Math 551: Homework Assignment #2
Solutions
1. Let p be a prime with p ≡ 3 (mod 4). Show that
p−1
! ≡ (−1)t (mod p)
2
where t is the number of positive integers less than p/2 which are quadratic nonresidues
modulo p.
Proof. Since ap = −1 when a is a quadratic nonresidue mod p (and otherwise equals 1),
then using Euler’s Criterion we have
1
2
(p − 1)/2
t
(−1) ≡
···
(mod p)
p
p
p
(p−1)/2
p−1
(p−1)/2 (p−1)/2
≡1
2
···
(mod p)
2
p − 1 (p−1)/2
!
(mod p)
≡
2
But
p−1 2
p−1 p−1
! ≡ 1 · 1 · 2 · 2···
2
2
2
(mod p)
p−1 p+1
(p−1)/2
≡ (−1)
1 · (p − 1) · 2 · (p − 2) · · ·
2
2
(mod p)
≡ (−1)(p−1)/2 (p − 1)!
≡ (−1)(p+1)/2
since Wilson’s Theorem gives (p − 1)! ≡ −1 (mod p). Then since p ≡ 3 (mod 4), we
know (p + 1)/2 is even, so that
p−1 2
! ≡ 1 (mod p),
2
and then
p−1
! ≡ ±1 (mod p).
2
Then since (p − 1)/2 is odd, we have
p − 1 (p−1)/2
p−1
t
(−1) ≡
!
≡
! (mod p).
2
2
2. Euler’s original statement of the Law of Quadratic Reciprocity was the following:
Suppose that p is an odd prime
a is an integer not divisible by p. If q is a prime
and
a
a
with p ≡ ±q (mod 4a), then p = q .
Prove that Euler’s statement is equivalent to the one we proved in class. That is, that
each implies the other.
Proof. Let (QR) be the statement of quadratic reciprocity we proved in class, and let
(Euler) be Euler’s statement, given above. Note in (Euler) that if p ≡ ±q (mod 4a) and
q | a then q | p, so if q 6= p we have q - a as well as p - a.
We first show that (QR) imples (Euler). If p = q then the conclusion of Euler’s
statement
is trivial, so suppose that p and q are distinct primes. If a = 1 then clearly
a
= aq , so we next consider a prime. If a = 2 and p ≡ ±q (mod 8), then p2 ≡ q 2
p
(mod 8), so
2
2
(p2 −1)/8
(q 2 −1)/8
= (−1)
= (−1)
=
.
p
q
Now suppose a is an odd prime and p ≡ q (mod 4a). Then p ≡ q (mod a), so using
(QR) we have
p q a
a−1 p−1
a−1 q−1
a
=
=
(−1) 2 2 =
(−1) 2 2 .
p
a
a
q
p−1
q−1
Since we also have p≡ q (mod
4), then 2 ≡ 2 (mod 2), so the powers of −1 on each
side are equal, thus ap = aq .
Next suppose p ≡ −q (mod 4a). Then p ≡ −q (mod a), so (QR) implies
p −q −1 q p−1 a−1
q−1 a−1
q+1 a−1
a−1
a
a
a
=
= (−1) 2
(−1) 2 2 =
=
(−1) 2 2 =
(−1) 2 2 .
p
a
a
a
a
q
q
Since p ≡ −q (mod 4), we have p−1
≡ q+1
(mod 2), so we have ap = aq .
2
2
Finally, suppose a is composite. Write a = r1 r2 · · · rm , where the ri are primes not equal
to p or q. Then p ≡ ±q (mod 4a) implies that p ≡ ±q (mod 4ri ) for each 1 ≤ i ≤ m.
Thus by our previous work,
a
r1
r2
rm
r1
r2
rm
a
=
···
=
···
=
.
p
p
p
p
q
q
q
q
Now we show that (Euler) implies (QR). Let p and q be distinct odd primes, and
without loss of generality assume p > q. Then p ≡ ±q (mod 4). First, suppose that
p ≡ q (mod 4). Then p = q + 4a for some integer a ≥ 1, so we can write p ≡ q (mod 4a).
Then using (Euler), we have
q + 4a
4a
4
a
a
p
=
=
=
=
,
q
q
q
q
q
q
and
q
p − 4a
−4a
−1
a
=
=
=
,
p
p
p
p
p
so
  q
if p ≡ q ≡ 1
a
a
−1
q
p
p
=
=
=
=
− q
q
q
p
p
p
if p ≡ q ≡ 3
p
(mod 4),
.
