MA2316 Introduction to Number Theory Assignment 9 Ewan Dalby ...

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.

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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

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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.

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