MATH3221: Numb er Theory
Philipp BRAUN
Homework until Test #2
Section 3.1 page 43, 1. It has been conjectur conjectured ed that there are infinitely infinitely many primes primes of the form n2 Exhibit five such primes.
Solution. Five such primes are e.g. 2 = 4 2 = 22 2, 7 = 9 52 2, 47 = 49 2 = 72 2 and 79 = 81 2 = 92 2.
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− 2 = 3 2 − 2, 23 = 25 − 2 =
2. Give an example to show that the following conjecture is not true: Every positive integer can be written in the form p + p + a a2 , where p where p is either a prime or 1, and a 0.
≥
Counterexample: 25. Since none of the numbers 25, 25, 25 = 24 24,, 25 22 = 21 21,, 25 32 = 16 16,, 25 42 = 9 are primes and a has to be less than 5 since 52 = 25 which forced p to be zero, there is neither a prime p such that there is a such that p + a + a2 = 25, nor there is a such that 1 + a + a2 = 25.
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3. Prove each of the assertions below: (a) Any prime of the form 3 n + 1 is of the form 6 m + 1 (b) Each integer of the form 3 n + 2 has a prime factor of this form. (c) The only prime of the form n 3 1 is 7. (d) The only prime p for which 3 p 3 p + + 1 is a perfect square is p = 5. 2 (e) The only prime of the form n 4 is 5.
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Proof. (a) Let p = 3n + 1 be any prime. prime. Since Since p is a prime by assumption, 3n 3n has to be even (since otherwise 3n 3n + 1 would be even and therefore divisible by 2, which contradicts the assumption that p is prime prime), ), i.e. 2 3n. Sinc Sincee 3 2, 2 n by Theorem 3.1 (since 2 is a prime). prime). That That means that there there is an integ integer er m such that n = 2m. Theref Therefore ore 3n + 1 = 3(2m 3(2m) + 1 = 6m 6m + 1, as we wanted to show. (b) Let m Let m = = 3n + 2 be any integer. If m is prime, we are done. Assume m Assume m is not a prime. Therefore there are integers a, b such that a , b > 1 and ab = 3n + 2. Consid Consider er the following following cases: Case I : a = 3k, b = 3l for integers k, l. Then Then ab = 3k 3l = 3(3kl 3(3kl), ), which is of the form 3z 3z and therefore cannot equal 3n 3n + 2. Case II : a = 3k, b = 3l + 1 (or vice versa) for integers k integers k,, l. Then ab Then ab = = 3k(3l (3l + 1) = 3(3 3(3kl kl + + k), which is also of the form 3z 3z . Case III : a = a = 3k + 1, b = 3l + 1 for intege integers rs k k,, l. Then ab Then ab = = (3k (3k + 1)(3 1)(3ll + 1) = 3(3kl 3(3kl + + k k + + l l)) + 1, which is of the form 3z 3z + 1.
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So none of these cases can occur, so at least one of the factors a, b has to be 3z 3z + 2 for an integer z. Either Either this this factor is a prime, prime, or we can use the same argumen argumentt to show that 3z 3z + 2 = (3x (3x + 2)y 2)y with integers x, y. Sinc Sincee a , b > 1, 3z + 2 < 3n + 2. Thus Thus in every step the absolute value of the factor of the form 3a 3a + 2 will be smaller, so the method has to terminate with a prime of that form. (c) Since n3 1 = (n2 + n + n + 1)(n 1)(n 1), (n (n 1) (n3 1). So n So n must be less than or equal to 2. For n = 1 we have 13 1 = 0, which which is not the prime. prime. So the only prime prime of the 3 3 form n form n 1 is 2 1 = 7. (d) Assume (3 p+1) p +1) = n = n2 for a prime p prime p and and an integer n integer n.. Then p Then p = = 13 (n2 1) = 13 (n +1)(n +1)(n 1 1 1). Since p Since p is a prime, 3 (n 1) = 1, which implies n implies n = = 4 and therefore p therefore p = = 3 5 3 = 5. (e) Since n Since n2 4 = (n +2)(n +2)(n 2), n 2), n 2 has to equal 1 (because otherwise n otherwise n 2 4 is no prime since (n (n 2) (n2 4)), i.e. n = 3, so the only prime of that form is 32 4 = 5.
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4. If p p
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≥ 5 is a prime number, show that p
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+ 2 is composite.
Proof. p2 + 2 = ( p +1)( p 1) + 3. Sinc Sincee p is a prime bigger than 3, p 3, p is not divisible by 3. But since one of three consecutiv consecutivee numbers numbers is divisible divisible by three, so is either ( p p 1) or ( p + p + 1). Since 3 3, by Theorem 2.2 (vii) 3 ( p + p + 1)( p 1)( p 1) + 3, so p2 + 2 is divisible by 3 and therefore therefore composite. composite.
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6. Establish each of the following statements: (a) Every integer of the form n 4 + 4, with n with n > 1, is composite. (b) If n n > 4 is composite, then n divides (n (n 1)!. (c) Any integer of the form 8 + 1, where n where n 1, is composite. (d) Each integer n > 11 can be written as the sum of two composite numbers.
