Number Theory

CHAPTER 3

Number Theory

1. Factors or not According to Carl Friedrich Gauss (1777-1855) mathematics is the “queen of sciences” and number theory is the “queen of mathematics”, where “queen” stands for elevated and beautiful. Number theory is mainly the study of the system of integers Z = 0, 1, 2,... and the consequences of the fact that division is not always possible within{Z.± E.g.± 15/3 } Z but 15/4 / Z. ∈ Let us start∈ with the well-known “long division” that has rather surprising consequences.

EXAMPLE 1.1. 300 396 090 006 223 88323 | 669 223 88323 | 66900 (= 300 223) 21423 · 2007 21423 20070 (= 90 223) 1353 · 1338 1353 1338 (= 6 223) 45 · 45 What is really being done and what has been achieved? On the right some zeros are ﬁlled in that are not written on the left. We now see that

88323 300 223 90 223 6 223 = 45. − · − · − · In effect, multiples of 223 were subtracted from 88323 until 223 could not be subtracted anymore without running into negative numbers. The mathematical content of “long division” is the following theorem and “long division” is just an efﬁcient way of ﬁnding q and r in the “Division Theorem”.

1 2 3. NUMBER THEORY

THEOREM 1.2. Division Theorem. Given integers a, b, b > 0, there exist unique integers q and r such that a = qb + r, 0 r < b ≤ The quotient q and the remainder r can be found by repeated subtractionof b. The division algorithm is just an efﬁcient way for computing q and r by repeated subtractions.

EXAMPLE 1.3. Let a = 567923 and b = 735. Find non-negative integers q and r such that 567923 = 735q + r and 0 r < 735. Answer: By Long Division≤ we ﬁnd q = 772 and r = 503.

EXAMPLE 1.4. Let a = 3257397 and b = 7593. Find non-negative integers q and r such that 3257397 = 7593q + r and 0 r < 7593. Answer: By Long Division we ﬁnd≤ q = 429 and r = 0.

DEFINITION 1.5. An integer a is even ifa = 2x for some integer x, i.e., r = 0 in Theorem ??. An integer b is odd if b = 2x + 1 for some integer x in Z, i.e., r = 1 in Theorem ??.

PROPOSITION 1.6. Every integer is either even or odd. The product of two odd integers is odd.

PROOF. Let a and b be odd integers. According to the deﬁnition of “odd integer” there exist integers x and y such that a = 2x + 1 and b = 2y + 1. Then ab = (2x + 1)(2y + 1)=(2x + 1)(2y)+(2x + 1) 1 · = 2(y(2x + 1)) + 2x + 1 = 2(y(2x + 1) + x) + 1. Here z = y(2x + 1) + x is an integer by the closure properties of Z, hence ab = 2z + 1is an odd number.

THEOREM 1.7. √2 is not rational.

PROOF. (Aristotle) By way of contradiction assume that √2 = a/b where either a or b 2 2 2 is odd. Then 2b = a , hence a is even and therefore a is even. This means that a = 2a′ 2 2 2 2 and substituting 2b = 4a′ . Thus b = 2a′ . This says that b is even, a contradiction.

DEFINITION 1.8. (1) Let a, f be integers. We say that f is a factor of a if

a = f someinteger or a = f a′ for some integer a′. · · (2) Let a and b be given integers. An integer f is a common factor ofaandbif f is a factor of a and a factor of b. (3) The greatest common factor of two integers a and b is the largest among the common factors of a and b. The greatest common factor of a and b is denoted by gcf(a,b). 1. FACTORS OR NOT 3

REMARK 1.9. In the literature it is much more common to say f divides a (or f divides a evenly) than to say that f is a factor of a. However, the ﬁrst gets confused with other uses of “divide” and therefore we will avoid its use.

REMARK 1.10. Suppose that a and b are positive integers. Then b is a factor of a if and only if in the Division Theorem a = qb+r the remainder r = 0. Therefore instead of saying b is a factor of a it is also said that b divides a evenly. Factoring Rules. Let a be a positive integer given in base 10 representation. Then the following rules are true. (1) 2 is a factor of a if its units digit is even. (2) 3 is a factor of a if 3 is a factor of the sum of the digits of a. (3) 4 is a factor of a if 4 is a factor of the number formed by the last two digits of a. (4) 9 is a factor of a if 9 is a factor of the sum of the digits of a.

