SOLUTIONS TO HOMEWORK 9
1. Suppose that p and q are prime numbers, and m is an integer.
(a). If p does not divide q − 1, prove that every element of Z/qZ is a pth power.
In particular, the equation
m ≡ xp (mod q)
always has a solution.
Ans: It’s clear that 0 is its own pth power, so we restrict our attention to non-zero
elements. Let a denote a primitive root mod q. Let m denote a non-zero element
of Z/qZ. We can write m ≡ ab (mod q) for some b. Because p is prime and does
not divide q − 1, gcd (p, q − 1) = 1. So, we can write px + (q − 1)y = 1 for some
integers x and y, and multiplying both sides by b, we get pw + (q − 1)z = b for
some w and z. Hence m ≡ ab = apw a(q−1)z ≡ apw = (aw )p (mod q), and so m is
indeed a pth power. (Note that z or w being negative is not a problem, since m is
invertible modulo q.)
(b). Suppose that p divides q − 1, and that m 6≡ 0 (mod q). Prove that
m ≡ xp
(q−1)/p
has a solution if and only if m
(mod q)
≡ 1 (mod q).
Ans: ⇒ We assume that there is an x such that m ≡ xp (mod q). Then
m(q−1)/p ≡ xq−1 ≡ 1
(mod q).
(q−1)/p
⇐ Now assume that m
≡ 1 (mod q). Write m ≡ ab (mod q), where a is
a primitive root. So by assumption, ab(q−1)/p ≡ 1 (mod q). So this implies that
q − 1 (the order of a) divides b(q − 1)/p, and hence p divides b. Thus m ≡ (ab/p )p
(mod q), as desired.
2. Suppose that p and q are distinct odd primes. Prove that there is never a
“primitive root modulo pq”; that is, that there is no element of (Z/pqZ)× having
order equal to (p − 1)(q − 1).
Ans: If xb ≡ 1 (mod pq) for a non-zero natural number b, then neither p nor q
can divide x, since they don’t divide 1. In particular, because it’s relatively prime
to p, we have xp−1 ≡ 1 (mod p). Now, note that p − 1 and q − 1 are both even. So
q − 1 divides (p − 1)(q − 1)/2, and hence (xp−1 )(q−1)/2 ≡ 1 (mod q). And of course
it’s also ≡ 1 (mod p). To save space, write y for x(p−1)(q−1)/2 . So, p divides y − 1
and q divides y − 1. Because these are distinct primes, pq also divides b − 1. Hence
our initial element x has order strictly less than (p − 1)(q − 1).
3. Suppose that p1 , . . . , pr are distinct prime numbers. Find the smallest positive
integer m leaving remainder p1 − 1 on division by p1 , p2 − 1 on division by p2 ,. . . ,
and pr − 1 on division by pr . Prove that your answer is correct.
Ans: It’s clear that p1 p2 · · · pr − 1 satisfies the stated division properties. Why
is it the smallest such positive integer? Call the above number x. Assume we had
another, smaller positive integer with the given properties; call it y. Because x and
y leave the same remainder upon division by pi for each i, we know that x − y is
1
2
SOLUTIONS TO HOMEWORK 9
actually divisible by each pi . Hence it’s divisible by their product. But it’s also less
than x which is less than their product, which is a contradiction.
4. Here is another way to state quadratic reciprocity. You can use quadratic
reciprocity as stated in the text to do this problem.
Suppose p and q are odd primes. Define p = (−1)(p−1)/2 . Prove that p p is a
square modulo q if and only if q is a square modulo p.
Ans: Let’s divide the problem into two cases:
Case 1: p ≡ 1 (mod 4), and hence p = p p.
⇒ So, p is a square modulo q and p ≡ 1 (mod 4), and thus quadratic reciprocity
says that q is a square modulo p.
⇐ Since q is a square modulo p, and since p ≡ 1 (mod 4), quadratic reciprocity
says that p is a square modulo q.
Case 2: p ≡ 3 (mod 4), and hence −p = p p.
We use the multiplicative property described at the bottom of page 162. Then
−1
p
−p
=
.
q
q
q
So, −p is a square modulo q if and only if −1
= pq . For q ≡ 1 (mod 4), the
q
last equality holds if and only if p is a square modulo q which holds by quadratic
reciprocity if and only if q is a square modulo p. For q ≡ 3 (mod 4), the last
equality holds if and only if p is not a square modulo q which holds by quadratic
reciprocity if and only if q is a square modulo p.
Alternative Ans: Here’s a solution a few students found which I like because it
covers all the cases at once: p
q
It suffices to show that pq
p = 1. Using quadratic reciprocity formulas found
on pages 162-166, we have:
q
p
p
q
p p
=
q
p
q
q
p
p−1 q−1
p
=
(−1) 2 2
q
p−1 p−1 q−1
(−1) 2
=
(−1) 2 2
q
p−1 q−1
p−1 q−1
2
= (−1) 2
(−1) 2 2
= 1 ∗ 1 or (−1) ∗ (−1)
= 1.
5. Problem 4 says that (for fixed p) the question of whether p p is a square modulo
q depends only on the class of q modulo p. How can this be consistent with the
results of problems 9.8.3 and 9.8.5 in the text, where the answers depend on the
class of q modulo 4p?
Ans: Those problems in the book involve p p = −3 and 5. For the case of −3,
our problem 4 says that −3 is a square modulo q for an odd prime q if and only if
q ≡ 1 (mod 3). Problem 9.8.3 in the book says that −3 is a square modulo q for
an odd prime q if and only if q ≡ 1 (mod 12) or q ≡ 7 (mod 12). So it seems to
SOLUTIONS TO HOMEWORK 9
3
be omitting the cases q ≡ 4 (mod 12) and q ≡ 10 (mod 12), but of course, in those
cases, q is even. The case p p = 5 is practically identical.