LECTURE 9: QUADRATIC RESIDUES AND THE LAW OF QUADRATIC RECIPROCITY


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1 LECTURE 9: QUADRATIC RESIDUES AND THE LAW OF QUADRATIC RECIPROCITY 1. Bsic roerties of qudrtic residues We now investigte residues with secil roerties of lgebric tye. Definition 1.1. (i) When (, m) 1 nd x n (mod m) hs solution, then we sy tht is n nth ower residue modulo m. (ii) When (, m) 1, we sy tht is qudrtic residue modulo m rovided tht the congruence x 2 (mod m) is soluble. If the ltter congruence is insoluble, then we sy tht is qudrtic nonresidue modulo m. Theorem 1.2. Suose tht is rime number nd (, ) 1. Then the congruence x n (mod ) is soluble if nd only if 1 (n, 1) 1 (mod ). Proof. Let g be rimitive root modulo. Then for some nturl number r one hs g r (mod ). If 1 (n, 1) 1 (mod ), then g r( 1) (n, 1) 1 (mod ). But since g is rimitive, the ltter congruence cn hold only when ( 1) r( 1) (n, 1), whence (n, 1) r. But by the Eucliden Algorithm, there exist integers u nd v with nu + ( 1)v (n, 1), so on writing r k(n, 1), we obtin g k(n, 1) (g ku ) n (g 1 ) kv (g ku ) n (mod ). Thus is indeed n nth ower residue under these circumstnces. On the other hnd, if the congruence x n (mod ) is soluble, then 1 (n, 1) (x 1 ) n/(n, 1) 1 (mod ), on mking use of Fermt s Little Theorem. This comletes the roof of the theorem. Exmle 1.3. Determine whether or not 3 is 4th ower residue modulo 17. Observe tht on mking use of Theorem 1.2, the congruence x 4 3 (mod 17) is soluble if nd only if 3 16/4 1 (mod 17), tht is, if 81 1 (mod 17). Since this congruence is not stisfied, one finds tht 3 is not 4th ower residue modulo 17. 1
2 2 LECTURE 9 Definition ( ) 1.4. When is n odd rime number, define the Legendre symbol by +1, when is qudrtic residue modulo, 1, when is qudrtic nonresidue modulo, 0, when. Theorem 1.5 (Euler s criterion). When is n odd rime, one hs ( 1)/2 (mod ). Proof. If ( 1)/2 1 (mod ), then the desired conclusion is n immedite consequence of Theorem 1.2. The conclusion is lso immedite when. It remins to consider the sitution in which ( 1)/2 1 (mod ). Let be n integer with (, ) 1, write r ( 1)/2, nd note tht in view of Fermt s Little Theorem, one hs r (mod ), whence r ±1 (mod ). Then if r 1 (mod ), one necessrily hs r 1 (mod ). Thus, in the sitution in which ( 1)/2 1 (mod ), wherein Theorem 1.2 estblishes tht is qudrtic nonresidue modulo, one hs ( 1)/2 1 (mod ), nd so the desired conclusion follows once gin. This comletes the roof of the theorem. Theorem 1.6. Let be n odd rime number. Then (i) for ll integers nd b, one hs ( ) ( ) ( ) b b ; (ii) whenever b (mod ), one hs (iii) whenever (, ) 1, one hs 2 1 nd (iv) one hs ( ) 1 1 nd ( ) b ; 2 b ( ) b ; ( ) 1 ( 1) ( 1)/2. Proof. These conclusions re essentilly immedite from Theorem 1.5. exmle, the ltter theorem shows tht ( ) ( ) ( ) b b (b) ( 1)/2 ( 1)/2 b ( 1)/2 (mod ), nd so the conclusion of rt (i) of the theorem follows on noting tht since is odd, one cnnot hve 1 1 (mod ). Prts (ii) nd (iv) re trivil from the lst observtion, nd rt (iii) follows from Fermt s Little Theorem. For
3 LECTURE 9 3 Note: ( The ) number of solutions of the congruence x 2 (mod ) is given by 1 +. For when (, ) 1 nd the congruence is soluble, one hs two distinct solutions nd In the corresonding cse in which the congruence is insoluble, one hs ( 1) 0. When (, ) > 1, ( one ) the other hnd, one hs the single solution x 0 (mod ), nd then The bove observtion rovides mens of nlysing the solubility of qudrtic equtions. For if (, ) 1 nd > 2, then the congruence x 2 +bx+c 0 (mod ) is soluble if nd only if (2x + b) 2 b 2 4c (mod ) is soluble, tht is, if nd only if either b 2 4c 0 (mod ), or else ( ) b 2 4c 1. The number of solutions of the congruence is therefore recisely ( ) b 2 4c 1 +. ( ) It is cler from the multilictive roerty of tht it suffices now to ( ) ( ) ( ) q 2 comute for odd rime numbers q nd in order to clculte in generl. 2. The lw of qudrtic recirocity We now come to one of the most beutiful results of our course the Lw of Qudrtic Recirocity, which Guss clled the ureum theorem ( golden theorem ). Euler ws the first to mke conjectures equivlent to Qudrtic Recirocity, but he ws unble to rove it. Legendre lso worked on this roblem very seriously nd develoed mny vluble ides, in rticulr, he lso introduced the Legendre symbol. Finlly, Guss gve comlete roof of the Lw of Qudrtic Recirocity in 1797, when he ws 19. Now there re over 200 different roofs of this fundmentl result. Theorem 2.1 (Lw of Qudrtic Recirocity; Guss). Let nd q be distinct odd rime numbers. Then ( ) ( ) q ( 1) ( 1)(q 1)/4. q