Recall that gcf(a,b) is the largest among the common factors of a and b. Therefore gcf(a,b) can be found in the following way which is instructive but not very efﬁcient.

REMARK 1.11. Finding greatest common factors Let a and b be given integers. (1) List the positive factors of both a and b. (2) List the common (positive) factors of a and b. (3) Pick the largest in the list of common factors.

EXAMPLE 1.12. Finding greatest common factors. Let a = 12 and b = 28. (1) The positive factors of 12 are 1,2,3,4,6,12. (2) The positive factors of 28 are 1,2,4,7,14,28. (3) The common factors are 1,2,4. (4) The greatest common factor is 4.

EXAMPLE 1.13. Let a = 240 and b = 330. The positive factors 0f 240 are 1,2,3,4,5,6,8,10,12,15,16,20,24,30,40,48,60,80,120,240, and the positive factors of 330 are 1,2,3,5,6,10,11,15,22,30,33,55,66,110,165,330. The positive common factors of a and b are 1,2,3,5,6,10,15,30, and the greatest common factor of a and b is 30, gcf(240,330) = 30. The following lemmas will be used over and over again and should be memorized.

LEMMA 1.14. Let a,b,c be integers and suppose that c = a + b. If a number f is a factor of two of the three integers a,b,c, then f is a factor of the third.

LEMMA 1.15. Let a andb be integers. If f isa factorofa, then f is afactorof ab. 4 3. NUMBER THEORY

EXERCISE 1.16. (1) Rewrite the number 1239584 in the form as 1239584 = 12395 100 + 84. Use Lemma ?? and Lemma ?? to show that 4 is a factor of 1239584· because 4 is a factor of 84. (2) Rewrite the number 1239582 in the form as 1239582 = 12395 100 + 82. Use Lemma ?? and Lemma ?? to show that 4 is NOT a factor of 1239582· because 4 is NOT a factor of 82.

The Euclidean Algorithm is a very beautiful and efﬁcient way of ﬁnding the greatest common factor of two integers. It is bases on the following fact.

LEMMA 1.17. Let a and b be integers and a = bq + r for some integers q and r. Then the common factors of a and b are exactly the same as those of b and r. In particular,

gcf(a,b) = gcf(b,r).

EXAMPLE 1.18. Let a = 56371 and b = 3476. Then a = 16 b + 755, hence · gcf(56371,3476) = gcf(3476,755).

Further, 3476 = 4 755 + 456, so · gcf(56371,3476) = gcf(3476,755) = gcf(755,456).

Further, 755 = 1 456 + 299, 456 = 1 299 + 157, 299 = 1 157 + 122, 157 = 1 122 + 35, hence · · · ·

gcf(56371,3476) = gcf(755,456) = gcf(456,299) = gcf(299,157) = gcf(157,122) = gcf(122,35).

It is clear that 35 has the positive factors 1,5,7,35 and of these only 1 is a factor of 122. Hence gcf(56371,3476) = 1.

THEOREM 1.19. Let a and b be (positive) integers. Then there exist integers u and v such that gcf(a,b) = ua + vb. Consequently, every common factor of a and b is a factor of gcf(a,b). The integers u, v and gcf(a,b) can be found efﬁciently using the Euclidean Algorithm described below.

