Math 261 Exercise sheet 2

Transcription

1 Math 261 Exercse sheet 2 Verson: September 25, 2017 Answers are due for Monday 25 September, 11AM. The use of calculators s allowed. Exercse 2.1 Fnd all solutons x, y Z to the followng equatons: 1. 3x + 5y = 2, 2. 18x + 30y = 2016, 3. 18x + 30y = Soluton Snce 3 and 5 are coprme, ths equaton has solutons. An obvous one s x = 1, y = 1, and we know that we get all the other ones by addng a multple of 3 5 on one sde, and compensatng on the other sde. So the solutons are exactly the x = 1 + 5t, y = 1 3t (t Z). 2. Ths tme 18 and 30 are not coprme, but ther gcd, whch s 6, dvdes 2016, so we can get an equvalent equaton by dvdng everythng by 6: 3x + 5y = 336. We spot a soluton, for nstance y = 60, x = 12; and the other solutons dffer by multples of 3 5. So fnally the solutons are x = t, y = 60 3t (t Z). 3. Ths tme, gcd(18, 30) = 6 does not dvde 2017, so the equaton has no solutons. 1

2 Exercse 2.2: The lcm 1. Let I and J be two deals of Z. Apply the defnton of an deal to prove that the ntersecton I J s also an deal of Z. 2. Let a and b be postve ntegers. By the prevous queston, az bz s an deal of Z. Prove that ths deal s not the zero deal. Why does ths mply that there exsts a postve nteger c such that az bz = cz? 3. The nteger c defned n the prevous queston s called the lowest common multple of a and b, and s denoted by lcm(a, b). Explan ths name. 4. Let p 1,, p n be the prmes that dvde ether a or b, so that we may wrte n a = p u, b = wth non-negatve ntegers u 1,, u n and v 1,, v n. Express lcm(a, b) n terms of the p, the u and the v. 5. Deduce that gcd(a, b) lcm(a, b) = ab. Remark: Ths means that t s possble to compute the lcm by computng the gcd through the Eucldan algorthm. When a and b are large, ths s much more effcent than computng the factorzaton of a and b. Soluton 2.2: 1. We know that every deal of Z contans at least 0. So 0 I, and 0 J, so that 0 I J; as a result, I J s not empty. Let x and y be elements of I J. Then snce x and y are n I, x + y s n I; smlarly, snce x and y are n J, x + y s also n J. So x + y I J. Fnally, let n Z, and let x I J. Then nx I, and nx J, so nx I J. Ths shows that I J s an deal of Z. 2. We have ab az bz, so az bz {0}. Snce every deal of Z s of the form cz for some c Z, and snce az bz s an deal by the prevous queston, there exsts c Z such that az bz = cz. But snce ab az bz, we cannot have c = 0, and snce cz = ( c)z, we may assume that c > Clearly, c s the smallest postve element of the deal cz. But cz = az bz s the set of ntegers that are both a multple of a and of b. The number c s thus the smallest such common multple, whence ths name. 4. The formula az bz = cz says that the common multples of a and b are precsely the multples of c. Now ab = n pu +v s such a common multple, so c dvdes t; therefore c = n pw for some ntegers w u + v. In order for c to be a multple both of a and b, we need w to be both u and v for each. Then c wll be the smallest common multple precsely when w = max(u, v ), so that c = p max(u,v ). 2 p v

3 5. Recall that gcd(a, b) = n all u and v, so that pmn(u,v ). Now, mn(u, v) + max(u, v) = u + v for gcd(a, b) lcm(a, b) = p mn(u,v )+max(u,v ) = p u +v = ab. Exercse Use the Eucldan algorthm to determne f 47 s nvertble mod 111, and to fnd ts nverse f t s. 2. Solve the equaton 47x 5 (mod 111) n Z/111Z. Soluton We know that 47 s nvertble mod 111 f and only f 47 and 111 are coprme. If they are, we need to look for u and v Z such that 47u + 111v = 1; ndeed, u wll then be an nverse of 47 (mod 111). To fnd u and v, we ether spot them drectly 1, or we use the Eucldan algorthm. Ths algorthm wll also tell us f the gcd of 47 and 111 s not 1, so let s apply t: 111 = = = = So the gcd s 1, so 47 s nvertble mod 111. To fnd t, we wrte 1 = = 13 3(17 13) = = 4( ) 3 17 = = ( ) = , whence 26 mod 111 s the nverse of 47 mod Snce 47 s nvertble mod 111, the only soluton s x = = 5 26 = (mod 111). 1 It can happen sometmes, but here there are no obvous canddates 3