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n
Proof. (a) n4 + 4 = (n2 + 2n +2)(n +2)(n2 2n + 2), and and for n for n > 1, each of these factors is bigger than 1, hence n4 + 4 is composite. (b) If n > 4 is composite, then it is a square number (then 2 n < n since n > 4, and therefore n therefore n 1 2 . . . n . . . 2 n . . . (n 1) = (n (n 1)!) or it can be written as a product of a prime 1 < 1 < p < n and an integer n < m < n. n. Since (n (n 1)! = 1 2 p p . . . m . . . (n 1), pm (n 1)!, i.e. n (n 1)!. (c) 8n + 1= 1=(2 (2n )3 + 1 = (2 (2n + 1)((2 1)((2n )2 (2n ) + 1). Since Since both factor factorss are bigger bigger than than 1 for n n 1, 8 + 1 is not prime and therefore composite. (d) In case n > 11 is even, there is k > 5 such that n = 2k . Therefore Therefore n = 6 + 2(k 2(k 3), where 6 = 2 3 is a composite number and 2 (k 3) is a composite number (since k 3 > 1 > 1 since k since k > 5). In case n case n is odd, there is k is k > 5 such that n that n = = 2k + 1. Theref Therefore ore n = 9 + 2(k 2(k 4), where 9 = 3 3 and 2(k 2(k 4) are composite numbers (since k (since k 4 > 1 > 1 because k because k > 5).
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Section 3.2 page 49, 1. Determine whether the integer 701 is prime by testing all primes p divisors. Do the same for integer 1009.
≤ √ 701 as possible
√ ≈
Solution. 701 26 26..47, i.e. we have to test the primes 2, 2, 3, 5, 7, 11 11,, 13 13,, 17 17,, 19 and 23 as possible divisors of 701: 701/ 701 /2 = 350. 350.5 / Z 2 701. 701. 701/ 701 233.6 / Z 3 701. 701. /3 = 233. 701/ 701 /5 = 140. 140.2 / Z 5 701. 701. 701/ 701 /7 100 100..14 / Z 7 701. 701. 701/ 701 /11 = 63. 63.72 / Z 11 11 701. 701. 701/ 701 /13 53 53..92 / Z 13 13 701. 701. 701/ 701 /17 41 41..24 / Z 17 17 701. 701. 701/ 701 36..89 / Z 19 19 701. /19 36 701. 701/ 701 /23 30 30..48 / Z 23 23 701. 701. Hence 701 is a prime. Since 1009 31 31..76, we have to test the primes 2, 2, 3, 5, 7, 11 11,, 13 13,, 17 17,, 19 19,, 23 23,, 29 and 31 as possible divisors of 1009: 1009/ 1009 /2 = 504. 504.5 / Z 2 1009. 1009. 1009/ 1009 /3 = 336. 336.3 / Z 3 1009. 1009. 1009/ 1009 /5 = 201. 201.8 / Z 5 1009. 1009. 1009/ 1009 /7 144 144..14 / Z 7 1009. 1009. 1009/ 1009 /11 = 91. 91.72 / Z 11 11 1009. 1009. 1009/ 1009 /13 77 77..62 / Z 13 13 1009. 1009. 1009/ 1009 /17 59 59..35 / Z 17 17 1009. 1009. 1009/ 1009 /19 53 53..11 / Z 19 19 1009. 1009. 1009/ 1009 43..87 / Z 23 23 1009. /23 43 1009. 1009/ 1009 /29 34 34..79 / Z 29 29 1009. 1009. 1009/ 1009 /31 32 32..54 / Z 31 31 1009. 1009. Hence 1009 is a prime.
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2. Employi Employing ng the Sieve Sieve of Eratosthenes, Eratosthenes, obtain all the primes primes betw b etween een 100 and 200.
Solution. 100 101 102 101 102 103 110 1 110 111 11 112 112 113 120 1 120 121 21 122 122 12 12 3 130 131 132 132 13 13 3 140 1 140 141 41 142 142 14 14 3 150 151 152 152 15 15 3 160 1 160 161 61 162 162 163 170 1 170 171 71 172 172 173
104 114 124 134 144 154 164 174
105 106 107 108 105 106 107 108 109 115 1 115 116 16 117 117 118 118 119 1 19 125 126 125 126 127 128 127 128 129 1 29 135 136 135 136 137 138 137 138 139 145 1 145 146 46 147 147 148 148 149 155 156 155 156 157 158 157 158 159 1 59 165 166 165 166 167 168 167 168 169 1 69 175 1 175 176 76 177 177 178 178 179
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180 181 182 182 18 18 3 184 785 785 1 186 86 187 187 188 188 189 1 89 190 191 192 191 192 193 194 195 195 196 196 197 198 197 198 199 (x means x is divisible by 2, 2, 3, or 5, x means that x is divisible by 7, 7, 11 11,, or 13.) Thus the primes between 100 and 200 are 101, 101, 103 103,, 107 107,, 109 109,, 113 113,, 127 127,, 131 131,, 137 137,, 139 139,, 149 149,, 151 151,, 157,, 163 157 163,, 167 167,, 173 173,, 179 179,, 181 181,, 191 191,, 193 193,, 197 and 199.
≤ √ n, show that n > 1 is either a prime or the product of
3. Given Given that p n n for all primes p two primes.
3
Proof. If n n was the product of three or more primes, at least one of them had to be less than or equal to n. Since by assumption none of the primes p n divides n, this is a contradiction. Hence n Hence n is a prime or a product of two primes.
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≤ √
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4. Establish the following facts: (a) p is p is irrational for any prime p. p . (b) If a a > 0 and a is rational, then (c) For n For n 2, n is irrational.