EXAMPLE 1.20. Let a = 569321 and b = 347. Then gcf(69321,347) = gcf(a,b) = ua+vb where gcf(a,b) = 1,u = 36 and v = 59065. − PROOF. 1. FACTORS OR NOT 5

(1) 569321 = 1 569321 + 0 347 (2) 347 = 0 · 569321 + 1 · 347 (3)=(1) 1640 (2) 241 = 1 · 569321 1640 · 347 (4)=(−2) 1 · (3) 106 = 1 · 569321 −+ 1641 · 347 (5)=(3) − 2 · (4) 29 = −3 · 569321 4922 · 347 (6)=(4) − 3 · (5) 19 = 10 · 569321 −+ 16407 · 347 (7)=(−5) · (6) 10 = −13 · 569321 21329 · 347 (8)=(6) − (7) 9 = 23 · 569321 −+ 37736 · 347 (9)=(7) − (8) 1 = −36 · 569321 59065 · 347 − · − ·

The long divisions used:

569321 = 1640 347 + 241 347 = 1 · 241 + 106 241 = 2 · 106 + 29 106 = 3 · 29 + 19 29 = 1 · 19 + 10 19 = 1 · 10 + 9 10 = 1 · 9 + 1 ·

The same process can be done in a short form that only lists the essential data.

action a b (1) 569321 1 0 (2) 347 0 1 (3)=(1) 1640 (2) 241 1 - 1640 (4)=(−2) 1 · (3) 106 -1 + 1641 (5)=(3) − 2 · (4) 29 3 - 4922 (6)=(4) − 3 · (5) 19 -10 + 16407 (7)=(5)-(6)− · 10 13 - 21329 (8)=(6)-(7) 9 -23 + 37736 (9)=(7)-(8) 1 36 - 59065

EXAMPLE 1.21. Let a = 3675 and b = 791. Then gcf(a,b) = 7 = 48 3675 223 791. · − · 6 3. NUMBER THEORY

PROOF. action a b (1) 3675 1 0 (2) 791 0 1 (3)=(1) 4 (2) 511 1 -4 (4)=(2) − 1 · (3) 280 -1 +5 (5)=(3) − 1 · (4) 231 2 -9 (6)=(4) − 4 · (5) 49 -3 + 14 (7)=(5) − 1 · (6) 35 14 -65 (8)=(6) − 1 · (7) 14 -17 + 79 (9)=(7) − 2 · (8) 7 48 -223 − ·

DEFINITION 1.22. An integer m is a multiple of the integer a if a is a factor of m. An integer m is a common multiple of the integers a and b if m is a multiple of both a and b. The smallest positive common multiple of a and b is the least common multiple ofaandb denoted by lcm(a,b). Given two fractionsa/bandc/d, the least common denominator of the fractions is lcm(b,d).

EXAMPLE 1.23. Let a = 35 and b = 21. Some multiples of a (there are inﬁnitely many of them) are a = 35,70,105,140,175,... and some multiples of b are b = 21,42,63,84,105,126,... We can now see that lcm(35,21) = 105. It is easy to see that gcf(35,21) = 7, and we note the curious fact that gcf(a,b)lcm(a,b) = 7 105 = 735 = ab. · THEOREM 1.24. For any postive integers a and b, gcf(a,b)lcm(a,b) = ab.

REMARK 1.25. To ﬁnd the least common multiple of two integers a and b, we use the Euclidean algorithm to ﬁnd gcf(a,b) and then compute a b lcm(a,b) = gcf(·a,b) . EXAMPLE 1.26. 240 330 (1) lcm(240,330) = 30· = 240 11 = 2640. (See Example ??). 56371 3476 · (2) lcm(56371,3476) = 1· = 56371 3476 = 195945596. (See Example ??). (3) lcm(569321,347) = 569321 347 = 197554387· . (See Example ??). 3675 791 · (4) lcm(3675,791) = · = 3675 113 = 415275. (See Example ??). 7 · 2.THEFUNDAMENTALTHEOREMOFARITHMETIC 7

EXAMPLE 1.27. 1 1 11 6 17 240 + 330 = 2640 + 2640 = 2640 . 2. The Fundamental Theorem of Arithmetic Given a positive integer, say 113, it can always be factored as 113 = 1 113, in general a = 1 a = a 1. This is an uninteresting “trivial” factorization. · · · DEFINITION 2.1. An (positive) integer a is composite if it can be factored as a = b c where neither b nor c is equal to one. In other words, the (positive) integer a is composite· if it is the product of two positive integers that are both smaller than a. A positive integer that is not composite - can only be factored trivially - is a prime number or simply a prime.

