MA2316 Introduction to Number Theory
Assignment 9
Ewan Dalby, Kimara Lynch, John Madden, Max LaVictorie, Keith
Glennon
3.1
Q.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 p−1 and that the nonresidues are
congruent to g, g 3 , , g p−2 . Thus there are equally many residues and non-residues
for an odd prime.
g is a primtive root, so hgi ={1, 2....., p − 1}
Let a = g {2k} , k ∈ 1, 2, ....., p−1
2
2k p−1
g
≡ g 2k( 2 ) = g kp · g −k ≡ g k · g −k ≡ 1
p
∴ g {2k} is a quadratic residue mod p, ∀k ∈ 1, 2, , p−1
2
Let b = g 2k−1 , k ∈ {1, 2, ....., (p − 1)/2} .
2k−1 1−p
1−p
p−1
p−1
g
≡ g (2k−1)( 2 ) = g kp · g −k · g ( 2 ) ≡ 1 · g ( 2 ) ≡ g ( 2 ) (g p−1 )−1 ≡
p
g(
p−1
2
) ≡ ±1.
But 1 < p−1
2 < p − 1 , and g is a primitve root of p
p−1
p−1
p−1
So g
≡ 1 ⇒ g 2 6≡ 1 ⇒ g 2 ≡ −1
∴ g 2k−1 is a quadratic nonresidue mod p, ∀k ∈ 1, 2, , p−1
So there are
2
quadratic residues and p−1
quadratic
nonresidues
mod
p.
2
p−1
2
Q. 13 Denote quadratic residues by r, non residues by n. Prove that r1 r2
and n1 n2 are residues and that rn is a non residue for any odd prime p. Give
a numerical example to show that the product of two non residues is not necessarily a quadratic residue modulo 12.
r
= 1 is a quadratic residue
p n
p = −1 is a non quadratic residue
r 1 r2
= rp1 · rp2 = 1 · 1 = 1
p
∴ r1 r2 is a quadratic residue.
n1 n2
= np1 · np2 = −1 · −1 = 1
p
∴ n1 n2 is a quadratic residue.
rn
= pr · np = 1 · −1 = −1
p
∴ rn is a non residue.
The only quadratic residues are 0, 1, 4, 9. So 2, 3, 6 are non residues and 2·3 = 6
Q.20 Let p be an odd prime. Prove that if there is an integer x such that
1
p|(x2 + 1) then p ≡ 1(mod4);
p|(x2 − 2) then p ≡ 1 or 7(mod8);
p|(x2 + 2) then p ≡ 1 or 3(mod8);
p|(x4 + 1) then p ≡ 1(mod8);
Show that there are infinitely many primes of each of the forms 8n + 1, 8n +
3, 8n + 5, 8n + 7
p | (x2 + 1) ⇒ x2 ≡ −1 mod p ⇒ −1
= (−1)p−1/2 = 1 ⇒ p ≡ 1 mod 4.
p
p | (x2 − 2) ⇒ x2 ≡ 2 mod p ⇒ p2 = 1 ⇒ p ≡ 1 or 7 mod 8.
−1
= p2
=1
p | (x2 + 2) ⇒ x2 ≡ −2 mod p ⇒ −2
p
p
⇒ p ≡ 1 or 3 mod 8.
Note p ≡ 3 mod 8 works because p2 = −1
= −1.
p
a
2
p | (x4 +1) ⇒ x4 ≡ −1 mod p ⇒ −1
=
1
⇒
∃a
st
a
≡
−1
mod
p
and
p
p =
ap−1/2 = (−1)p−1/4 = 1 ⇒ p ≡ 1 mod 8.
Next suppose there exists only finitely many primes(say k) of the form 8n + 1.
Let x = p1 p2 . . . pk where pi is of the form 8n + 1.
Consider q a prime divisor of x4 + 1.
Clearly q is not one of the pi but q ≡ 1 mod 8 which contradicts the our assumption.
So there exist infinitely many primes of the form 8n + 1.
Now assume there exists only finitely many primes(again k) of the form
8n + 7.
Let x = p1 p2 . . . pk where pi is of the form 8n + 7.
Then all prime divisors of x2 − 2 are of the form 8n + 1 or 8n + 7, but
x2 − 2 ≡ 7 mod 8
⇒ ∃ a prime q | (x2 − 2), q ≡ 7 mod 8.
However q again can not be one of the pi ⇒ there are infinitely many primes of
the form 8n + 7.
We use a similar argument to show there are infinitely many primes of the form
8n + 3 looking for prime divisors of x2 + 2.
Finally if there are only k primes of the form 8n + 5 let x = 2 ∗ p1 p2 . . . pk where
pi is of the form 8n + 5.
Then x2 + 1 ≡ 5 mod 8 and it only has prime factors of the forms 8n + 1 and
8n + 5
⇒ ∃ a prime q | (x2 + 1), q ≡ 5 mod 8 but q is not one of the pi
⇒ there are infinitely many primes of the form 8n + 5.
3.2
Q1.Verify that x2 ≡ 10 mod 89 is solvable
89 is an
odd2 prime
5
= 89
· 89
5
89 ≡ 1(mod8) ⇒ 89
=
10
89
−1
5
= 1,as 5 is an odd prime congruent to 1 mod 4.