√
√ ≥ √ n
√ a must be an integer. n
n
√
√
Proof Proof.. (a) Ass Assum umee p is ration rational. al. Then Then there there are integer integerss a, b with p = ab and Square the equation and multiply multiply both sides with b to get b2 p = gcd( gcd(a, b) = 1. Square p = a2 . That means p means p a2 and since p since p is is a prime p prime p a by Theorem 3.1. Therefore there is an integer c integer c such such 2 2 2 2 2 that a = cp = cp.. Applying that to our first equation, we have b p = p = (cp) cp) = c p , or b = c 2 p. p. That means p means p b2 , and again by Theorem 3.1 p b, so p so p gcd( gcd(a, b), a contradic contradiction. tion. p (b) Assume a is rational, i.e. a = q for relatively prime integers p, q . Then Then a = pq . Since p Since p and and q q are are relatively prime, p prime, p n and q and q n are relatively prime, too. That a That a is is an integer n forces q forces q to be 1, so q so q = = 1, i.e. a = p = p,, an integer. (c) Since 2n > n for n 2, 2n > n. But But 2n = 2, so n < 2. On the other other hand hand n n > 1 since 1 = 1 < n. By part (b), n is either an integer or irrational, so n has to be irrational as there is no integer between 1 and 2.
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5. Show that any composite composite three-digit three-digit number number must have a prime factor factor less than or equal to 31.
Proo Proof. f. Let Let c be any composite composite three-di three-digit git number. number. Since Since c is com composi posite, te, there there are integers a, b such that c = ab. ab . Assume Assume a and b are both greater greater than than 31, i.e. at least 32. 5 5 10 Then ab Then ab 32 32 = 2 2 = 2 = 1024, which contradicts to the fact that c that c is is a three-digit number by assumption.
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Section 3.3 page 57, 1. Verify that the integers 1949 and 1951 are twin primes.
√
Solution Solution.. Since Since 1951 44 44..17, we have have to chec check k if 1949 1949 or 1951 1951 are are divi divisi sibl blee by 2, 3, 5, 7, 11 11,, 13 13,, 17 17,, 19 19,, 23 23,, 29 29,, 31 31,, 37 37,, 41 41,, or 43. Obviously, both numbers are odd and therefore not divisible by 2. Since the sum of the digits of 1949, 1+9+4+9=23 and the sum of the digits of 1951, 1+9+5+1=16 are both not divisible by 3, both numbers are not divisible by 3. Since Since both numbers numbers do not end with “0” or “5”, they are not divisibl divisiblee by 5. Using Using the same method as in exercise 1 on page 49, we have 1949/ 1949 /7 278 278..43 / Z 7 1949, 1949, 1951 1951/ /7 278 278..71 / Z 7 1951 1951 1949/ 1949 /11 = 177. 177.18 / Z 11 11 1949, 1949, 1951 1951/ /11 = 177. 177.36 / Z 11 11 1951 1951 1949/ 1949 /13 149 149..92 / Z 13 13 1949, 1949, 1951 1951/ /13 150 150..08 / Z 13 13 1951 1951 1949/ 1949 /17 114 114..65 / Z 17 17 1949, 1949, 1951 1951/ /17 114 114..76 / Z 17 17 1951 1951 1949/ 1949 /19 102 102..58 / Z 19 19 1949, 1949, 1951 1951/ /19 102 102..68 / Z 19 19 1951 1951 1949/ 1949 /23 84 84..74 / Z 23 23 1949, 1949, 1951 1951/ /23 84 84..83 / Z 23 23 1951 1951 1949/ 1949 /29 67 67..21 / Z 29 29 1949, 1949, 1951 1951/ /29 67 67..28 / Z 29 29 1951 1951 1949/ 1949 /31 62 62..87 / Z 31 31 1949, 1949, 1951 1951/ /31 62 62..94 / Z 31 31 1951 1951 1949/ 1949 /37 = 52. 52.675 / Z 37 37 1949, 1949, 1951 1951/ /37 = 52. 52.729 / Z 37 37 1951 1951 1949/ 1949 /41 = 47. 47.53658 / Z 41 41 1949, 1949, 1951 1951/ /41 = 47. 47.58536 / Z 41 41 1951 1951 1949/ 1949 /43 45. 45 .33 / Z 43 1949 1949,, 1951 1951/ /43 45. 45 .37 / Z 43 1951. 1951. Therefore Therefore 1949 and 1951 both are prime. Since 1951 1949 = 2, they are twin primes.
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2. (a) If 1 is added to a product of twin primes, prove that a perfect square is always obtained. (b) Show that the sum of twin primes p and p + p + 2 is divisible divisible by 12, provided provided that p that p > 3.
Proof. (a) Assume p Assume p and and p p +2 are twin primes. Then their product is p is p(( p +2) = p = p 2 + 2 p. p. 2 2 If we add 1, we have p + 2 p 2 p + + 1, which is equal to ( p ( p + + 1) , which obviously is a perfect square. (b) The sum of the twin primes p and p + 2 is p + ( p + 2) = 2 p 2 p + + 2. Sinc Sincee p is a prime p + 2) greater than 3, p is of the form 3k 3k + 2 for an integer k (if it was of the form 3k 3k, it would be divisible by 3, which contradicts the assumption that p that p is a prime, and if it was of the form 3k 3k + 1, p 1, p + 2 = 3k + 1 + 2 = 3(k + 1) would would not be a prime). prime). So 2 p + 2 is of the form form 2(3k 2(3k + 2) + 2 = 6k 6 k + 4 + 2 = 6k + 6 = 6k + 6 = 6(k 6(k + 1). Since Since p is a prime, k is odd (because if k is even, then so is 3k 3 k + 2, which contradicts that p is a prime greater than 3). Hence k + 1 is even, i.e. there is an integer n such that k + 1 = 2n 2n. That means that 2 p + p + 2 = 6(k 6(k + 1) = 6 2 n = 12 12n n, so the sum of the primes p primes p and p + p + 2 is divisible by 12.