LEMMA 2.2. Let p be a prime. Then p has exactly two positive factors, namely 1 and p. Let a be any other integer. Then gcf(a, p) = 1 or gcf(a, p) = p depending on whether p is a factor of a or not.

EXAMPLE 2.3. The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97 are all prime numbers. The number 123 is composite because 123 = 3 41. The number 3127 is composite because 3127 = 53 59. The number 1111 is composite· because 1111 = 11 101. · · PROPOSITION 2.4. Let n be a positive integer and suppose that n = a b · where a and b are again positive integers. If b √n, then a √n. ≥ ≤ PROOF. If b √n and also a > √n then a b > √n √n = n while a b = n by hypothesis. ≥ · · · COROLLARY 2.5. If the positive integer p has no factors that are less than or equal to √p then p is a prime number.

PROOF. If p had a factor larger than √n then it would have a factor √n and these have all been eliminated. ≤

COROLLARY 2.6. If the positive integer p has no prime factors which are less than or equal to √p then p is a prime number. PROOF. If p has no prime factors √n then there is no composite factor n either, since such a factor would in turn contain≤ a prime factor of n. ≤

THEOREM 2.7. There exist inﬁnitely many primes.

PROOF. This is an example of a proof by contradiction. We make an assumption, then derive an impossibility from it and conclude that the assumption was false. Assume that there are only ﬁnitely many primes. List them in increasing order: (2.8) 2,3,5,7,11,13,17,...,P, 8 3. NUMBER THEORY

so P is the largest prime. Study the number n obtained by multiplying all the primes and adding 1:

n = 2 3 5 7 11 13 17 p P+ 1. · · · · · · ··· ··· Now 2 is not a factor of n since 2 is not a factor of 1; 3 is not a factor of n since 3 is not a factor of 1; 5 is not a factor of n since 5 is not a factor of 1; in general, p is not a factor or n because p is not a factor of 1. So n is either a prime itself different from any one in the list (??) or it is a product of primes, each of which is not contained in the list (??). This means that the list (??) did not contain all the primes after all, so the assumption that there are only ﬁnitely many primes is false. Therefore, there are inﬁnitely many primes.

THEOREM 2.9. Fundamental Theorem of Arithmetic. Every integer a > 1 can be factored uniquely as

a = pn1 pn2 pnk , 1 2 ··· k

ni 1, p1 < p2 < < pk, pi primes. ≥ ··· We will indicate a proof later. For the moment we observe that every integer a > 1 is a product of primes for the following reason. If a is itself a prime it is considered a “product of primes”. If a is composite, then a = a1 a2 and both a1 and a2 are smaller than a. Now · look at a1 and a2. Factor them if they are composite, and keep going until you arrive at primes. The process have to come to a halt because integers cannot become smaller and smaller forever.

EXAMPLE 2.10. 155771 = 539 289 =(11 49) (17 17) = 11 (7 7) 17 17 = 11 7 7 17 17 = 72 11 172. · · · · · · · · · · · · · · Before getting into the uniqueness proof, let us see what it can do for us. The uniqueness says that however we arrive at a factorization into a product of primes, the result is always the same.

EXAMPLE 2.11. Let a = 23 52 7. Now suppose that b is a factor of a and a = b c. Let n1 nk m1 · m·k · b = p1 pk and c = p1 pk , where p1 < p2 < < pk, and 0 ni,mi, be the prime ··· ··· ··· n1+≤m1 nk+mk factorizations of b and c as in Theorem ??. Nowa = b c = p1 pk . By the uniqueness of the prime factorization we must have · ···

a = 23 52 7 = pn1+m1 pnk+mk . · · 1 ··· k

This means that p1 = 2,n1 + m1 = 3, p2 = 5,n2 + m2 = 2, p3 = 7, and n3 + m3 = 1. We conclude that k = 3,n1 3,n2 2 and n3 1. ≤ ≤ ≤ 2.THEFUNDAMENTALTHEOREMOFARITHMETIC 9

