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18.781 Homework 8
Due: 8th April 2014
Q 1 (3.1(10)). Prove that if p is an odd prime, then x2 ≡ 2(mod p) has solutions if and only if p ≡ 1(mod 8)
or if p ≡ 7(mod 8).
2
Proof. By Theorem 3.3, p2 = (−1)(p −1)/8 . As p is an odd prime, p = 8k + r for r ∈ {1, 3, 5, 7}. p2 − 1 =
2
2
64k 2 + 16kr + r2 − 1, so (−1)(p −1)/8 = (−1)(r −1)/8 . (r2 − 1)/8 is even for r ∈ {1, 7} and odd for r ∈ {3, 5}.
So 2 is a square modulo p if and only if p ≡ 1 or 7 (mod 8).
Q 2 (3.1(11)). Let g be a primitive root of an odd prime p. Prove that the quadratic residues modulo p are
congruent to g 2 , g 4 , g 6 , · · · , g p−1 and that the nonresidues are congruent to g, g 3 , g 5 , · · · , g p−2 . Thus there
are equally many residues and nonresidues for any odd prime.
Proof. Let a be a residue class such that (a, p) = 1. Then a ≡ g i (mod p) for a unique i ∈ {0, 1, · · · , p − 1}.
Let x(mod p) be a solution to x2 ≡ a(mod p). Then x ≡ g k (mod p) for some k ∈ {0, 1, · · · , p − 1}, and we
have g 2k ≡ g i (mod p). This implies 2k ≡ i(mod p − 1). As p − 1 and 2k are both even, this forces i to be
even. Conversely, if i = 2k, then g k (mod p) satisfies x2 ≡ a(mod p).
Q 3 (3.1(20)). Let p be an odd prime. Prove that if there is an integer x such that
p|(x2 + 1) then p ≡ 1(mod 4)
p|(x2 − 2) then p ≡ 1 or 7(mod 8)
p|(x2 + 2) then p ≡ 1 or 3(mod 8)
p|(x4 + 1) then p ≡ 1(mod 8)
Show that there are infinitely many primes of each of the forms 8n + 1, 8n + 3, 8n + 5, 8n + 7.
2
Proof.
p | x + 1 for some integer x if and only if −1 is a square modulo p and this is true if and only if
−1
= (−1)(p−1)/2 = 1. So p | x2 + 1 for some integer x if and only if (p − 1)/2 is even, or in other words,
p
if and only if p ≡ 1(mod 4).
p | x2 − 2 for some integer x if and only if the equation x2 ≡ 2(mod p) is solvable and by Problem 3.1(10),
this is possible if and only if p ≡ 1 or 7 (mod 8).
2
p | x +2 for
some
integer x if and only if −2 is a square modulo p and this is true if and only
if
−2
p
=
−1
p
2
2
p
= (−1)(p−1)/2 (−1)(p
2
−1)/8
= 1. So p | x2 + 2 for some integer x if and only if
(p − 1)/2 ≡ (p − 1)/8(mod 2) and looking at congruence classes of p modulo 8 (p = 8k + r for some
r ∈ {1, 3, 5, 7}), we see that this is true precisely when p ≡ 1 or 3(mod 8).
p | x4 + 1 for some integer x if and only if the equation x4 ≡ −1(mod p) is solvable and by Theorem 2.37,
this is posisble if and only if (−1)(p−1)/(4,p−1) ≡ 1(mod p). As p is odd, this is equivalent to saying p | x4 + 1
for some integer x if and only if (−1)(p−1)/(4,p−1) = 1, or equivalently, if and only if (p − 1)/(4, p − 1) =
(p−1)/2
(2,((p−1)/2)) is even.
This first forces (p−1)/2 to be even (if (p−1)/2 is odd, then (2, ((p−1)/2)) = 1 and
is odd). So if p = 8k + r, then r ∈ {1, 5}.
1
(p−1)/2
(2,((p−1)/2))
= (p−1)/2
If p = 8k + 5, then (p − 1)/2 = 4k + 2 and (2, ((p − 1)/2)) = 2, so (p − 1)/(4, p − 1) = 2k + 1 which is
odd. If p = 8k + 1, then (p − 1)/2 = 4k and (2, ((p − 1)/2)) = 2, so (p − 1)/(4, p − 1) = 2k which is even.
This shows that p | (x4 + 1) for some integer x if and only if p ≡ 1(mod 8).
Suppose there are only finitely many primes of the form 8n + 1. Let them be {p1 , p2 , . . . , pk }. Let p be a
prime factor of (2p1 p2 . . . pk )4 + 1. It has to be odd, and by what we have showed above must be of the form
8n + 1. But p 6= pi for any i, as pi - (2p1 p2 . . . pk )4 + 1, which is a contradiction. So there must be infinitely
many primes of the form 8n + 1.
Suppose there are only finitely many primes of Q
the form 8n + 3. Let them be {p1 , p2 , . . . , pk }. Let
a = (2p1 p2 . . . pk )2 + 2 = 2 ∗ (2(p1 p2 . . . pk )2 + 1) = 2 qiαi for some odd primes qi and positive integers αi .