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3. Find all pairs of primes p and q satisfying q satisfying p p
− q = = 3.
Solution Solution.. Since Since the differenc differencee should should be 3, one of the primes primes has to be even and the
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other other has to be odd. Since Since the only even even prime is 2, the only of these pairs pairs is p = 5 and q = = 2. 5. In 1752, Goldbach Goldbach submitted submitted the followi following ng conjecture conjecture to Euler: Every Every odd integer integer can be 2 written in the form p + p + 2a 2 a , were p is either a prime or 1 and a 0. Show that the intege integerr 5777 refutes this conjecture.
≥
Solution. a has to be less than or equal to 53, since 2 54 2 = 5832 > 57 5777 77.. The The set 5777 2a2 0 a 53 = 5777 = 53 109 109,, 5775 = 5 1155 1155,, 5769 = 3 1923 1923,, 5759 = 13 443 443,, 5745 = 5 1149 1149,, 5727 = 3 1909 1909,, 5705 = 5 1141 1141,, 5679 = 3 1893 1893,, 5649 = 3 1883 1883,, 5615 = 5 1123 1123,, 5577 = 3 1859 1859,, 5535 = 5 1107 1107,, 5489 = 11 499 499,, 5439 = 3 1813 1813,, 5385 = 5 1077 1077,, 5327 = 7 761 761,, 5265 = 5 1053 1053,, 5199 = 3 1733 1733,, 5129 = 23 223 223,, 5055 = 5 1011 1011,, 4977 = 3 1659 1659,, 4895 = 5 979 979,, 4809 = 3 1603 1603,, 4719 = 3 1573 1573,, 4625 = 5 925 925,, 4527 = 3 1509 1509,, 4425 = 5 885 885,, 4319 = 7 617 617,, 4209 = 3 1403 1403,, 4095 = 5 819 819,, 3977 = 41 97 97,, 3855 = 5 771 771,, 3729 = 3 1243 1243,, 3599 = 59 61 61,, 3465 = 5 693 693,, 3327 = 3 1109 1109,, 3185 = 5 637 637,, 3039 = 3 1013 1013,, 2889 = 3 963 963,, 2735 = 5 547 547,, 2577 = 3 859 859,, 2415 = 5 483 483,, 2249 = 13 173 173,, 2079 = 3 693 693,, 1905 = 5 381 381,, 1727 = 11 157 157,, 1545 = 5 309 309,, 1359 = 3 453 453,, 1169 = 7 167 167,, 975 = 5 195 195,, 777 = 7 111 111,, 575 = 5 115 115,, 369 = 3 123 123,, 159 = 3 53 contains neither a prime nor 1, so 5777 cannot be written in the wanted form.
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7. A conjecture of Lagrange (1775) asserts that every odd integer greater than 5 can be written as a sum p sum p 1 + 2 p 2 p2 , where p where p 1 , p2 are both primes. Confirm this for all odd integers through 75.
Solu Solutio tion. n. 7 = 3 + 2 2, 9 = 5 + 2 2, 11 = 7 + 2 2, 13 = 7 + 2 3, 15 = 5 + 2 5, 17 = 7 + 2 5, 19 = 5 + 2 7, 21 = 7 + 2 7, 23 = 13 + 2 5, 25 = 3 + 2 11 11,, 27 = 5 + 2 11 11,, 29 = 7 + 2 11 11,, 31 = 5+ 2 13 13,, 33 = 7+ 2 13 13,, 35 = 13+2 11 11,, 37 = 11+2 13 13,, 39 = 13+2 13 13,, 41 = 7 + 2 17 17,, 43 = 5 + 2 19 19,, 45 = 7 + 2 19 19,, 47 = 13 13 + 2 17 17,, 49 = 3 + 2 23 23,, 51 = 5 + 2 23 23,, 53 = 7 + 2 23 23,, 55 = 17+ 17+ 2 19 19,, 57 = 19+ 19+ 2 19 19,, 59 = 37+ 37+ 2 11 11,, 61 = 3+ 2 29 29,, 63 = 5+ 2 29 29,, 65 = 7 + 2 29 29,, 67 = 5 + 2 31 31,, 69 = 7 + 2 31 31,, 71 = 13 + 2 29 29,, 73 = 47 + 2 13 13,, 75 = 13 + 2 31.
