Chapter 9. Quadratic Residues. 9.1 Introduction. 9.2 Prime moduli

Transcription

1 Chater 9 Quadratic Residues 9.1 Introduction Definition 9.1. We say that a Z is a quadratic residue mod n if there exists b Z such that a b 2 mod n. If there is no such b we say that a is a quadratic non-residue mod n. Examle: Suose n = 10. We can determine the quadratic residues mod n by comuting b 2 mod n for 0 b < n. In fact, since ( b) 2 b 2 mod n, we need only consider 0 b [n/2]. Thus the quadratic residues mod 10 are 0, 1, 4, 9, 6, 5; while 3, 7, 8 are quadratic non-residues mod 10. Proosition 9.1. If a, b are quadratic residues mod n then so is ab. Proof. Suose a r 2, b s 2 mod. ab (rs) 2 mod. 9.2 Prime moduli Proosition 9.2. Suose is an odd rime. the quadratic residues corime to form a subgrou of (Z/) of index 2. Proof. Let Q denote the set of quadratic residues in (Z/). If θ : (Z/) (Z/) denotes the homomorhism under which r r 2 mod 9 1

2 then ker θ = {±1}, im θ = Q. By the first isomorhism theorem of grou theory, Thus Q is a subgrou of index 2: kerθ im θ = (Z/). Q = 1 2. Corollary 9.1. Suose is an odd rime; and suose a, b are corime to. 1. 1/a is a quadratic residue if and only if a is a quadratic residue. 2. If both of a, b, or neither, are quadratic residues, then ab is a quadratic residue; 3. If one of a, b is a quadratic residue and the other is a quadratic nonresidue then ab is a quadratic non-residue. 9.3 The Legendre symbol Definition 9.2. Suose is a rime; and suose a Z. We set 0 if a = 1 if a and a is a quadratic residue mod 1 if if a is a quadratic non-residue mod. ( ) ( ) 2 1 Examle: = 1, 3 4 ( ) 0 Proosition a b mod = = 3. b = ( a ) ( ) b. ( ) 1 = 1, = 1, 4 ( ) 1 = 0, = 1; ( ) b ; ( ) 3 = 1. 5 Proof. (1) and (2) follow from the definition, while (3) follows from the Corollary above. 9 2

3 9.4 Euler s criterion Proosition 9.4. Suose is an odd rime. a ( 1)/2 mod. Proof. The result is obvious if a. Suose a. ( a ( 1)/2 ) 2 = a 1 1 mod, by Fermat s Little Theorem. It follows that ±1 mod. Suose a is a quadratic residue, say a r 2 mod. a ( 1)/2 b 1 1 mod by Fermat s Little Theorem. These rovide all the roots of the olynomial Hence if a is a quadratic non-residue. f(x) = x ( 1)/2 1. a ( 1)/2 1 mod 9.5 Gauss s Lemma Suose is an odd rime. We usually take r [0, 1] as reresentatives of the residue-classes mod But it is sometimes more convenient to take r [ ( 1)/2, ( 1)/2], ie { /2 < r < /2}/ Let P denote the strictly ositive residues in this set, and N the strictly negative residues: P = {1, 2,..., ( 1)/2}, N = P = { 1, 2,..., ( 1)/2}. Thus the full set of reresentatives is N {0} P. Now suose a (Z/). Consider the residues ap = {a, 2a,..., 1 2 a}. Each of these can be written as ±s for some s P, say ar = ɛ(r)π(r), 9 3