By what we have showed before, each qi is either ofQthe form 8n + 1 or 8n + 3. We claim that there is at
least one prime factor p of the form 8n + 3. If not, qiαi ≡ 1(mod 8) (as a product of numbers congruent
to 1(mod 8) is still 1(mod 8)) and a ≡ 2(mod 8). But (2p1 p2 . . . pk )2 ≡ 4(mod 8) as p2j ≡ 1(mod 8) for every
j, and a = (2p1 p2 . . . pk )2 + 2 ≡ 6(mod 8), so this is a contradiction. At least one of the prime factors of a,
say p, must be of the form 8n + 3. But p 6= pj for any j, as pj - (2p1 p2 . . . pk )2 + 2, which is a contradiction.
So there must be infinitely many primes of the form 8n + 3.
Suppose there are only finitely many primes of Q
the form 8n + 7. Let them be {p1 , p2 , . . . , pk }. Let
a = (4p1 p2 . . . pk )2 − 2 = 2 ∗ (8(p1 p2 . . . pk )2 − 1) = 2 qiαi for some odd primes qi and positive integers αi .
By what we have showed before, each qi is either Q
of the form 8n + 1 or 8n + 7. We claim that there is at
least one prime factor p of the form 8n + 7. If not, qiαi ≡ 1(mod 8) (as a product of numbers congruent to
1(mod 8) is still 1(mod 8)) and a ≡ 2(mod 8). But (4p1 p2 . . . pk )2 ≡ 0(mod 8) and a = (4p1 p2 . . . pk )2 − 2 ≡
−2 ≡ 6(mod 8), so this is a contradiction. At least one of the prime factors of a, say p, must be of the form
8n + 7. But p 6= pj for any j, as pj - (4p1 p2 . . . pk )2 + 2, which is a contradiction. So there must be infinitely
many primes of the form 8n + 7.
Suppose there are only
Q finitely many primes of the form 8n + 5. Let them be {p1 , p2 , . . . , pk }. Let
a = (2p1 p2 . . . pk )2 + 1 = qiαi for some odd primes qi and positive integers αi . By what we have showed
before, each qi is either of the form 8n + 1 or 8n + 5 (p ≡ 1(mod 4) implies
Q p ≡ 1 or 5(mod 8)). We claim
that there is at least one prime factor p of the form 8n + 5. If not, a = qiαi ≡ 1(mod 8) (as a product of
numbers congruent to 1(mod 8) is still 1(mod 8)). But (2p1 p2 . . . pk )2 ≡ 4(mod 8) as p2j ≡ 1(mod 8) for every
j, and a = (2p1 p2 . . . pk )2 + 1 ≡ 5(mod 8), so this is a contradiction. At least one of the prime factors of a,
say p, must be of the form 8n + 5. But p 6= pj for any j, as pj - (2p1 p2 . . . pk )2 + 1, which is a contradiction.
So there must be infinitely many primes of the form 8n + 5.
Q 4 (3.2(7)). Find all primes p such that x2 ≡ 13(mod p) has a solution.
Proof. p = 2 has a solution as 12 ≡ 13(mod 2) and p = 13 has a solution given by x = 0(mod 13). From
now on we will look for odd primes p, different from13 such that x2 ≡ 13(mod p) has a solution. We are
looking for odd primes p, different from 13 such that 13
= 1. As 13 ≡ 1(mod 4), by quadratic reciprocity,
p
p
p
13
= 1. 13
= p6 (mod 13). Looking at residue classes of p modulo 13, we see that
= 1 if and only if 13
p
p6 ≡ 1(mod 13) if and only if p ≡ 1, 3, 4, 9, 10, 11(mod 13). (It is enough to test for p ≡ 1, 2, 3, 4, 5, 6(mod 13)
as r6 ≡ (−r)6 (mod 13) which implies primes congruent to r modulo 13 satisfy 13 | p6 − 1 if and only if
primes congruent to 13 − r modulo 13 satisfy 13 | p6 − 1). x2 ≡ 13(mod p) has a solution if and only if p = 2
or p = 13 or p ≡ 1, 3, 4, 9, 10, 11(mod 13).
Q 5 (3.2(9)). Find all primes q such that 5q = −1.
Proof. q is odd. Clearly q 6= 5. By quadratic reciprocity, 5q = −1 if and only if 5q = −1. Now,
q
2
2
5 = q (mod 5). Looking at residue classes of q(mod 5), we see that q ≡ −1(mod 5) if and only if q ≡ 2
or 3(mod 5).
Q 6 (3.2(13)). Prove that there are infinitely many primes of each of the forms 3n + 1 and 3n − 1.
2
Proof. We first look for odd
primes p, different from 3, such that x2 ≡ −3(mod p) is solvable. This equation
is solvable if and only if −3
= 1. Using quadratic reciprocity and multiplicativity of the Legendre symbol,
p
we see
(p−1)/2 p
(p−1)/2
−3
3
= −1
= p(mod 3)
p
p
p = (−1)
3 (−1)
= 1 if and only if p ≡ 1(mod 3).