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Sections 4.1 & 4.2 page 67, 1. Prove Prove each of the following following assertions: assertions: (a) If a a b (mod n (mod n)) and m and m n, then a then a b (mod m (mod m)) (b) If a a b (mod n (mod n)) and c and c > 0, then ca cb (mod cb (mod cn cn)) (c) If a a b (mod n (mod n)) and the integers a,b, a,b, n are all divisible by d > 0, then a/d
≡ ≡ ≡
|
≡ ≡
≡ b/d (mod b/d (mod n/d n/d))
Proof. (a) a (a) a b (mod n (mod n)) n (a b). Since m Since m n, by Theorem 2.2 (iv) m (iv) m (a b), or in other words a b (mod m). (b) a (b) a b (mod n (mod n)) n (a b). Thus there is an integer d integer d such such that nd that nd = = a a b. Multiply both sides with c with c to to get cnd get cnd = = c c((a b) = ca cb. cb. Since c Since c > 0 and n and n > 0, cn 0, cn > 0, and since cn (ca cb), cb), ca cb (mod cb (mod cn cn). ). (c) a b (mod n) n (a ( a b). Therefore Therefore there there is an integer integer c such that cn = cn = a b. If a, b and n all are divisible by d, then cn/d then cn/d = = (a b)/d = /d = a/d b/d, b/d, where cn where cn/d, /d, a/d, a/d, b/d all are integers. So n/d (a/d b/d). b/d). Since d > 0, n/d > 0 > 0 and therefore a/d b/d (mod b/d (mod n/d). n/d).
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2. Give an example to show that a 2
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≡b
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(mod n (mod n)) need not imply a
≡
≡ b (mod n (mod n). ).
Solution Solution.. Let a = 2, b = 3, n = 5. 5. Then Then a2 = 4, b2 = 9, and 4 9 (mod 5) (because 9 4 = 5, where 5 is obviously a multiple of 5), but 2 3 (mod 5) (because 3 2 = 1, and 1 is not a multiple of 5).
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3. If a a
−
≡ b (mod n (mod n), ), prove that gcd g cd((a, n) = gcd( gcd (b, n).
Proof. a b (mod n) n (a b) there is an integer c such that cn = a b. Let d be any divisor of both a and n. Then Then there are intege integers rs k, l such that dk = a and dl = dl = n n,, which means cdl means cdl = dk = dk b, or in other words b words b = dk = dk cdl = cdl = d d((k cl), cl), so d divides b. Especia Especially lly gcd( gcd(a, n) gcd g cd((b, n). Similarly Similarly we can show show that gcd( gcd(b, n) gcd g cd((a, n), which forces the gcd’s to be equal.
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4. (a) Find the remainders when 2 50 and 4165 are divided by 7. (b) What is the remainder when the following sum is divided by 4? 1 5 + 25 + 35 +
510
5
· · · + 99
+ 10 10005
Solution Solution.. (a) 250 = 2 , and 25 32 5 (mod 7). By Theo Theore rem m 4.2 (f), 250 510 5 (mod 7). 510 = 52 , and 52 25 4 (mod 7). Aga Again in by Theore Theorem m 4.2 (f ), 250 45 (mod 7). 45 = (25 )2 , and 25 5 (mod 7), as we already saw. Hence 250 52 25 4 (mod 7), i.e. the remainder when dividing 250 by 7 is 4. Since 41 ( 1) (mod 7), 4165 ( 1)65 (mod 7) by Theorem 4.2 (f). Since ( 1)65 = 1, 4165 1 6 (mod 7), so the remainder when dividing 4156 by 7 is 6. (b) The even numbers to the fifth power have a factor 25 and therefore are divisible by 4 (i.e. (i.e. congru congruen entt 0 (mod 4)). The other other numbers numbers are alterna alternatin tingg either either congru congruen entt 1 or congruent -1, and so are their fifth powers. Since the sum of numbers is congruent to the sum of any numbers their summands are congruent to by Theorem 4.2 (d), 15 + 25 + 35 +
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≡ ≡
≡ ≡
≡ ≡ ≡
≡−
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7
≡
≡
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· · · + 995 + 1005 ≡ 1 + 0 − 1 + · · · − 1 + 0 ≡ 0 (mod 4), i.e.
the remainder remainder when dividing dividing
the sum by 4 is 0.
5. Prove that the integer 53 103 +10353 is divisible by 39, and that 111 333 +333111 is divisible by 7.
Proof. 53103 + 10353 ( 1)103 + 153 1 + 1 0 (mod 3) (by Theorem 4.2 (f)), and 53103 +10353 1103 + ( 1)53 (mod 13), so both 3 and 13 divide 53103 +10353 . Since 3 and 13 are relatively prime (in fact, they are both prime numbers), the fist statement is true by Collary 2 of Theorem 2.4. For the second statement, note that 111333 + 333111 ( 1)333 + 4111 1 + (43 )37 1 + (64)37 = 1 + 137 1 + 1 0 (mod 7).
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6. For n For n 1, use congruence theory to establish each of the following divisibility statements: (a) 7 52 + 3 25 −2 . (b) 13 3 +2 + 42 +1 . (c) 27 25 +1 + 5 +2 . (d) 43 6 +2 + 72 +1 .