4 where ɛ(r) = ±1. It is easy to see that the ma is injective; for π : P P π(r) = π(r ) = ar ±ar mod = r ±r mod = r r mod, since s and s are both ositive. Thus π is a ermutation of P (by the igeon-hole rincile, if you like). It follows that as r runs over the elements of P so does π(r). Thus if we multily together the congruences ar ɛ(r)π(r) mod we get on the left, and a ( 1)/2 1 2 ( 1)/2 on the right. Hence But ɛ(1)ɛ(2) ɛ(( 1)/2)1 2 ( 1)/2 a ( 1)/2 ɛ(1)ɛ(2) ɛ(( 1)/2) mod. a ( 1)/2 mod, by Euler s criterion. Thus we have established Theorem 9.1. Suose is an odd rime; and suose a Z. Consider the residues a, 2a,..., a( 1)/2 mod, choosing residues in [ ( 1)/2, ( 1)/2]. If t of these residues are < 0 then = ( 1) t. Remarks: 1. Note that we could equally well choose the residues in [1, 1], and define t to be the number of times the residue aears in the second half ( + 1)/2, ( 1). 2. The ma a ( 1) t is an examle of the transfer homomorhism in grou theory. Suose H is an abelian subgrou of finite index r in the grou G. We know that G is artitioned into H-cosets: G = g 1 H g r H. 9 4

5 If now g G then gg i = g j h i for i [1, r]. Now it is easy to see the argument is similar to the one we gave above that the roduct h = h 1 h r is indeendent of the choice of coset reresentatives g 1,..., g r, and the ma τ : G S is a homomorhism, known as the transfer homomorhism from G to S. If G is abelian which it is in all the cases we are interested in we can simly multily together all the equations gg i = g j h i, to get 9.6 Comutation of τ(g) = g r. ( ) 1 Proosition 9.5. If is an odd rime then ( ) { 1 1 if 1 mod 4, = 1 if 1 mod 4. Proof. The result follows at once from Euler s Criterion a ( 1)/2 mod. But it is instructive to deduce it by Gauss s Lemma. We have to consider the residues 1, 2,..., ( 1)/2 mod. All these are in the range N = [ ( 1)/2, ( 1)/2]. t = ( 1)/2; all the remainders are negative. Hence ( ) 1 = ( 1) ( 1)/2 { 1 if 1 mod 4, = 1 if 1 mod 4. It follows that Examle: According to this, ( ) ( ) 2 1 = =

6 (since 3 1 mod 4), ie 2 is a quadratic non-residue mod 3. Again ( ) ( ) 12 1 = = 1, since 13 1 mod 4. Thus 12 is a quadratic residue mod13. In fact it is easy to see that = 5 2 mod Comutation of ( ) 2 Proosition 9.6. If is an odd rime then ( ) { 2 1 if ±1 mod 8, = 1 if ±3 mod 8. Proof. We have to consider the residues 2, 4, 6,..., ( 1) mod. We have to determine the number t of these residues in the first half of [1, 1], and the number in the second. We can describe these two ranges as {0 < r < /2} and {/2 < r < }. Since it follows that Suose where r = 1, 3, 5, 7. Thus /2 < 2x < /4 < x < /2 t = /2 /4. = 8n + r, /2 = 4n + r/2, /4 = 2n + r/4. t r/2 + r/4 mod 2. The result follows easily from the fact that 0 for r = 1 1 for r = 3 r/2 = 2 for r = 5 3 for r = 7, while r/4 = { 0 for r = 1, 3 1 for r = 5,

7 Examle: Since 71 1 mod 8, ( ) 2 = 1, 71 Can you find the solutions of x 2 2 mod 71? Again Since 19 3 mod 8, ( ) 2 = So by Euler s criterion, mod 19. Checking, = = mod Comosite moduli Proosition 9.7. Suose m, n are corime; and suose a is corime to m and n. a is a quadratic residue modulo mn if and only if it is a quadratic residue modulo m and modulo n Proof. This follows at once from the Chinese Remainder Theorem. For Conversely, suose a r 2 mod mn = a r 2 mod m and a r 2 mod n. a r 2 mod m and a s 2 mod n. By the Chinese Remainder Theorem, we can find t such that t r mod m and t s mod n; and then t 2 r 2 a mod m and t 2 s 2 b mod n. 9.9 Prime ower moduli Proosition 9.8. Suose is an odd rime; and suose a Z is corime to. a is a quadratic residue mod e (where e 1) if and only if it is quadratic residue mod. Proof. The argument we gave above for quadratic residues modulo still alies here. 9 7