So −3
p
Suppose there are only finitely many primes of the form 3n + 1. Let them be p1 , p2 , . . . , pk . Consider
a = (2p1 p2 . . . pk )2 + 3. a is odd, so all its prime factors are also odd. Let p be any prime factor of a. As
x2 ≡ (−3)(mod p) has a solution given by the integer 2p1 p2 . . . pk , we see that p ≡ 1(mod 3). But p 6= pi for
any i as pi - a (pi ≥ 7), which is a contradiction. So there must be infinitely many primes of the form 3n + 1.
Suppose ther are only finitely many primes of the form 3n − 1. Let them be 2, p1 , p2 , . . . , pk . Consider
a = 3p1 p2 . . . pk + 2. Then a ≡ (−1)(mod 3) and a is odd. There has to be at least one prime factor p of
a such that p ≡ −1(mod 3) (this is because a product of numbers congruent to 1(mod 3) is still 1(mod 3)).
Let p be such a prime factor of a. But p 6= pi for any i as pi - a (pi ≥ 5), which is a contradiction. So there
must be infinitely many primes of the form 3n − 1.
−23
83
Q 7 (3.3(1)). Evaluate:
51
71
,
71
73
,
−35
97
,
.
Proof. 3, 5, 7, 23, 83, 17, 71, 73, 97 are all prime.
−23
83
(−1)(−1)
−1
83
2
(7 −1)/8
=
=
3
71
71
73
=
73
71
=
−1
97
=
= −
23
83
83
23
=
14
23
=
2
23
=
7
23
2
= (−1)(23
−1)/8
23
7
(−1)
= (−1)
2
7
=
= −1
51
71
−35
97
23
83
17
71
= (−1)
2
71
5
97
71
3
71
17
2
= (−1)(71
7
97
=
97
5
= (−1)
−1)/8
97
7
2
3
3
17
=
17
3
=
2
3
= −1
=1
=
2
5
6
7
= (−1)(−1) = 1
Q 8 (3.3(3)). Which of the following congruences are solvable?
(a) x2 ≡ 11(mod 61)
(b) x2 ≡ 42(mod 97)
(c) x2 ≡ −43(mod 79)
(d) x2 ≡ 31(mod 103)
Proof. As 61, 97, 79, 103 are all prime, the solvability of these congruences can be tested by computing the
corresponding Legendre symbol.
3
6
2
(112 −1)/8
= 61
(−1) 11
=
11 = 11 = 11
11 = (−1)
3
This tells us that the equation x2 ≡ 11(mod 61) is not solvable.
11
61
2
3
= −1
3 7
97 2
(972 −1)/8 97
= 97
= 13 67 = −1
97
97 = (−1)
3
7
This tells us that the equation x2 ≡ 42(mod 97) is not solvable.
42
97
43 79
36
= −1
79
79 = 43 = 43 = 1
This tells us that the equation x2 ≡ (−43)(mod 79) is solvable.
−43
79
5
2
(312 −1)/8
= (−1) 103
= (−1) 10
31
31 = (−1) 31
31 = (−1)(−1)
This tells that the equation x2 − 31 = 0(mod 103) is not solvable.
31
103
31
5
= (−1)
1
5
= −1
Q 9 (3.3(10)). Let k be odd. Prove that x2 ≡ k(mod 2) has exactly one solution. Furthermore, x2 ≡
k(mod 22 ) is solvable if and only if k ≡ 1(mod 4), in which case there are two solutions.
3
Proof. The only residue classes modulo 2 are 0(mod 2) and 1(mod 2). If k is odd, then 12 ≡ 1 ≡ k(mod 2).
If k is odd, either k ≡ 1(mod 4) or k ≡ 3(mod 4). There are four residue classes modulo 4 and of these
the square classes are the classes of 0(mod 4) and 1(mod 4) (This is true as 32 ≡ 12 ≡ 1(mod 4) and
02 ≡ 22 ≡ 0(mod 4)). So x2 ≡ k(mod 4) is solvable if and only if k ≡ 1(mod 4) and in this case the two
solutions are 1(mod 4) and 3(mod 4).
Q 10 (3.4(8)). Show
that the discriminant of the quadratic form (h1 x + k1 y)(h2 x + k2 y) is the square of
h1 h2 . Deduce that if h1 , h2 , k1 , k2 are all integers, then the discriminant is a perfect
the determinant k1 k2 square, possibly zero.
Proof.
(h1 x + k1 y)(h2 x + k2 y) = h1 h2 x2 + (h1 k2 + h2 k1 )xy + k1 k2 y 2
The discriminant is
h
(h1 k2 + h2 k1 )2 − 4(h1 k2 h2 k1 )2 = (h1 k2 − h2 k1 )2 = 1
k1
2
h2 k2 If h1 , h2 , k1 , k2 are all integers, then the discriminant
is the square of the integer (h1 k2 − h2 k1 ) which is itself
h1 h2 .
the determinant of the integer matrix k1 k2 4