4 Exercse 2.4: An unsolvable dophantne equaton Prove that the equaton x 3 + y 3 + z 3 = 31 has no soluton wth x, y, z Z. Hnt: try solvng the equaton mod 9. Soluton 2.4: Let us make a table of the cubes n Z/9Z: x x So the only cubes n Z/9Z are 1,0 and 1. As a result, a sum of 3 cubes can be 3, 2, 1, 0, 1, 2 or 3 (mod 9), but t can never be 4 nor 4. Snce 31 4 (mod 9), the equaton x 3 + y 3 + z 3 = 31 has no soluton n Z/9Z; therefore, t has no soluton n Z ether. Exercse 2.5: Prmes mod 6 1. Let p be a prme number whch s nether 2 nor 3. Prove that ether p 1 (mod 6) or p 1 (mod 6). 2. Prove that there are nfntely many prmes p such that p 1 (mod 6). Hnt: Suppose on the contrary that there are fntely many, say p 1,, p k. Let N = 6p 1 p k 1, and consder a prme dvsor of N. 3. Why does the same proof fal to show that there are nfntely may prmes p such that p 1 (mod 6)? 4. Drchlet s theorem on prmes n arthmetc progressons, whch s way beyond the scope of ths course, states that for all coprme postve ntegers a and b, there are nfntely many prmes p such that p a (mod b); n partcular, there are n fact nfntely many prmes p such that p 1 (mod 6). Why, n the statement of ths theorem, s t necessary to assume that a and b are coprme? Soluton 2.5: 1. If p s nether 2 nor 3, then p 6, so p and 6 are coprme. But the only nvertbles n Z/6Z are ±1 (ether see t by nspecton of all 6 elements, or use the fact that φ(6) = 2), so we must have p ±1 (mod 6). 2. Suppose that p 1,, p k are the only such prmes, and let N = 6p 1 p k 1. Clearly, nether 2 nor 3 dvde N (snce N 1 mod 6), so the prmes dvdng N are all ±1 (mod 6) by the prevous queston. If they were all +1 (mod 6), then N, ther product, would also be +1 (mod 6), whch s not the case. So at least one of them, say p, s 1 (mod 6). But ths p cannot be one of p 1,, p k, else we would have p (6p 1 p k N) = 1. We therefore have reached a contradcton. 3. We could suppose by contradcton that p 1,, p k are the only prmes 1 (mod 6), and consder a prme dvsor of N = 6p 1 p k 1 (or N = 6p 1 p k + 1). Such a prme could not be any of the p for the same reason as above, but there s no reason why t would have to be 1 (mod 6); ndeed, nothng prevents the dvsors of N from beng all 1 (mod 6). So we are stuck. 4

5 4. If p a (mod b), then p = bx + a for some x Z, so gcd(a, b) p; and obvously, f gcd(a, b) > 1, ths can only happen for at most one prme p (exactly one f gcd(a, b) = p s tself prme, and none else). The exercse below has been added for practce. not mandatory. It s not mandatory, and not worth any ponts. The soluton wll be made avalable wth the solutons to the other exercses. Exercse 2.6: Let n N. Dvsblty crtera 1. Prove that 3 n ff. 3 dvdes the sum of dgts of n. 2. Prove that 9 n ff. 9 dvdes the sum of dgts of n. 3. Fnd a smlar crteron to test whether 11 n. Soluton 2.6: The key s the followng observaton: f n 0, n 1, n 2, are the dgts of n from rght to left, so that n = n n n 2 + = n 10, and snce 10 1 (mod 9), we have n = n 10 n 1 = n (mod 9); n other words, n s congruent to the sum of ts dgts (mod 9). In partcular, ths congruence also holds mod 3 snce 3 9. So we have and 9 n n 0 (mod 9) n 0 (mod 9) 9 n 3 n n 0 (mod 3) n 0 (mod 3) 3 n. For dvsblty by 11, we notce that 10 1 (mod 11), so that n = n 10 n ( 1) = n 0 n 1 + n 2 n 3 + (mod 11). As a result, 11 n f and only f the expresson n 0 n 1 + n 2 n 3 +, whch s called the alternate sum of dgts of n, s dvsble by 11. Examples: For n = 261, we have = 9, so For n = 1452, we have = 0, so We also see that and that but