Using this we can list the positive factors of a = 23 52 7 = 1400 (there are 4 3 2 = 24 of them) as follows: · · × × 20 50 70 = 1, 20 50 71 = 7, 20 51 70 = 5, · · · · · · 20 51 71 = 35, 20 52 70 = 25, 20 52 71 = 175, · · · · · · 21 50 70 = 2, 21 50 71 = 14, 21 51 70 = 10, · · · · · · 21 51 71 = 70, 21 52 70 = 50, 21 52 71 = 700, · · · · · · 22 50 70 = 4, 22 50 71 = 28, 22 51 70 = 20, · · · · · · 22 51 71 = 140, 22 52 70 = 100, 22 52 71 = 700, · · · · · · 23 50 70 = 8, 23 50 71 = 56, 23 51 70 = 40, · · · · · · 23 51 71 = 280, 23 52 70 = 200, 23 52 71 = 1400. · · · · · · COROLLARY 2.12. Let a = pn1 pn2 pnk , 1 2 ··· k ni 1, p1 < p2 < < pk, pi primes. ≥ ··· Then the positive factors of a are exactly the integers f = pi1 pi2 pik , 1 2 ··· k where 0 i j ni. There are (n1 + 1)(n2 + 1) (nk + 1) such factors. ≤ ≤ ··· EXAMPLE 2.13. How many positive factors does 120 posses? We factor 120 = 23 3 5. There are 4 2 2 = 16 factors. · · × × EXAMPLE 2.14. (1) The number a = 5274132176 has 3 5 3 7 = 315 different positive factors. (2) The number· · · 512 = 29 has 10 different positive factors. (3) The number 30030 = 2 3 5 7 11 13 · · · · · has 64 different positive factors. We come to a crucial result.

THEOREM 2.15. Let a,b,c be integers. If a is a factor of bc and gcf(a,b) = 1, then a isa factor of c.

PROOF. It is given that gcf(a,b) = 1. By Theorem ?? there are integers u,v such that 1 = ua + vb. Multiplying the equation by c we obtain c = uac + vab. Now a is a factor of uac trivially, and c is a factor of ab, so vab by hypothesis. By Lemma ?? a is a factor of c. 10 3. NUMBER THEORY

COROLLARY 2.16. Let p, p1, p2 primes. If p is a factor of p1 p2, then p = p1 or p = p2.

COROLLARY 2.17. Let p, p1,..., pn be primes. If p is a factor of p1 p2 pn, then p = p1 · ··· or p = p2,or... or p = pn. Corollary ?? is the result that is needed to prove the uniqueness part of the Fundamental Theorem of Arithmetic. We illustrate the formal proof by an example.

EXAMPLE 2.18. We ﬁnd that 10500 = 2 2 3 5 5 5 7 · · · · · · Somebody else come up with a “mysterious” factorization

n1 n2 nk (2.19) 10500 = p p p , p1 < p2 < < pk. 1 2 ··· k ··· Therefore we have that 2 2 3 5 5 5 7 = pn1 pn2 pnk · · · · · · 1 2 ··· k Take a factor pi of the right hand side which is then also a factor of the left hand side. By Corollary ?? pi = 2 or pi = 3 or pi = 5 or pi = 7. Hence (??) must look like 10500 = 2n1 3n2 5n3 7n4 and we have 2 2 3 5 5 5 7 = 2n1 3n2 5n3 7n4 · · · · · · Now2 is a factor of theleft hand side soit isa factorof the right hand side by Corollary ??. This means that n1 1 and we can cancel 2 to get ≥ 2 3 5 5 5 7 = 2n1 13n2 5n3 7n4 · · · · · − The prime 2 is still a factor of the left hand side, so of the right hand side, and we must have n1 1 1. This means we can cancel another 2 and obtain − ≥ 3 5 5 5 7 = 2n1 23n2 5n3 7n4 . · · · · − Now there is no 2 on the left so there cannot be a 2 on the right and we conclude that n1 2 = 0 or n1 = 2. We next look at the prime 3 that appears on the left. It must appear − on the right also, so n2 1. Canceling the 3 we have ≥ 5 5 5 7 = 3n2 15n3 7n4 . · · · − We conclude that n2 = 1 because no 3 appears on the left and have now 5 5 5 7 = 5n3 7n4 . · · · We have a ﬁve on the left, so n3 1 enabling us to cancel a 5 and get ≥ 5 5 7 = 5n3 17n4 . · · − We still have a ﬁve on the left, so n3 1 1 enabling us to cancel another 5 to get − ≥ 5 7 = 5n3 27n4 . · − 2.THEFUNDAMENTALTHEOREMOFARITHMETIC 11

We still have a ﬁve on the left, so n3 2 1 enabling us to cancel another 5 to get − ≥ n3 3 n4 7 = 5 − 7 .

