M3P14 EXAMPLE SHEET 1 SOLUTIONS

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1 M3P14 EXAMPLE SHEET 1 SOLUTIONS 1. Show tht for, b, d itegers, we hve (d, db) = d(, b). Sice (, b) divides both d b, d(, b) divides both d d db, d hece divides (d, db). O the other hd, there exist m d such tht (, b) = m + b. The d(, b) = md + db. Sice (d, db) divides both d d db, d d(, b) is lier combitio of d d db, we hve (d, db) divides d(, b). Thus (d, db) d d(, b) re equl. 1b. Let,, b be itegers d suppose tht divides, b. Show tht (,) (,) divides b. Let = (,), d =. Both re itegers. Sice divides b, we hve k = b for some iteger k, d dividig by (, ) we fid tht k = b; tht is, divides b. But (, ) = ((, ), (, ) ) = (, )(, ) by prt 1, so (, ) = 1. Thus divides b by result from lecture. 2. Express 18 s iteger lier combitio of 327 d 120. We hve: 327 = 2(120) = 1(87) = 2(33) = 1(21) = 1(12) = 1(9) + 3 Therefore: 3 = 12 9 = 12 (21 12) = 2(12) 21 = 2(33 21) 21 = 2(33) 3(21) = 2(33) 3(87 2(33)) = 8(33) 3(87) = 8(120 87) 3(87) = 8(120) 11(87) = 8(120) 11(327 2(120)) = 30(120) 11(327) 1

2 2 M3P14 EXAMPLE SHEET 1 SOLUTIONS We thus hve 18 = 180(120) 66(327). 2b. Fid, with proof, ll solutios to the lier diophtie equtio 110x + 68y = 14. We hve: 110 = 1(68) = 1(42) = 1(26) = 1(16) = 1(10) = 1(6) = 1(4) + 2 We hve (110, 68) = 2; uwidig the bove we fid 2 = 21(110) + 34(68). So 14 = 147(110) + 238(68) d oe solutio is x = 147, y = 238. The remiig solutios hve the form , for y iteger. 2c. Fid multiplictive iverse of 31 modulo 132. We must write 1 s lier combitio of 31 d 132. By Euclid s lgorithm, 132 = 4(31) = 3(8) = Thus: 1 = 8 7 = 8 (31 3(8)) = 4(8) 31 = 4(132 4(31)) 31 = 4(132) 17(31). I prticulr 17(31) is cogruet to 1 mod 132, so 17 is the desired iverse. 2d. Fid iteger cogruet to 3 mod 9 d cogruet to 1 mod 49. Note tht 1 = 11(9) 2(49). Thus 98 is cogruet to 1 mod 9 d 0 mod 49, d 99 is cogruet to 0 mod 9 d 1 mod 49. Thus 3( 98)+99 = 195 is cogruet to 3 mod 9 d 1 mod 49. 2e. Fid, with proof, the smllest oegtive iteger such tht 1 (mod 3), 4 (mod 5), d 3 (mod 7). We hve 1 = 2(3) 5, so 4(2(3)) 5 = 19 is cogruet to 1 mod 3 d 4 mod 5. So is 4. We thus kow tht 4 (mod 15) by the uiqueess i the Chiese Remider Theorem. We hve 1 = 15 2(7), so 2(4(7)) + 3(15) = 11 is cogruet to 4 (mod 15) d 3 (mod 7). Thus is cogruet to 11 (mod 105) by the Chiese remider theorem. Sice 94 is the smllest positive iteger cogruet to 11 mod 105, we must hve = Let m d be itegers. Show tht the gretest commod divisor of m d is the uique positive iteger d such tht:

3 M3P14 EXAMPLE SHEET 1 SOLUTIONS 3 d divides both m d if x divides both m d, the x divides d. (I other rigs, we will tke these properties to be the defiitio of gretest commo divisor.) It is cler tht (m, ) divides both m d. Coversely, suppose x divides both m d. Writig (m, ) = m + b for some itegers d b we see tht x divides (m, ) s well. For uiqueess, suppose there were other iteger d with these properties. The d divides m d, so it divides (m, ). Similrly, (m, ) divides m d so it divides d. Thus d = ±(m, ), d if d is positive it must equl (m, ). 4. Lest Commo Multiples 4. Let d b be ozero itegers. Show tht there is uique positive iteger m with the followig two properties: d b divide m, d If is y umber divisible by both d b, the m. The umber m is clled the lest commo multiple of d b. We first show tht m is uique. Let be other positive iteger with the sme properties. The d b divide by the first property (of ), so by the secod property (of m) we hve m. Similrly, by the first property (of m) d the secod property (of ) we hve m. So m = ±, but both re positive, so m =. Existece of m will follow from prt 2b, below: 4b. Show tht the lest commo multiple of d b is give by b. It is cler tht b b is positive. It is divisible by becuse is iteger, d similrly is divisible by b. We must thus show tht if d b divide, the so does b. If, b divide, the d b divide. But d b b re reltively prime, so their product divides 2. Thus b divides. 5. Let m d be positive itegers, d let K be the kerel of the mp: Z/mZ Z/mZ Z/Z tht tkes clss mod m to the correspodig clsses modulo m d. Show tht K hs (m, ) elemets. Wht re they? If lies i K, the 0 (mod m) d 0 (mod ). Thus both m d divide, so by questio 4, the lest commo multiple d = m (m,) of m d m divides. O the other hd it is cler tht y multiple of (m,) lies i K. Thus K cosists precisely of these multiples. Modulo m these multiples fll ito the clsses [d], [2d],... [d(m, )] = [0]. There re thus (m, ) such clsses, so K hs (m, ) elemets s climed. 6. Show tht the equtio x b (mod ) hs o solutios if b is ot divisible by (, ), d exctly (, ) solutios i Z/ otherwise.

4 4 M3P14 EXAMPLE SHEET 1 SOLUTIONS If x b (mod ), the x b = k for some iteger k. Sice both x d k re divisible by (, ), b must be s well. Suppose tht b is divisible by (, ). The solvig x b = k is the sme s solvig cx d = km, where c = (,), d = b (,), m = (,). Sice c d m re reltively prime, this equtio hs uique solutio y modulo m. Thus y x tht is cogruet to y modulo m is solutio. Sice there re (, ) cogruece clses modulo tht re cogruet to give cogruece clss mod m, we re doe. 7. For positive iteger, let σ() deote the sum d,d>0 d of the positive divisors of. Show tht the fuctio σ() is multiplictive. We must show tht σ(m) = σ()σ(m) wheever m d re reltively prime. Let d be divisor of m, d set e = (d, m), f = d e. The e d f re itegers, d e divides m. Sice (e, ) = (d, m, ) = 1, e d re reltively prime. Thus f divides by 1b bove, d f d m re reltively prime. The mp tht tkes divisor d to the correspodig e, f is bijectio betwee divisors of m d pirs cosistig of divisor of m d divisor of. We thus hve: d = ef. d m,d>0 e m,e>0 f,f>0 Fctorig the right-hd side, we fid tht it is equl to the product: e f = σ(m)σ() s climed. e m,e>0 f,f>0 8. Let p be prime, d be y iteger. Show tht p2 +p+1 is cogruet to 3 modulo p. Note tht if p does ot divide, the p 1 1 (mod p), so p (mod p). If p does divide, the d p re both 0 (mod p). Thus p (mod p) for ll. Now p2 = ( p ) p p (mod p). Thus p2 +p+1 = p2 p 3 (mod p). 9. Let be squrefree positive iteger, d suppose tht for ll primes p dividig, we hve (p 1) ( 1). Show tht for ll itegers with (, ) = 1, we hve (mod ). Sice is squrefree, it is the product of the primes tht divide it. Therefore, by the uiqueess prt of the Chiese remider theorem, it suffices to show tht (mod p) for ll p dividig. Sice (, ) = 1, p does ot divide, d so p 1 1 (mod p). Sice (p 1) ( 1), we hve = 1 + k(p 1) for some k. The = ( p 1 ) k (mod p) s climed.

5 M3P14 EXAMPLE SHEET 1 SOLUTIONS Let be positive iteger. Show tht the sum d,d>0 Φ(d) is equl to. For y positive iteger less th or equl to, (, ) is positive divisor d of. It follows (sice is the umber of positive itegers less th or equl to ) tht we hve: = #{ : 1, (, ) = d }. d,d>0 Therefore, it suffices to show tht the umber of positive itegers betwee 1 d with (, d) = d is Φ(d). But multiplictio by d gives bijectio betwee the set of positive itegers b with 1 b d d (b, d) = 1, d the set of positive itegers with 1 d (, ) = d. Thus these sets both hve Φ(d) elemets. NOTE: this c lso be doe by iductio o the umber of primes dividig.

The Structure of Z p when p is Prime

The Structure of Z p when p is Prime LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z