MATH 118 HW 7 KELLY DOUGAN, ANDREW KOMAR, MARIA SIMBIRSKY, BRANDEN LASKE Prt 1. Let be odd rime d let Z such tht gcd(, 1. Show tht if is qudrtic residue mod, the is qudrtic residue mod for y ositive iteger. Proof. This ws rove i the first roblem of the lst homework, but we will give differet roof here. Suose tht b (mod. We wt to show tht there is iteger x such tht x (mod for ech ositive iteger. We do this by iductio. The bse cse, 1, is cler becuse we re give tht x b is solutio. For the iductive ste, suose tht c (mod. We wt to show tht there is x such tht x (mod +1. Sice c (mod, we kow tht c + k for some iteger k. Cosider itegers x of the form c + l. We hve x c + lc + l + (lc + k + l + (lc + k (mod +1 sice + 1, so 0 (mod +1. For x to be cogruet to mod +1, we just eed +1 to divide (lc + k, i.e. we eed to divide lc + k. But this is esy to rrge we just choose l to be iteger cogruet to k (c 1 mod, which is ossible sice is odd d c is reltively rime to (sice gcd(, 1 d c (mod. The we will hve lc + k 0 (mod, d the x (mod +1 s desired. Prt. Sectio 3.3 Problem 1. Cosider the cogruece x (mod α with rime, 1, β b, (b, 1. Prove tht if β α the the cogruece is solvble, d tht if β < α the the cogruece is solvble if d oly if β is eve d x b (mod α β is solvble. 1
MATH 118 HW 7 Proof. Cse 1: If β α, show x (mod α hs solutio. β b. If β α, the α β. It follows tht α β b. Therefore, β b 0 (mod α. It follows to fid solutio such tht x 0 (mod α. Tke x 0, so x 0 (mod α. Therefore, x (mod α is solvble. Cse : If β < α, rove tht x (mod α is solvble β is eve d x b (mod α β is solvble. Let x (mod α be solvble. The there exists z such tht z + t α for some t Z z β b + t α β (b + t α β Note tht (b, 1 d α > β, so (b + α β. Therefore β is the highest ower of such tht β z. It follows tht β must be eve. Let β k. Defie z k (b + t α k where s Z sice k z. s z k s b + t α k b + t α β s b (mod α β Thus, s is solutio of x b (mod α β. Therefore x b (mod α β is solvble. Let β be eve, d let x b (mod α β be solvble, where (b, 1. The there exists y such tht y b + t α k where β k. y k b k + t α (y k + t α Set x y k. Thus x (mod α hs solutio.
Prt 3. MATH 118 HW 7 3 Usig Chiese Remider Theorem d first rt of roblem coclude tht if m is odd d gcd(, m 1, the is qudrtic residue mod m if d oly if is qudrtic residue mod for every rime dividig m. Proof. Suose tht the rime fctoriztio of m is m k i1 i i. The CRT sys tht the cogruece x (mod m hs solutio if d oly if the cogrueces x (mod i i hve simulteous solutio for i 1,..., k. Sice i is odd d (, i 1, by rt 1 of this roblem we kow tht x (mod i i hs solutio if d oly if is qudrtic residue mod i. Therefore x (mod m hs solutio if d oly if is qudrtic residue mod i for ech i, s desired. Prt 4. Let m be odd d let Z. Prove tht the cogruece x (mod m hs solutio if d oly if for ech rime dividig m, oe of the followig coditios holds, where α m d β : 1. β α.. β < α, β is eve, d / β is qudrtic residue mod. Proof. Agi, suose tht the rime fctoriztio of m is m k i1 i i. The CRT sys tht the cogruece x (mod m hs solutio if d oly if the cogrueces x (mod i i hve simulteous solutio for i 1,..., k. Sice i is odd, by rt of this roblem we kow tht x (mod i i hs solutio if d oly if the coditio listed i the roblem holds for ech i. This gives the desired result. Nive 3.3 Problem 14. Suose tht is rime ( 1 (mod 4 d tht + b with odd d ositive. Show tht 1. Proof. Sice is rime, is Legedre symbol, which we c lso view s Jcobi symbol. Sice is odd d 1 (mod 4, we hve ( + b b 1. For the lst ste we must ote tht (, b 1, sice if l, l b the l ( + b which is cotrdictio sice is rime.
Problem 17. MATH 118 HW 7 4 Problem 17. Let be odd rime, d ut s(, 1 Show tht s(0, 1. ( (+ Proof. We hve s(0, 1. By defiitio of the Legedre Symbol, 1 for ll vlues of for 1 1. For, the Legedre symbol is zero thus s(0, 1 1. Problem 17b. Show tht 1 s(, 0. Proof. We hve ( + s(, 1 1 1 ( + 1 1 b by the chge of vribles b + (mod 1 b1 b by the multilictivity of the Legedre symbol ( Now 1 the vlue ( 1 b1 1 ( 1 ( b1 ( b. 0 sice there re ( 1/ qudrtic residues givig 1, lus ( 1/ qudrtic o-residues givig the vlue ( 1, lus ( 0 0..
MATH 118 HW 7 5 Problem 17c. Show tht if (, 1 the s(, s(1,. Proof. I the sum s(, (+ 1, use the chge of vribles b 1 (mod, so tht b (mod, to rewrite b(b + s(, b1 b(b + 1 b1 b(b + 1 b1 b(b + 1 sice 1 b1 s(1, by defiitio. Problem 17d. Coclude tht s(, 1 if (, 1. Proof. Combiig the revious rts, we fid 0 s(, by rt b 1 ( 1s(1, + s(0, by rt c ( 1s(1, + ( 1 by rt. Therefore s(1, 1, d hece by rt c, s(, 1 for ll such tht (, 1.