We ﬁnally get n3 = 3 and n4 = 1 showing that the “mysterious” factorization is identical with ours.

PROPOSITION 2.20. Common Factors. Suppose

n1 n2 n3 nk a = p1 p2 p3 ... pk is the factorization of the integer a into a product of primes, and

m1 m2 m3 mk b = p1 p2 p3 ... pk is the factorization of the integer b into a product of primes, then the common factors of a and b are all integers E1 E2 E3 Ek f = p1 p2 p3 ... pk where Ei is less than or equal to both ni and mi. Therefore the greatest common factor of a and b is e1 e2 e3 ek gcf(a,b) = p1 p2 p3 ... pk where ei is the lesser of ni and mi.

PROPOSITION 2.21. Multiples. (1) The positive multiples of an integer a are the integers a b where b is any positive integer. · (2) If n1 n2 n3 nk a = p1 p2 p3 ... pk is the factorization of a into a product of primes, then making the exponents ni larger produces multiples, also bringing in additional primes produces multiples. (3) The multiples of a are all integers of the form m = pm1 pm2 pm3 ... pmk qs1 qs2 qs3 1 2 3 k 1 2 3 ··· where mi ni and si 0. ≥ ≥ PROPOSITION 2.22. Common Multiples. Suppose

n1 n2 n3 nk a = p1 p2 p3 ... pk is the factorization of the integer a into a product of primes, and

m1 m2 m3 mk b = p1 p2 p3 ... pk is the factorization of the integer b into a product of primes, then the common multiples of a and b are all integers m = pM1 pM2 pM3 ... pMk qs1 qs2 qs3 1 2 3 k 1 2 3 ··· 12 3. NUMBER THEORY where Mi is greater or equal to both ni and mi and si 0. Therefore the least common multiple ofaandbis ≥ M1 M2 M3 Mk lcm(a,b) = p1 p2 p3 ... pk where Mi is the greater of ni and mi.

THEOREM 2.23. Let a and b be positive integers. Then lcm(a,b) gcf(a,b) = a b. · · EXAMPLE 2.24. Let a = 3452136173232292 and let b = 53132172292 Then lcm(a,b) = 3453136173232292 and gcf(a,b) = 52132172292 and hence lcm(a,b) gcf(a,b)=(3453136173232292)(52132172292) = ab. · 3. Exercises

EXERCISE 3.1. For each of the following values of a and b ﬁnd the unique integers q and r such that a = qb + r, 0 r < b. ≤ (1) a = 723,b = 23. (2) a = 1582,b = 231. (3) a = 123456789,b = 12345. (4) a = 12345,b = 123456789. (5) a = 0,b = 13. (6) a = 365,904,b = 2376.

EXERCISE 3.2. Decide whether or not the following numbers have factors of 2, 3, 4, 5 and 9. 12345,182734,293045,29480,298754,123456789.

EXERCISE 3.3. For the following integers a and b list the positive factors and ﬁnd the greatest common factor gcf(a,b). (1) a = 115,b = 225. (2) a = 1111,b = 333. (3) a = 237,b = 1659. (4) a = 17 23,b = 17 19. · · 3. EXERCISES 13

EXERCISE 3.4. Find the greatest common factor of the following integers a and b using the method of Euclid. (1) a = 543,b = 113. (2) a = 4563,b = 981. (3) a = 451,b = 85. (4) a = 1111,b = 11111. (5) a = 12345,b = 1234.

EXERCISE 3.5. Find the greatest common factor of the following integers a and b and ﬁnd integers u, v such that gcf(a,b) = ua + vb. (1) a = 543,b = 113. (2) a = 4563,b = 981. (3) a = 451,b = 85. (4) a = 1111,b = 11111. (5) a = 12345,b = 1234.