4 4 LECTURE 9 Rewriting the exression on the right hnd side of the lst eqution in the she ( ) ( ) q ( 1) 1 2 ( 1) 1 2 (q 1), q ( ) ( ) q we see tht unless nd q re both congruent to 3 modulo 4. q We give the roof of qudrtic recirocity which is due to Eisenstein. It is bsed on the following wy to comute the Legendre symbol: Lemm 2.2 (Eisenstein). For n odd rime nd (, ) 1, ( 1) ( 1)/2 k1 2k/. Proof. Let E {2, 4,..., 1}. For every x E, we write x x/ + r x, 0 r x <. We observe tht ech of the numbers ( 1) rx r x is congruent to n element of E. This is cler when r x is even, nd when r x is odd, ( 1) rx r x r x (mod ) where r x E. We lso clim tht if ( 1) rx r x ( 1) ry r y (mod ), then x y. Indeed, if r x r y (mod ), then x y (mod ), nd it follows tht x y (mod ). If r x r y (mod ), then x y (mod ), nd x y (mod ), nd (x+y), but x+y 2( 1), so tht this is imossible. Hence, we conclude tht nd x E x x E On the other hnd, ( Since {( 1) rx r x (mod ) : x E} E, ( 1) rx r x ( 1) x E rx r x (mod ). r x x ( 1)/2 x E x E x E ) ( 1)/2 (mod ), we deduce tht ( 1) x E rx. x E x (mod ). Finlly, we observe tht r x x/ (mod 2). This imlies the lemm. ( ) ( ) q Proof of Qudrtic Recirocity. We shll use the formuls for nd q rovided by the Einsenstein Lemm. The min ide of the roof is tht the sums ( 1)/2 (q 1)/2 2kq/ nd 2k/q k1 cn be interreted geometriclly. k1
5 LECTURE 9 5 We my think bout 2kq/ s the number of oints (x, y) with x 2k nd y equl to ositive integer t most 2kq/. Then the sum ( 1)/2 k1 2kq/ is equl to the number of integrl oints with even xcoordinte contined in the interior of tringle ABD. Note tht there re no integrl oints on the line AB (why?). We note tht the number of integrl oints contined in the interior of rectngle AF BD nd lying on fixed integrl verticl line is equl to q 1, thus, even. This imlies tht the number of integrl oints in KHBD with even xcoordinte is equl modulo 2 to the number of integrl oints in HJB with even xcoordinte. We lso observe tht the trnsformtion (x, y) ( x, q y) send the integrl oints with even xcoordintes contined in HJB to the integrl oints with odd xcoordintes contined in AHK. Finlly, we conclude tht he number of integrl oints with even xcoordinte contined in the interior of tringle ABD is congruent modulo 2 to the sum of the integrl oints with odd xcoordintes contined in AHK lus the integrl oints with even xcoordintes contined in AHK, nmely, it is recisely the number of integrl oints contined in AHK. We obtin tht ( 1)/2 k1 2kq/ v 1 (mod 2), where v 1 is the number of integrl oints contined in AHK. The sme rgument gives tht (q 1)/2 k1 2k/q v 2 (mod 2), where v 2 is the number of integrl oints contined in ALH. In view of Eisenstein s Lemm, ( ) ( ) q ( 1) v 1+v 2, q
6 6 LECTURE 9 but v 1 + v 2 is recisely the number of integrl oints in ALHK. Hence, v 1 + v 2 1 q This comletes the roof. Theorem 2.3. For ny odd rime, ( ) 2 ( 1) (2 1)/8. Proof. We use tht ( 1)/2 k1 When 8m ± 1, then ( 1)/2 k1 4k/ {k N : /4 < k < /2} /2 /4. 4k/ 4m ± 1/2 2m ± 1/4 0 ( 2 1)/8 (mod 2). When 8m ± 3, then ( 1)/2 k1 4k/ 4m ± 3/2 2m ± 3/4 2m ± 1 1 ( 2 1)/8 (mod 2). Exmle 2.4. Determine the vlue of ( ) 3. By Qudrtic Recirocity we hve ( ) 3 ( ) ( 1) (3 1)( 1)/4 ( 1) ( 1)/2, 3 nd by Euler s criterion, on the other hnd, ( ) 1 ( 1) ( 1)/2. Thus we see tht ( ) 3 ( 1 ) ( ) 3 ( ) ( ( 1) ( 1)/2 ( 1) ( 1)/2. 3 3) But ( ) 1 ( 1, when 1 (mod 3), 3 ( ) 3) 2 1, when 2 (mod 3). 3 Thus we deduce tht ( ) { 3 1, when 1 (mod 3), 1, when 2 (mod 3).
7 LECTURE 9 7 One cn use this evlution to show tht the only ossible rime divisors of x 2 + 3, for integrl vlues of x, re 3 nd rimes with 1 (mod 3). From here, n rgument similr to tht due to Euclid shows tht there re infinitely mny rimes congruent to 1 modulo 3. ( ) 21 Exmle 2.5. Determine the vlue of. 71 Alying the multilictive roerty of the Legendre symbol, followed by qudrtic recirocity, one finds tht ( ) ( ) ( ) ( ) ( ) ( 1) (71 1)(3 1)/4+(71 1)(7 1)/ ( ) ( ) ( ) ( ) ( ) ( ) 21 So 1, nd hence 21 is not qudrtic residue modulo
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