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n
n
Proof. (a) For n For n = = 1, the statement is true since 7 49 = 25+ 25+ 24 = 25+ 25+ 3 8 = 5 2 + 3 23 . Assume the statement is true for n = k. Then Then 52k + 3 25k 2 0 (mod 7). 52(k+1) + 3 25(k+1) 2 52 52k + 25 3 25k 2 25 52k + 32 3 25k 2 25(52k + 3 25k 2 ) + 7 3 25k 2 (mod (mod 7). Sinc Sincee 52k + 3 25k 2 0 (mod 7), 25(52k + 3 25k 2 ) 0 (mod 7) by Theorem 4.2 (e). (e). Since Since 7 is a factor factor,, 7 3 25k 2 0 (mod 7). Theref Therefore ore the statemen statementt is true for n = k = k + + 1 by Theorem 4.2 (d). (b) For n = 1, we have 13 91 = 27 + 64 = 3 3 + 43 , which is true since 13 7 = 91. Assume Assume the statement statement is true for n for n = = k k.. Then 3(k+1)+2 + 42(k+1)+1 3 3k+2 + 42 42k+1 3(3k+2 + 4 2k+1 ) + 13 42k+1 (mod 13). By induction induction assumption, assumption, 3k+2 + 4 2k+1 0 (mod 13), so 3(3k+2 + 42k+1 ) 0 (mod 13) by Theorem 4.2(e). Since 13 is a factor of 13 42k+1 , 13 42k+1 0 (mod 13). By Theorem 4.2 (d), the statement is true for n = k = k + + 1. (c) For n = 1, it is stated that 27 189 = 64 + 125 = 26 + 53 , which is true since 27 7 = 189 189.. Assu Assume me the state stateme ment nt is true true for for n = k. Then 25(k+1)+1 + 5(k+1)+2 25 25k+1 + 5 5k+2 32 25k+1 + 5 5k+2 5(25k+1 + 5 k+2 ) + 27 25k+1 (mod (mod 27). 27). By assumption and since 27 is a factor of the second summand, for the same reasons as in parts (a) and (b), 5(25k+1 + 5 k+2 ) + 27 25k+1 0 (mod 27), hence the statement is true for n for n = = k k + + 1. (d) Since 43 559 = 216 + 343 = 63 + 73 because 43 13 = 559, the statement is true for n = 1. Assume Assume the statem statemen entt is true true for n = k. Simi Similar lar to parts parts (a), (a), (b) (b) and and (c), ( 2( 2 2 2 2 k +1)+2 k +1)+1 k+2 k+1 k+2 k+1 k+2 k+1 6 +7 6 6 +7 7 6 6 +49 7 6( 6 +7 )+43 72k+1 0 + 0 0 (mod 43), thus the statement is true for n for n = k = k + + 1. −
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· · ·
−
−
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≡ · ·
−
−
· ·
·
·
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−
≡
−
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·
≡
≡ ·
≡
−
·
· ·
· · ≡
|
· ·
·
≡ ·
·
≡
·
−
·
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·
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·
·
≡ ·
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≡ ·
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8
≡ ·
·
≡
7. For n For n
≥ 1, show that (−13)
+1
n
≡ (−13)
n
+ ( 13)
−
−1
n
(mod 181).
Proof Proof.. For n = 1, the statement is true because ( 13)2 169 12 131 + ( 13)0 (mod 181). ). Assu Assume me the stat statem emen entt is true for n = k . Then ( 13)(k+1)+1 (modk 181 ( 13) ( 13) + ( 13)k 1 ( 13)k+1 + ( 13)(k+1) 1 (mod 181).
− −
−
−
−
≡−
−
9
−
−
≡
≡− ≡− −
≡
Section 4.3 page 73, 1. Use the binary exponentiation algorithm to compute both 19 53 (mod 503) and 141 47 (mod 1537).
Solution. 1953 = 1932 1916 194 19, so 1953 1932 1916 130321 19 1932 (194 )4 44 19 1932 (44)4 44 19 1932 3, 748 748,, 096 44 19 (1916 )2 243 44 19 2432 243 44 19 ˙ 59049 243 4419 (198 243) (44 19) 48114 836 329 333 406 (mod 503) 14147 = 14132 1418 1414 1412 141, so 14147 14132 1418 (1412 )2 19881 141 14132 1418 (1437)2 1437 141 14132 (1414 )2 2, 064 064,, 969 1437 141 14132 (778)2 778 1437 141 (1418 )4 605284 778 1437 141 (1243)4 1243 778 1437 141 (1, (1, 545 545,, 049)2 1243 778 1437 141 3642 1243 778 1437 141 132496 1243 778 1437 141 (314 1243) (778 1437) 141 390302 1, 117 117,, 986 141 1441 587 141 845867 141 517 141 72897 658 (mod 1537).
·
· · ≡ · · ≡ · · · · ≡ · · · ≡ · · · · ·
·
· · ·
·
·
· · ·
·
≡ · ≡ ≡ ·
≡ · · · · ≡ · · · ≡ · ≡ · ≡ · · · · · ≡ · · · · ≡ · · · · ≡ · ≡ · ≡
· ≡ · · ≡ · ≡ · · ≡ · · · · · · · · ≡
· · ≡ · · ≡ ·
· · ≡ · · · · ≡ ·
2. Prove the following statements: (a) For any integer a, a , the units digit of a of a 2 is 0, 1, 4, 5, 6, or 9. (b) Any one of the integers 0, 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9 can occur as units digit of a of a 3 . (c) For any integer a, a , the units digit of a of a 4 is 0, 1, 5, or 6. (d) The units digit of a triangular number is 0 , 1, 3, 5, 6, or 8.
