(q2 − 1) . . . (qk − 1) ≤ 256 we conclude that
get k = 1. Moreover, either e = 0 or e = 1 and so
n = 257 or 2 · 257.
Suppose next that qk 6= 257. Since (3 − 1)(5 − 1)(17 − 1) = 21+2+4 = 27 < 28 = 256 we
see that e 6= 0. Moreover we can choose q1 , . . . qk to be any of the eight subsets of {3, 5, 17}
and get a unique choice for e such that φ(n) = 256. This gives eight more solutions:
29 ,
28 · 3,
27 · 5,
25 · 17,
26 · 3 · 5,
24 · 3 · 7,
23 · 5 · 17,
22 · 3 · 5 · 17
So there altogether 10 positive integers with φ(n) = 256.
#5. Find all Carmichael number of the form 7 · 19 · p, p a prime.
By Lemma 4.8 and it converse Theorem 6.15, 7 · 19 · p is a Carmichael number if and
only if p 6= 7, 19 and (7 · 19 · p) − 1 is divisible by 6 = 7 − 1, 18 = 19 − 1 and p − 1. Note
that this is the case if and only if
7 · 19 · p ≡ 1
(mod 18) and 7 · 19 · p ≡ 1
(mod p − 1)
Since 19 ≡ 1 (mod 18) and p ≡ 1 (mod p − 1), this is equivalent to
7·p≡1
(mod 18) and 7 · 19 ≡ 1
(mod p − 1)
We have 7 · 5 = 35 ≡ −1 (mod 18). So −5 is the unique inverse of 7 modulo 17. So we
obtain
p ≡ −5
(mod 18) and p − 1 | 7 · 19 − 1 = 132 = 22 · 3 · 11
The first condition gives
p = 13, 31, 49, 67 or p > 67
So
p − 1 = 12, 48, 66 or p − 1 > 66
The second condition shows that p − 1 6= 48 and if p − 1 > 66 = 132
2 , then p − 1 = 132
and p = 133. Since 133 is not a prime, this latter case is impossible. So p − 1 = 12 or 66
and p = 13 or 67. So the Carmichael numbers of the form 7 · 19 · p are
7 · 19 · 13 and 7 · 19 · 67
#6. Let p be prime with p ≡ −1 (mod 6). Prove that
(p − 4)! ≡
p+1
6
3
(mod p)
By Wilson Theorem (Corollary 4.5)
(p − 1)! ≡ −1
(mod p)
Note that p ≡ −1 (mod 6) implies that p ≥ 5. So we conclude
(p − 4)!(p − 3)(p − 2)(p − 1) ≡
−1
(mod p),
(p − 4)!(−3)(−2)(−1)
≡
−1
(mod p),
6(p − 4)!
≡
1
(mod p),
6(p − 4)!
≡ p+1
(mod p).
Since p ≡ −1 (mod p), 6 is relatively prime to p and 6 divides p + 1. So we can divide
by 6 and obtain:
(p − 4)! ≡
p+1
6
4
(mod p).