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Math 7 Spring 2016
Instructor: Serin Hong
PROBLEM SET 4 SOLUTIONS
1. (10 pts) Prove, without using a computer, that 213 − 1 and 216 + 1 are both prime numbers.
Solution. Let us first prove that 213 − 1 is a prime number. Since 213 − 1 = 8191 is between 902 = 8100
and 912 = 8281, it is not a prime if and only if it has a prime factor q less than 91. By a proposition proved
in class we must have q ≡ 1 (mod 26). Hence we can check primality of 213 − 1 by testing q = 53, 79. Since
213 − 1 = 53 · 154 + 29 = 79 · 103 + 54, we conclude that 213 − 1 is a prime.
Now we prove that 216 + 1 is a prime number. Since 216 + 1 is easily seen to be between 216 = (28 )2 = 2562
and (28 + 1)2 , it is not a prime if and only if it has a prime factor q less than 28 + 1 = 257. By a
proposition proved in class we must have q ≡ 1 (mod 32). Hence we can check primality of 216 + 1 by
testing q = 97, 161, 199. Since 216 + 1 = 97 · 675 + 62 = 161 · 407 + 10 = 199 ∗ 329 + 66, we conclude that
216 + 1 is a prime.
2. (10 pts) Let p be a prime number. Suppose that a and n are positive numbers such that gcd(a, p) = 1 and
gcd(n, p − 1) = 1. Determine the number of solutions of the congruence xn ≡ a (mod p).
Solution. Let g be a primitive root modulo p. Since gcd(a, p) = 1, we can find an integer l such that
a ≡ g l (mod p). In addition, since x ≡ 0 (mod p) is not a solution, we can write x ≡ g k (mod p)
for some integer k. Now the congruence can be written as g kn ≡ g l (mod p), which is equivalent to
kn ≡ l (mod p − 1). This congruence can be solved in k by k ≡ ln−1 (mod p − 1) as gcd(n, p − 1) = 1,
yielding a unique solution x ≡ g k (mod p) of the congruence xn ≡ a (mod p).
3. (15 pts) Let a and m be relatively prime positive integers. Prove that a is a primitive root modulo m if
and only if aφ(m)/p 6≡ 1 (mod m) for all prime factors p of φ(m).
Solution. If a is a primitive root modulo m, the order of a modulo m is φ(m). Then for any prime factor
p of φ(m), we have aφ(m)/p 6≡ 1 (mod m) since φ(m)/p < φ(m).
For the converse, suppose that aφ(m)/p 6≡ 1 (mod m) for all prime factors p of φ(m). Let h be the order
of a modulo m, and let φ(m) = pe11 pe22 · · · perr be a prime factorization. Since h divides φ(m), h is of the form
Math 7 Spring 2016
Instructor: Serin Hong
h = pf11 pf22 · · · pfrr where fi ≤ ei for each i = 1, 2, · · · , r. If fi < ei for some i, then by compairing exponents
we see that h divides φ(m)/pi . This implies that aφ(m)/p ≡ 1 (mod m), which is a contradiction. Hence we
must have fi = ei for each i = 1, 2, · · · , r, or equivalently h = φ(m). This proves that a is a primitive root
modulo m, as desired.
4. (a) (5 pts) Prove that 3 is a primitive root modulo 17.
Solution. Since 17 is a prime, φ(17) = 17 − 1 = 24 . The first 8 powers of 3 (mod 17) are given by
31 ≡ 3, 32 ≡ 9, 33 ≡ 10, 34 ≡ 13, 35 ≡ 5, 36 ≡ 15, 37 ≡ 11, 38 ≡ −1 (mod 17).
In particular, 3φ(17)/2 = 38 6≡ 1 (mod 17), so by Problem 3 we deduce that 3 is a primitive root
modulo 17.
(b) (10 pts) Solve the congruence 8x5 ≡ 5 (mod 17).
Solution. Note that x ≡ 0 (mod 17) is not a solution. Since 3 is a primimtive root modulo 17,
we can write x ≡ 3k (mod 17) for some integer k > 0. As the first 8 powers of 3 (mod 17) are
computed in (a), we obtain 8 ≡ −9 ≡ 38 · 32 = 310 (mod 17) and 5 ≡ 35 (mod 17). Now the
congruence 8x5 ≡ 5 (mod 17) can be written as 310+5k ≡ 35 (mod 17), which is equivalent to
10 + 5k ≡ 5 (mod 16). This congruence can be solved by k ≡ 5−1 · (5 − 10) ≡ −1 ≡ 15 (mod 16).
Hence the congruence 8x5 ≡ 5 (mod 17) is solved by x ≡ 315 = 38 · 37 ≡ (−1) · 11 ≡ 6 (mod 17).
(c) (10 pts) Find all integers x such that 7x ≡ 4 (mod 17).
Solution. From the first 8 powers of 3 (mod 17) computed in (a) we find
7 ≡ −10 ≡ 38 · 33 = 311 (mod 17),
4 ≡ −13 ≡ 38 · 34 = 312 (mod 17).
Then the congruence 7x ≡ 4 (mod 17) can be written as 311x ≡ 312 (mod 17), which is equivalent
to 11x ≡ 12 (mod 16). This congruence is solved by x ≡ 11−1 · 12 ≡ 3 · 12 ≡ 4 (mod 16).