EXERCISE 3.6. Find the least common multiple of the following integers a and b by listing common multiples and looking for the least one among these. (1) a = 543,b = 113. (2) a = 56,b = 98. (3) a = 45,b = 85. (4) a = 111,b = 111. (5) a = 123,b = 234.

EXERCISE 3.7. Find the least common multiple of the following integers a and b using Theorem ??. (1) a = 543,b = 112. (2) a = 4563,b = 981. (3) a = 451,b = 85. (4) a = 1111,b = 11111. (5) a = 12345,b = 1234.

EXERCISE 3.8. Compute the following sums and differences using least common denomi- nators. 28 25 (1) + . 543 213 23 12 (2) + . 323 437 61 51 (3) . 245 − 300 14 23 (4) + . 289 221 14 3. NUMBER THEORY 1 1 (5) . 2599 − 12769

EXERCISE 3.9. Which ones of the following numbers are prime and which ones are composite? 211,373,453,565,463,371,637,343,111111,3487 + 539.

EXERCISE 3.10. How many positive factors do the following integers possess? 24, 67, 69, 2445, 1111, 999,

56, p7 where p is a prime, 7n, pn where p is a prime.

EXERCISE 3.11. Using your calculator ﬁnd integers q and r such that 1238459 = q4593 + r, 0 r < 4593. ≤ EXERCISE 3.12. Let x and y be (unknown) integers that are related by the equality x 130 = 75y. − Why are the following statements true? (1) 5 is a factor of x. (2) 3 is not a factor of x. (3) If y has a factor of 13, then x has a factor of 13. (4) If y is even, then x is also even. (5) If y is odd, then x is also odd. (6) If gcf(13,y) = 1, then 13 cannot be a factor of x.

EXERCISE 3.13. Verify the following statements. (1) odd odd = odd. (2) even× even = even. (3) even × odd = even. (4) odd +× odd = even. (5) even + even = even.

EXERCISE 3.14. Let a = 25 42 19 b where b is some unknown positive integer such that gcf(2 5 19,b) = 1. Which× × of the× following numbers are factors of a and which ones are not? × × 16,80,95,361,100,76.

EXERCISE 3.15. The integer 378 has 16 different positive factors and a partial list of factors is 1,2,3,6,7,9,14,18,21,63,378. Complete the list. 4. SOLUTION TO EXERCISES 15

4. Solution to exercises ?? Double check your answers by computing qb + r. It has to come to be a.

?? 12345 has factors 3 and 5, 182734 has factor 2, 293045 has factor 5, 29480 has factors 2, 4 and 5, 298754 has a factor 2, 123456789 has a factor of 9.

?? gcf(115,225) = 5, 115 = 5 23, 225 = 3252. gcf(1111,333) = 1, 1111 = 11× 101, 333 = 3237. gcf(237,1659) = 237, 237 = 3 × 79, 1659 = 3 7 79. gcf(17 23,17 19) = 17. × × × · · ?? gcf(543,113) = 1; gcf(4563,981) = 9; gcf(451,85) = 1; gcf(1111,11111) = 1; gcf(12345,1234) = 1.