Proof. Let a be an integer. (a) By Divisi Division on Alg Algori orithm thm,, a is of the form 10k 10k + r, wher wheree k is an integer and r 2 2 2 2 2 0, 1, , 9 . a = (10k (10k + r ) = 10 k + 2 10 r + r , and since 10 is a factor of both 102 k2 and 2 10 r , a2 r 2 (mod 10). Considering the values r can have, we get: in case r case r = 0: a2 02 0 (mod 10), in case r case r = 1: a2 12 1 (mod 10), in case r case r = 2: a2 22 4 (mod 10), in case r case r = 3: a2 32 9 (mod 10), in case r case r = 4: a2 42 16 6 (mod 10), in case r case r = 5: a2 52 25 5 (mod 10), in case r case r = 6: a2 62 36 6 (mod 10), in case r case r = 7: a2 72 49 9 (mod 10), in case r case r = 8: a2 82 64 4 (mod 10), and finally in case r = 9: a2 92 81 1 (mod 10). So we see that in each case a2 is congruent to either 0, 0, 1, 4, 5, 6, or 9 (mod 10), which means that one of these is the units digit of a a 2 . (b) Using the same method as in (a), we have a3 r3 (mod 10), because by Binomial Theorem the other summands of the binomial extension of (10k (10k + r + r))3 have a factor 10. Considering the cases, we see that in case r case r = 0: a3 03 0 (mod 10), in case r case r = 1: a3 13 1 (mod 10),
{
··· } · ·
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
∈
· ·
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
≡ ≡ ≡ ≡ ≡ ≡
≡
≡ ≡ ≡ ≡
10
in case r case r = 2: a3 23 8 (mod 10), in case r case r = 3: a3 33 27 7 (mod 10), in case r case r = 4: a3 43 64 4 (mod 10), in case r case r = 5: a3 53 125 5 (mod 10), in case r case r = 6: a3 63 216 6 (mod 10), in case r case r = 7: a3 73 343 3 (mod 10), in case r case r = 8: a3 83 512 2 (mod 10), and finally in case r = 9: a3 93 729 9 (mod (mod 10). 10). So we see that that each each of the the numbe umbers rs 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 occur as units digit of a a 3 . (c) We use again the same method to obtain in case r case r = 0: a4 04 0 (mod 10), in case r case r = 1: a4 14 1 (mod 10), in case r case r = 2: a4 24 16 6 (mod 10), in case r case r = 3: a4 34 81 1 (mod 10), in case r case r = 4: a4 44 256 6 (mod 10), in case r case r = 5: a4 54 625 5 (mod 10), in case r case r = 6: a4 64 1296 6 (mod 10), in case r case r = 7: a4 74 2401 1 (mod 10), in case r case r = 8: a4 84 4096 6 (mod 10), and finally in case r = 9: a4 94 6561 1 (mod 10). So the only possibilities for the units digit of a4 are 0, 0, 1, 5, and 6. (d) Let a be a triang triangula ularr numbe number. r. By problem problem 1.(a) of section section 2.1, there is an integer integer n n(n+1) such that a that a = 2 = n 2+n . By Division Division Algorith Algorithm, m, n n is of the form 10k 10k + r + r,, where k is
≡ ≡ ≡ ≡ ≡ ≡ ≡
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ 2
2
an integer and r and r 0, 1, , 9 . So a So a = (10k+r)2+10k+r = 10 k +2 10 2r+r +10k+r . Not that that a multiple of 5 ends either with 0 or with 5, so 5k 5k 0 (mod 10) or 5k 5k 5 (mod 10). 0 +0+10k In case r = 0: a 0 + 5k 5k (mod 10), so a so a [0], [0], [5] 2 1 +1+10k in case r case r = 1: a 1 + 5k 5 k (mod 10), so a so a [1], [1], [6] 2 2 +2+10k in case r case r = 2: a 3 + 5k 5 k (mod 10), so a so a [3], [3], [8] 2 k in case r case r = 3: a 3 +3+10 6 + 5k 5 k (mod 10), so a so a [6], [6], [1] 2 4 +4+10k in case r case r = 4: a 10 + 5k 5k 0 + 5k 5 k (mod 10), so a [0], [0], [5] 2 k in case r case r = 5: a 5 +5+10 15 + 5k 5k 5 + 5k 5 k (mod 10), so a [5], [5], [0] 2 6 +6+10k in case r case r = 6: a 21 + 5k 5k 1 + 5k 5 k (mod 10), so a [1], [1], [6] 2 7 +7+10k in case r case r = 7: a 28 + 5k 5k 8 + 5k 5 k (mod 10), so a [8], [8], [3] 2 8 +8+10k in case r case r = 8: a 36 + 5k 5k 6 + 5k 5 k (mod 10), so a [6], [6], [1] , and finally 2 9 +9+10k in case r case r = 9: a 45 + 5k 5k 5 + 5k 5k (mod 10), so a so a [0], [0], [5] . In every case a 2 is element of one of the congruence classes [0], [0], [1], [1], [3], [3], [5], [5], [6], [6], and [8] (mod 10). This implies the statement.
∈{
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
2
2 2
2
2
2
2
2 2
2
··· } ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
2
≡
≡ ≡ ≡ ≡ ≡ ≡
11
∈{ ∈{ ∈{ ∈{
2
·
} } } }
·
∈{ ∈{ ∈{ ∈{ ∈{ ∈{
2
≡
} } } } } }
9
3. Find the last two digits of the number 9 9 .
Solution. 99 (92 )4 9 814 9 14 9 9 (mod 9 9 k, and therefore 99 = 910k+9 . So 99 910k+9 910k 100). Since 99 93 93 93 729 729 729 29 29 (mod 100), (99 9)k 99 (89 9)k 89 (801)k 89 9 the last two digits of 99 are 89.