5. (10 pts) Let p be an odd prime number. For a given positive integer n, prove that
p
X
j=1
jn ≡



−1 (mod p)
if p − 1 | n


 0 (mod p)
otherwise.
Math 7 Spring 2016
Instructor: Serin Hong
Solution. If p − 1 divides n, then by Fermat’s little theorem j n ≡ 1 (mod p) for j = 1, 2, · · · , p − 1. Hence
p
X
j n ≡ 1 · (p − 1) + 0 = p − 1 (mod p).
j=1
Now suppose that p − 1 does not divide n. Let g be a primitive root modulo p. Using the fact that
{1, 2, · · · , p − 1} ≡ {g, g 2 , · · · , g p−1 } (mod p) we evaluate the sum modulo p as follows:
p
X
j=1
jn ≡
p−1
X
j=1
jn ≡
p−1
X
g kn =
k=1
g np − g n
(mod p).
gn − 1
By Fermat’s little theorem we have g np − g n ≡ 0 (mod p). On the other hand, since n is not divisible by
g np − g n
p − 1, we have g k − 1 6≡ 0 (mod p). Hence we deduce that n
is divisible by p, completing the proof.
g −1
6. (a) (10 pts) Let a and k be positive integers, with a > 1. Prove that φ(ak − 1) is divisible by k.
(Hint: What is the order of a modulo ak − 1?)
Solution. Note that ak ≡ 1 (mod ak − 1). On the other hand, for j = 1, 2, · · · , k − 1 we have
1 < aj < ak − 1. In particular, aj 6≡ 1 (mod ak − 1) for j = 1, 2, · · · , k − 1. Hence we deduce that
the order of a modulo ak − 1 is k. This implies that k divides φ(ak − 1), as desired.
(b) (10 pts) For any two relatively prime positive integers m and n, prove that gcd(2m − 1, 2n − 1) = 1.
(Hint: If a prime p divides both 2m − 1 and 2n − 1, consider the order of 2 modulo p.)
Solution. Suppose for contradiction that 2m − 1 and 2n − 1 have a common prime factor p. Then
we have 2m ≡ 1 (mod p) and 2n ≡ 1 (mod p). Therefore the order of 2 modulo p must divide both
m and n. Since m and n are relatively prime, the order of 2 modulo p must be 1. This implies that
21 ≡ 1 (mod p), which is clearly impossible. Hence we conclude that gcd(2m − 1, 2n − 1) = 1.
(c) (10 pts) For any positive integer n > 1, show that 2n − 1 is not divisible by n.
(Hint: Consider the order of 2 modulo p, where p is the smallest prime factor of n.)
Solution. Suppose for contradiction that there exists a positive integer n which divides 2n − 1. Let
p be the smallest prime factor of n. Then p also divides 2n − 1, so we have 2n ≡ 1 (mod p). Take
h to be the order of 2 modulo p. The congruence 2n ≡ 1 (mod p) implies that h divides n. On the
other hand, we know that h divides p − 1. Hence h is a common divisor of p − 1 and n. However,
p − 1 and n cannot have a common prime factor since p is the smallest prime factor of n. Hence we
must have h = 1, implying that 21 ≡ 1 (mod p). This yields a contradiction as desired.
Math 7 Spring 2016
Instructor: Serin Hong
Extra Credit Problem.
Recall that a positive number m is called a Carmichael number if it satisfies the following two conditions:
(i) m is a composite number;
(ii) am−1 ≡ 1 (mod m) for all integers a which are relatively prime to m.
(15 pts) Prove that m is a Carmichael number if and only if m is a product of at least two distinct primes
p1 , p2 , · · · , pr such that pi − 1 divides m − 1 for i = 1, 2, · · · , r.
Solution. Suppose first that m = p1 p2 · · · pr where r ≥ 2 and pi ’s are distinct primes such that pi − 1
divides m − 1. Let a be an integer relatively prime to m. Then a is relatively prime to the prime factors
pi ’s, so Fermat’s little theorem says that api −1 ≡ 1 (mod pi ). Since m is divisible by each pi − 1, we obtain
am ≡ 1 (mod pi ) for i = 1, 2, · · · , r. Now the Chinese remainder theorem yields am ≡ 1 (mod m), proving
that m is a Carmichael number.
For the converse, suppose that m is a Carmichael number with a prime factorization m = pe11 pe22 · · · perr .
Choose a primitive root gi modulo pei i for i = 1, 2, · · · , r. By the Chinese remainder theorem, there exists
an integer a such that a ≡ gi (mod pei i ) for i = 1, 2, · · · , r. Note that a is relatively prime to m since it is
relatively prime to each pei i . Since m is a Carmichael number, we have am−1 ≡ 1 (mod m) and therefore
am−1 ≡ 1 (mod pei i ) for i = 1, 2, · · · , r. On the other hand, a is taken to be a primitive root modulo pei i .
Hence we see that m − 1 is divisible by φ(pei i ) = (pi − 1)piei −1 for i = 1, 2, · · · , r. If ei > 1 for some i, then
pi divides both m and m − 1, which is a contradiction. Thus we must have ei = 1 for i = 1, 2, · · · , r, and
m − 1 is now divisible by φ(pi ) = pi − 1 for i = 1, 2, · · · , r.