?? In the following the triples ( , , ) are the rows of the tables in the text. (543,1,0) 4(113,0,1)=(91,∗1,∗ ∗4) (113,0,1) − (91,1, 4)=(22, 1−,5) (91,1, 4)− 4(22,−1,5)=(3−,5, 24) (22, 1−,5)− 7(3,5−, 24)=(1, 36−,173). Check:− 36− 543 +−173 113 =− 1. − × × (4563,1,0) 4(981,0,1)=(639,1, 4) (981,0,1) −(639,1, 4)=(342, 1−,5) (639,1, 4−) (342, −1,5)=(297−,2, 9) (342, 1−,5)− (297,−2, 9)=(45, 3,−14) (297,−2, 9)− 6(45, 3−,14)=(27−,20, 93) (45, 3,−14)− (27,20−, 93)=(18, 23−,107) (27,−20, 93)− (18, 23−,107)=(9−,43, 200) Check: 43− 4563− −200 981 = 9. − × − × (451,1,0) 5(85,0,1)=(26,1, 5) (85,0,1) −3(26,1, 5)=(7, 3−,16) (26,1, 5−) 3(7, −3,16)=(5−,10, 53) (7, 3,−16)− (5,10−, 53)=(2, 13−,69) (5,−10, 53)− 2(2, −13,69)=(−1,36, 191) − − − − (11111,1,0) 10(1111,0,1)=(1,1, 10) − − (12345,1,0) 10(1234,0,1)=(5,1, 10) (1234,0,1) −246(5,1, 10)=(4, 246− ,2461) (5,1, 10) −(4, 246,2461− )=(1,−247, 2471) Check:− 247− 12345− 2471 1234 = 1− × − × 16 3. NUMBER THEORY

?? These problems are very time consuming and unpleasant. The lesson is that with more mathematics life gets much easier. The answers given are obtained in the “advanced fash- ion”. ?? gcf(543,112) = 1, lcm(543,112) = 543 112/gcf(543,112) = 60816, gcf(4563,981) = 9; lcm(4563,981) = 4563× 981/gcf(4563,981) = 497367 gcf(451,85) = 1; 451 85 = 38335 × gcf(1111,11111) = 1;× lcm(1111,11111) = 1111 11111 = 12344321 gcf(12345,1234) = 1; lcm(12345,1234) = 12345× 1234 = 15233730 × ?? 28 25 2171 543 + 213 = 12851 23 12 733 323 + 437 = 7429 61 51 387 245 300 = 4900 14 − 23 573 289 + 221 = 3757 1 1 = 90 2599 − 12769 293687 ?? 211 = 211 is prime; 373 = 373 is prime; 453 = 3 151 is composite; 565 = 5 113 is composite; 463 = 463 is prime; 371 = 7 53 is× composite 637 = 7213 is composite;× 343 = 73 is composite; 111111 = 3 7 11 ×13 37 is comp[osite; 3487+539 = 4026 = 2 3 11 61 is composite. × × × × ??×1238459× × 4593 = 269.6.... So q = 269 and r = 1238459 269 4593 = 2942. ?? ÷ − × (1) 5 is a factor of x because 5 is a factor of 75y and 130. (2) 3 isnotafactorof x because if it were, then 3 would be a factor of x and 75, hence of 75y and x, hence of 130 which is not true. (3) If y has a factor of 13, then x has a factor of 13. True. (4) If y is even, then x is also even. True. (5) If y is odd, then x is also odd. True. (6) If gcf(13,y) = 1, then 13 cannot be a factor of x. True. Assume to the contrary that 13 is a factor of x. Then 13 is a factor of x and 130 hence of 75y. Since gcf(13,y) = 1, we know that 13 is not a factor of y so it would have to be a factor of 75, which is false. ?? (1) odd odd = odd. This was done in class. (2) even× even = even. Take two even numbers x and y. Being even they are of the × form x = 2x′, y = 2y′. Hence xy = 2x′ 2y′ = 4x′y′, so xy even has a factor 4 which is more than having a factor 2. · (3) even odd = even. Let x be even and y be odd. Then x = 2x and y = 2y + 1. × ′ ′ Hence xy = 2x′y which is even. (4) odd + odd = even. True. (5) even + even = even. True. 4. SOLUTION TO EXERCISES 17

?? Let a = 25 42 19 b where b is some unknown positive integer such that Since gcf(2 ×5 19×,b)× = 1 any integer having only factors 2, 5, and 19 must be a factor of 25 42 ×19 =× 52 24 19. So × × × × 16 : Yes,80 Yes,95 Yes,361 = 192 No,100 Yes,76 = 4 19 Yes. × ?? Look for the complementary factors 378/1 = 378, 378/2 = 189, 378/3 = 126, 378/6 = 63, 378/7 = 54, 378/9 = 42, 378/14 = 27, 378/18 = 21. Hence the complete list is 1,2,3,6,7,9,14,18,21,27,42,54,63,126,189,378.