Divisibility. c = bf = (ae)f = a(ef) EXAMPLE: Since 7 56 and , the Theorem above tells us that

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1 Divisibility DEFINITION: If a and b ae integes with a 0, we say that a divides b if thee is an intege c such that b = ac. If a divides b, we also say that a is a diviso o facto of b. NOTATION: d n means n is divisible by d o d divides n. EXAMPLE: We have: 4 12, since 12 = , since 15 = THEOREM 1: If a,b, and c ae integes with a b and b c, then a c. Poof: Because a b and b c, thee ae integes e and f such that ae = b and bf = c. Hence, and we conclude that a c. c = bf = (ae)f = a(ef) EXAMPLE: Since 7 56 and , the Theoem above tells us that THEOREM2: Ifa,b,c,m, andnaeintegeswithc 0, andifc aandc b, thenc (ma+nb). Poof: Since c a and c b, by the definition above we have fo some k 1,k 2 Z. Theefoe Consequently, we see that c (ma+nb). a = c k 1, b = c k 2 ma+nb = mc k 1 +nc k 2 = c(mk 1 +nk 2 ). EXAMPLE: Since and 11 33, the Theoem above tells us that 11 divides = 319 EXAMPLE: Let k be an intege numbe. Show that 8 (n 2 1) if n = 4k +1. Poof: We have n 2 1 = (4k +1) 2 1 = 16k 2 +8k +1 1 = 16k 2 +8k = 8(2k 2 +k) Theefoe 8 (n 2 1) by the definition above. EXAMPLE: Let a and b be intege numbes. Show that (a b) (a 2 b 2 ) if a b 0. Poof: We have a 2 b 2 = (a b)(a+b) Theefoe (a b) (a 2 b 2 ) by the definition above. REMARK. Fo moe examples, see Appendix I. 1

2 THEOREM 3 (DIVISION ALGORITHM): Fo any integes a and b with b 0 thee exist unique integes q and such that a = bq +, whee 0 < b The integes q and ae called the quotient and the eminde, espectively. Poof: Conside the set S of all integes of the fom a bk whee k is an intege, that is, S = {a bk k Z} Let T be the set of all nonnegative integes in S. T is nonempty, because a bk is nonnegative wheneve k is an intege with k a/b (if b > 0) o k a/b (if b < 0) By the well-odeing popety, T has a least element = a bq (Theseaethevaluesfoq and specifiedinthetheoem.) Weknowthat 0byconstuction, and it is easy to see that < b. In fact, if b, then ( > b = a bq b = a b q + b ) = a b(q ±1) 0 b which contadicts the choice of = a bq as the least nonnegative intege of the fom a bk. Hence, 0 < b. To show that these values fo q and ae unique, assume that we have two equations a = bq and a = bq with 0 1 < b and 0 2 < b. By subtacting the second of these equations fom the fist, we find that 0 = b(q 1 q 2 )+( 1 2 ) Hence, we see that 2 1 = b(q 1 q 2 ) This tells us that b divides 2 1. Because 0 1 < b and 0 2 < b, we have b < 2 1 < b Hence, b can divide 2 1 only if 2 1 = 0 o, in othe wods, if 1 = 2 Because bq = bq and 1 = 2, we also see that q 1 = q 2 This shows that the quotient q and the eminde ae unique. 2

3 EXAMPLE: Let a = 49 and b = 4, then 49 = so the quotient is 12 and the eminde is 1. Note that we can also wite 49 as , but in this case 13 is not a eminde, since it is not less than 4. Similaly, if a = 49 and b = 4, then 49 = ( 4) ( 12)+1 if a = 49 and b = 4, then 49 = 4 ( 13)+3 if a = 49 and b = 4, then 49 = REMARK: Let b > 0. Because the quotient q is the lagest intege such that bq a and it follows that Fo example, if a = 49 and b = 4, then = a bq q = a/b, = a b a/b if a = 49 and b = 4, then q = 49/4 = = 12, = = = 1 q = 49/4 = = 13, = 49 4 ( 13) = = 3 Similaly, if b < 0, then Fo example, if a = 49 and b = 4, then q = a/b, = a b a/b q = 49/( 4) = = 12, = 49 ( 4) ( 12) = = 1 if a = 49 and b = 4, then q = 49/( 4) = = 13, = 49 ( 4) 13 = = 3 DEFINITION: If the emainde when n is divided by 2 is 0, then n = 2k fo some intege k, and we say that n is even, wheeas if the emainde when n is divided by 2 is 1, then n = 2k+1 fo some intege k, and we say that n is odd. Thanks to the Division Algoithm fo any intege numbe n thee ae only two possibilities: n = 2k o n = 2k +1 theefoe any intege numbe is eithe even o odd. EXAMPLE: Let b = 3 in the Division Algoithm. Since 0 < 3, then fo any intege numbe n we have only thee possibilities: n = 3k, n = 3k +1, o n = 3k +2 3

4 EXAMPLE: Let k be an intege numbe. Then 3 k 2 2. Poof: Assume to the contay that thee is an intege numbe k such that 3 k 2 2. By the definition of divisibility, k 2 2 = 3m ( ) fo some intege m. On the othe hand, by the Division Algoithm, thee ae only thee possibilities: k = 3q, k = 3q +1, o k = 3q +2 whee q is an intege. We show that fo k 2 2 we have only two possibilities: k 2 2 is eithe 3+1 o 3 +2 whee is an intege (which gives us a contadiction with ( ) by the Division Algoithm). Indeed, if k = 3q, then if k = 3q +1, then k 2 2 = 9q 2 2 = 9q = 3(3q 2 1)+1 = 3 +1 k 2 2 = (3q +1) 2 2 = 9q 2 +6q +1 2 = 9q 2 +6q 1 if k = 3q +2, then = 9q 2 +6q 3+2 = 3(3q 2 +2q 1)+2 = 3 +2 k 2 2 = (3q +2) 2 2 = 9q 2 +12q +4 2 = 9q 2 +12q +2 = 3(3q 2 +4q)+2 = 3 +2 REMARK. Fo moe examples, see Appendix II. 4