8 Lemma 9.1. If θ : (Z/ e ) which then (Z/ e ) t imes is the homomorhism under t t 2 mod e ker θ = {±}. Proof. Suose a 2 1 = (a 1)(a + 1) 0 mod e. a 1 and a + 1 = 2a = a, which we have excluded. If a + 1 then e a 1; and if a 1 then e a + 1. Thus a ±1 mod e. It follows that the quadratic residues modulo e corime to form a subgrou of index 2 in (Z/ e ), ie just half the elements of (Z/ e ) are quadratic residues modulo e. Since just half are also quadratic residues modulo, the result follows. Remark: For an alternative roof, we can argue by induction of e. Suose a is a quadratic residue mod e, say ie a r 2 mod e, a = r 2 + t e. Set s = r + x e. s 2 = r 2 + 2x e + x 2 2e r 2 + 2x e mod e+1 a + (t + 2x) e mod e+1 a e mod e+1 if t + 2x 0 mod, ie x = t/2 mod, using the fact that 2 is invertible modulo an odd rime. 9 8

9 Corollary 9.2. The number of quadratic residues in (Z/ e ) is φ( e ) 2 = ( 1)e 1. 2 The argument above extends to moduli 2 e with a slight modification. Proosition 9.9. Suose is an odd rime; and suose a Z is corime to. a is a quadratic residue modulo e (where e 1) if and only if it is quadratic residue modulo. Proof. The argument we gave above for quadratic residues modulo still alies here. Lemma 9.2. If θ : (Z/ e ) (Z/ e ) t imes is the homomorhism under which t t 2 mod e then Proof. Suose ker θ = {±}. a 2 1 = (a 1)(a + 1) 0 mod e. a 1 and a + 1 = 2a = a, which we have excluded. If a + 1 then e a 1; and if a 1 then e a + 1. Thus a ±1 mod e. It follows that the quadratic residues modulo e corime to form a subgrou of index 2 in (Z/ e ), ie just half the elements of (Z/ e ) are quadratic residues modulo e. Since just half are also quadratic residues modulo, the result follows. Remark: For an alternative roof, we can argue by induction of e. Suose a is a quadratic residue mod e, say ie a r 2 mod e, a = r 2 + t e. Set s = r + x e. s 2 = r 2 + 2x e + x 2 2e r 2 + 2x e mod e+1 a + (t + 2x) e mod e+1 a mod e+1 9 9

10 if t + 2x 0 mod, ie x = t/2 mod, using the fact that 2 is invertible modulo an odd rime. Corollary 9.3. The number of quadratic residues in (Z/ e ) is φ( e ) ( 1)e 1 =. 2 2 The argument above extends to moduli 2 e with a slight modification. Proosition Suose a is an odd integer. a is a quadratic residue modulo 2 e (where e 3) if and only if a 1 mod 8 Proof. It is readily verified that 1 is the only odd quadratic residue modulo 8; 3,5 and 7 are quadratic non-residues. We show by induction on e that if a is an odd quadratic residue modulo 2 e then it is a quadratic residue modulo 2 e+1. For suose say a r 2 mod 2 e, Let a = r 2 + t2 e. s = r + t2 e 1. s 2 r 2 + t2 e mod 2 e+1 = a. Corollary 9.4. The number of quadratic residues in (Z/2 e ) (where e 3) is φ(2 e ) = 2 e 3. 4 Remarks: 1. It is easy to see that f (where f < e) is a quadratic residue modulo e if and only if f is even. This allows us to determine whether residues that are not corime to the modulus are quadratic residues or not. Thus the quadratic residues modulo 24 are 0, 1, 4, 7, 17, 23, while the quadratic residues modulo 36 are 0, 1, 4, 9, 17, 31 (noting that the quadratic residue modulo 4 are 0, 1). 2. The inductive argument above is an examle of Hensel s Lemma. In the simlest case this says that if f(x) Z[x] then any solution of f(a) 0 mod such that f (a) is corime to can be extended (in a unique way) to a solution of f(a) 0 mod e for all e