(mod 4)
Next consider the case p ≡ −q (mod 4). Then p = −q + 4a for some integer a ≥ 1, so
p ≡ −q (mod 4a). Then using (Euler) again,
p
−q + 4a
4a
a
=
=
=
,
q
q
q
q
and
q
4a − p
4a
a
=
=
=
,
p
p
p
p
so
p
a
a
q
=
=
=
.
q
q
p
p
Putting the two cases together gives us
 
− pq
if p ≡ q ≡ 3 (mod 4),
p
= q

q
otherwise.
p
By our discussion in class, this is equivalent to (QR).
3. Let p and q be odd prime numbers with q = 2p + 1.
(a) Prove that 2p is neither a quadratic residue, nor a primitive root modulo q.
Proof. First, we have (2p)2 = (q − 1)2 ≡ (−1)2 ≡ 1 (mod q). Since p ≥ 3 then q ≥ 7,
so 2 < q − 1 = φ(q), and hence 2p is not a primitive root modulo q. Next, since p is
odd it has
the form
p = 2k + 1, so q = 2(2k + 1) + 1 = 4k + 3, thus q ≡ 3 (mod 4).
2p
Hence q = −1
= −1, so 2p is a quadratic nonresidue modulo q. Further, by
q
problem 12 in Chapter 5 (done in class), every primitive root modulo q is a quadratic
nonresidue modulo q, and there are exactly q−1
− φ(q − 1) nonresidues that are not
2
q−1
primitive roots. Since q = 2p + 1, we have 2 − φ(q − 1) = p − φ(2p) = p − (p − 1) = 1.
So 2p is the only quadratic nonresidue that is not a primitive root modulo q.
(b) If p ≡ 1 (mod 4), prove that 2 is a primitive root modulo q.
Proof. By part (a), we only need to show that 2 is a quadratic nonresidue modulo
q that is incongruent to 2p modulo q. If p ≡ 1 (mod
4), then we can write q =
2(4k + 1) + 1 = 8k + 3, so q ≡ 3 (mod 8). Then 2q = −1, so 2 is a quadratic
nonresidue modulo q. Recall that 2p ≡ −1 (mod q). Since q > 3, 2 6≡ −1 (mod q),
so 2 is a primitive root modulo q.
(c) If p ≡ 3 (mod 4), prove that −2 is a primitive root modulo q.
Proof. If p ≡ 3 (mod
4), then q =2(4k + 3) + 1 = 8k + 7, so q ≡ 7 (mod 8), and q ≡ 3
2
= −1
= (−1)(1) = −1, so −2 is a quadratic nonresidue
(mod 4). Then −2
q
q
q
modulo q. Since q > 3, then −2 6≡ −1 (mod q), so −2 is a primitive root modulo
q.
(d) If a is an integer with 1 < a < q − 1, prove that q − a2 is a primitive root modulo q.
Proof. Recall that q ≡ 3 (mod 4). Then
q−a2
q
=
−a2
q
=
−1
q
= −1, so q − a2 is
a quadratic nonresidue mod q. If q − a2 ≡ 2p ≡ −1 (mod q), then a2 ≡ 1 (mod q),
hence a ≡ ±1 (mod q). This contradicts 1 < a < q − 1, so q − a2 is a primitive root
modulo q.
n
4. The nth Fermat number is Fn := 22 + 1. Show that if n ≥ 2, any prime divisor p of Fn
must be of the form 2n+2 k + 1. (Hint: first prove that ordp 2 = 2n+1 and that 2(p−1)/2 ≡ 1
(mod p).)
n
n
n+1
Proof. Suppose that p | 22 + 1. Then 22 ≡ −1 (mod p). Squaring, we have 22
≡1
n+1
j
(mod p), thus ordp 2 | 2 , so we must have ordp 2 = 2 for some 1 ≤ j ≤ n + 1. But
n
n
n
if j ≤ n then 22 = (2j )2 −j ≡ 1 (mod p), which contradicts 22 ≡ −1 (mod p). So we
must have ordp 2 = 2n+1 .
Now, since ordp 2 must divide p − 1, we have 2n+1 | (p − 1), so p=2n+1 m + 1 for some
integer m. Then since n ≥ 2, we have p ≡ 1 (mod 8), and thus
Euler’s Criterion, we have 2(p−1)/2 ≡ 1 (mod p).
, so 2n+1 k =
Finally, we must now have ordp 2 | p−1
2
p = 2n+2 k + 1.
p−1
2
2
p
= 1. Therefore by
for some integer k, hence