2
892 −1
2
= −1 8 = 1, Second Supplement to the Law of Quadratic Reciprocity,
89
p2 −1
2
8
p = −1
10
∴ 89 = 1
∴ 10 is a quadratic residue mod89andx2 ≡ 10 mod 89 has solutions
x2 ≡ 10 + 89k mod 89
x2 ≡ 10 + 89 · 10 mod 89
x2 ≡ 900 mod 89
x ≡ 30 mod 89
Also,x2 ≡ 10 + 89 · 39 mod 89
x2 ≡ 3481 mod 89
x ≡ 59 mod 89
Q2.Prove that if p and q are distinct primes of the form 4k+3 and if x2 ≡
p mod q has no solution then x2 ≡ q mod p has two solutions
p
q
= (−1) ∗
= (−1) ∗ (−1) = 1
p
q
so the equation x2 ≡ q mod p has two solutions as it has one solution x and
another distinct solution −x.
Q3.Prove that if a prime p is a quadratic residue of an odd prime q and p is
of the form 4k+1 the q is a quadratic residue of p
We have
Therefore
p−1 q−1
p
= 1 and (−1) 2 2 = 1
q
p−1 q−1
p
q
=
∗ (−1) 2 2 = 1
p
q
So q is a quadratic residue mod p.
Q4.Which of the following congruences are solvable?
(a) x2 ≡ 5 mod 227
227
2
5
=
=
= −1
227
5
5
so the congruence has no solutions.
(b) x2 ≡ 5 mod 229
5
229
4
=
=
=1
229
5
5
so the congruence has a solution.
(c) x2 ≡ −5 mod 227
−5
−1
227
2
=
∗
= (−1) ∗
=1
227
227
5
5
3
so the congruence has a solution.
(d) x2 ≡ −5 mod 229
−1
229
=1
so -1 is a quadratic residue and so is 5 so therefore −5 = −1 ∗ 5 is also.
(e) x2 ≡ 7 mod 1009
1009
1
7
=
=
=1
1009
7
7
so the congruence is a solution.
(f) x2 ≡ −7 mod 1009
−1
1009
=1
so -1 is a quadratic residue and so is 7 so therefore −7 = −1 ∗ 7 is also.
3.3
Q3. Which of the following congruences are solvable?
(a) x2 ≡ 11 mod 61
11
61
6
2
3
11
−1
=
=
=
∗
= (−1)∗
∗(−1) =
= −1
61
11
11
11
11
3
3
so the equation has no solutions.
(b) x2 ≡ 42 mod 97
42
7
3
2
97
97
−1
1
=
∗
∗
=
∗
=
∗
= −1
97
97
97
97
7
3
7
3
So again there are no solutions.
(c) x2 ≡ −43 mod 79
−43
79
=
−1
79
43
79
36
6 2
∗
= (−1) ∗
∗ (−1) =
=(
) =1
79
43
43
43
(d) x2 − 31 ≡ 0 mod 103
103
10
2
5
31
1
31
=−
=−
=−
∗
=
=
=1
103
31
31
31
31
5
5
So the congruences in parts (c) and (d) all have solutions.
Q5.
Prove that
p−1 X
j
j=1
p
= 0, p an odd prime
We had in class that an odd prime p has
an equal number of quadratic
P i p−1
p−1
residues and quadratic non-residues so
p = 2 ∗ 1 + 2 ∗ (−1) = 0
4
Extra Question: prove that the congruence ax2 +bx+c = 0 mod p with (a, p) =
1 has two solutions, one solution and zero solutions when the Legendre symbol
D
2
p = 1, 0, −1 respectively, where D = b − 4ac; do there exist solutions to the
2
congruence x + x + 1 = 0 mod 637 ?
Let p be an odd prime. For the equation
ax2 + bx + c = 0 we have solutions
√
−b± b2 −4ac
given by the quadratic formula: x =
.
2a
This formula is easily checked by finding the polynomial with these roots and
verifying that it is ax2 + bx + c = 0.
This formula also holds modulo p if (a, p) = 1 so solutions exist in Z/pZ iff
∃y∈ Z/pZ
st y 2 = D where D = b2 − 4ac.
D
p
= 1 then we have two solutions from the above formula since x + y ≡
= 1 ⇒ D/2a 6≡ 0 Mod p.
x − y Mod p ⇒ y ≡ 0 Mod p and D
p
D
If p = 0 then we have one solution: x = −b/2a
If D
= −1 then we have no solutions. To see this notice if there are
p
If
solutions, say ζ and ζ 0 , then ax2 + bx + c = a(x − ζ)(x − ζ 0 ) ⇒ a(ζ + ζ 0 ) =
2
−b, aζζ 0 = c ⇒ a√
(ζ − ζ 0 )2 = a(aζ 2 − 2aζζ 0 + aζ 02 ) = a(−bζ − c − 2c − bζ 0 − c) =
2
b − 4ac = D ⇒ D = a(ζ − ζ 0 ) ∈ Z/pZ ⇒⇐
In this case D = −3 and solutions exist mod 637 if they exist mod 49 and
mod 13 by the chinese
theorem.
remainder
−1
13
Modulo 13: −3
=
∗
= 1.
13 13
3
2 2
=
(
)
=
1.
In
particular 2 is a solution and trying lifts
Modulo 49: −3
7
7
of 2 we obtain 30 as a solution.
Therefore the equation has solutions.
5