5 Geatest Common Divisos DEFINITION: A common diviso of two integes a and b, which ae not both 0, is an intege c such that c a and c b. The geatest common diviso (gcd) of a and b, denoted by (a,b), is the lagest common diviso of integes a and b. EXAMPLE: The common divisos of 24 and 84 ae ±1, ±2, ±3, ±4, ±6, and ±12 Hence, (24,84) = 12. Similaly, looking at sets of common divisos, we find that (15,81) = 3, (100,5) = 5, (17,25) = 1, ( 17,289) = 17, etc. DEFINITION: The integes a and b, with a 0 and b 0, ae elatively pime if EXAMPLE: Since the numbes 12 and 25 ae elatively pime. (a,b) = 1 (12,25) = 1 5

6 Appendix I EXAMPLE: Let a be an intege numbe. Show that (a 2 +a+1) (a 3 1) if a 1. Poof: We have a 3 1 = (a 2 +a+1)(a 1) Theefoe (a 2 +a+1) (a 3 1) by the definition of divisibility. EXAMPLE: Let a and b be intege numbes. Show that (a+1) (ab+a+b+1) if a 1. Poof: We have ab+a+b+1 = a(b+1)+b+1 = (a+1)(b+1) Theefoe (a+1) (ab+a+b+1) by the definition of divisibility. EXAMPLE: Let a and b be intege numbes, not both zeo. Show that (a 2 +b 2 +ab) (a 4 + a 2 b 2 +b 4 ). Poof: We fist note that a 2 +b 2 +ab > 0 fo any eal numbes a and b. Indeed, if ab 0, then Similaly, if ab < 0, then We have a 2 +b 2 +ab a 2 +b 2 > 0 a 2 +b 2 +ab = (a 2 +2ab+b 2 ) ab = (a+b) 2 ab > (a+b) 2 0 a 4 +a 2 b 2 +b 4 = (a 4 +2a 2 b 2 +b 4 ) a 2 b 2 = (a 2 +b 2 ) 2 a 2 b 2 = (a 2 +b 2 ) 2 (ab) 2 = (a 2 +b 2 +ab)(a 2 +b 2 ab) Theefoe (a 2 +b 2 +ab) (a 4 +a 2 b 2 +b 4 ) by the definition of divisibility. 6

7 Appendix II EXAMPLE: Let k be an intege numbe. Then 4 k 2 3. Poof: Assume to the contay that thee is an intege numbe k such that 4 k 2 3. By the definition of divisibility, k 2 3 = 4m ( ) fo some intege m. On the othe hand, by the Division Algoithm, thee ae only two possibilities: k = 2q o k = 2q +1 whee q is an intege. We show that fo k 2 3 we have only two possibilities: k 2 3 is eithe 4+1 o 4 +2 whee is an intege (which gives us a contadiction with ( ) by the Division Algoithm). Indeed, if k = 2q, then if k = 2q +1, then k 2 3 = 4q 2 3 = 4q = 4(q 2 1)+1 = 4 +1 k 2 3 = (2q +1) 2 3 = 4q 2 +4q +1 3 = 4q 2 +4q 2 = 4q 2 +4q 4+2 = 4(q 2 +q 1)+2 = 4 +2 EXAMPLE: Let a and b be intege numbes. Then 4 a 2 +b 2 3. Poof: Assume to the contay that thee ae intege numbes a and b such that 4 a 2 +b 2 3. By the definition of divisibility, a 2 +b 2 3 = 4m ( ) fo some intege m. On the othe hand, by the Division Algoithm, fo any intege k we have only two possibilities: k = 2q o k = 2q +1 whee q is an intege. Theefoe, fo k 2 we also have only two possibilities: k 2 = 4q 2 = 4 o k 2 = 4q 2 +4q +1 = 4(q 2 +q)+1 =

8 whee is an intege. Fom this it follows that fo a 2 +b 2 3 we have only thee possibilities: o a 2 +b 2 3 = = = 4( )+1 = 4R+1 R a 2 +b 2 3 = = 4( )+2 = 4R+2 R a 2 +b 2 3 = = 4( )+3 = 4R+3 R whee R is an intege. This gives us a contadiction with ( ) by the Division Algoithm. EXAMPLE: Let a,b,c, and k be intege numbes. Then 8k +7 a 2 +b 2 +c 2. Poof (shot): Assume to the contay that thee ae intege numbes a,b,c, and k such that 8k +7 = a 2 +b 2 +c 2 On the othe hand, by the Division Algoithm, fo any intege k we have only eight possibilities: k = 8q, k = 8q +1,...,k = 8q +6, o k = 8q +7 whee q is an intege. Fom this one can deduce that fo k 2 we have only thee possibilities: k 2 = 8, k 2 = 8 +1, o k 2 = 8 +4 whee is an intege. Fom this it follows that thee is no combination of a 2,b 2, and c 2 such that a 2 +b 2 +c 2 = 4k +7. 8