10), i.e. 99 = 10 10k k + 9 for an integer · ≡ · ≡ 9 10 9 ≡ ≡ · 9 ≡ (9 ) · 9 ≡ (99 · 9) · 99 (mod ≡ · · ≡ · · ≡ · · 29 ≡ 841 · 29 ≡ 41 · 29 ≡ 1189 ≡ 89 · · ≡ · · ≡ · ≡ 1 · 89 ≡ 89 (mod 100). Therefore Therefore
≡
· ≡
k
k
k
4. Without performing the divisions, determine whether the integers 176,521,221 and 149,235,678 are divisible by 9 or 11.
Solu Solutio tion. n. Usin Usingg Theo Theore rem m 4.5 4.5,, we only have have to chec check k if the sum sum of the digi digits ts of the numbe umbers rs are are div divis isib ible le by by 9 to che checck divi divisi sibi bili litty by by 9. 1 + 7 + 6 + 5 + 2 + 1 + 2 + 2 + 1 = 27 27,, whic which h is divi divisi sibl blee by 9, 9, so 176,5 176,521, 21,22 221 1 is divi divisi sibl blee by 9. 9. 1 + 4 + 9 + 2 + 3 + 5 + 6 + 7 + 8 = 45, which is also divisible by 9, so 149,235,678 is divisible by 9, as well. To check divisibility by 11, we use Theorem 4.6: 1 7 + 6 5 + 2 1 + 2 2+1 = 3, which is not divisible divisible by 11, so 176,521 176,521,221 ,221 is not divisible by by 11. 1 4 +9 2 +3 5+ 6 7+8 = 9, which is also not divisible by 11, so 149,235,678 is not divisible by 11.
−
−
−
−
−
−
−
− −
7. Establish the following divisibility criteria: (a) An integer is divisible by 2 if and only if its units digit is 0 , 2, 4, 6, or 8. (b) An integer is divisible by 3 if and only if the sum of its digits is divisible by 3. (c) An integer is divisible by 4 if and only if the number formed by its tens and units digits is divisible by 4. (d) An integer is divisible by 5 if and only if its units digit is 0 or 5.
Proof. Proof. (a) An intege integerr n is divisible by 2 if and only if there is an integer k such that n = 2k. Let l be that number out of k, k 1, k 2, k 3, and k 4 that is divisible by 5. Since 2l 2l is divisible by 10, it is congruent 0 (mod 10). k is equal to one of the numbers l, l + 1, 1 , l + 2, 2 , l + 3, 3 , and l + 4. In case k = l, l , n 2k 2l 0 (mod 10), in case k = l + l + 1, n 2k 2(l 2(l + 1) 2l + 2 2 (mod 10), in case k case k = l = l + + 2, n 2k 2(l 2(l + 2) 2l + 4 4 (mod 10), in case k = l + l + 3, n 2k 2(l 2(l + 3) 2l + 6 6 (mod 10), and finally in case k = l + 4, n 2k 2 k 2(l 2( l + 4) 2l 2 l + 8 8 (mod 10), so the units digit of n is one of the numbers 0, 0, 2, 4, 6, and 8. Trivially rivially,, if n ends with 0, 0, 2, 4, 6, or 8, it is even and therefore divisible by 2. (b) Let n be an integ integer. er. There There are m Z, a0 , a1 , . . . , am 0, 1, . . . , 9 such that n =
−
≡ ≡
≡
≡ ≡
≡
−
−
≡ ≡ ≡
≡ ≡ ≡ ≡
≡
∈ m
10m am +
−
k
≡
≡ ≡
∈ {
≡
}
≡
m
10a1 + a + a0 . Let P ( P (x) = a x . Then P (10) P (10) = n = n and P (1) P (1) = a , the · · · + 10a =0 =0 sum of the digits of n. Sinc Sincee 1 ≡ 10 (mod 3), P (1) P (1) ≡ P (10) P (10) (mod 3) by Theorem 4.4, k
m
k
k
which implies the statement. (c) Let n be an intege integer. r. There There are m Z, a0 , a1 , . . . , am 0, 1, . . . , 9 such that n = m 10 am + + 10a 10a1 + a + a0 . Sinc Sincee 100 = 4 25, 100 4 25 0 25 0 (mod 4). Therefore
···
∈ ·
∈{ ≡ · ≡ · ≡
12
}
100 10m 0 10m 0 (mod 4) for any m 0; in other words 10k 0 (mod 4), for k = m + m + 2. Therefore Therefore 10m am + + 10a 10a1 + a + a0 10a 10 a1 + a + a0 (mod 4). That That is why n 0 (mod 4) 10a 10 a1 + a + a0 0 (mod 4). Note that 10a 10a1 + a + a0 is the number formed by the tens and units digits of n to obtain the statement. (d) Let n be an integ integer. er. There There are m Z, a0 , a1 , . . . , am 0, 1, . . . , 9 such that n = m k k k 10 am + + 10a 10 a1 + a + a0 . Sinc Sincee 10 = 2 5, 10 2 5 0 (mod 5), if k > 0. 0. He Henc ncee 10m am + + 10a 10a1 + a + a0 a0 (mod 5). The statement is true since a since a 0 0 (mod 5) if and only if a a 0 = 0 or a or a 0 = 5.
·
≡ ·
≡
⇔
···
≡
··· ···
≥ ≡
∈ ·
≡
8. For any integer a, a , show that a 2
≡
∈ { ≡ · ≡
≡
}
≡
− a + 7 ends in one of the digits 3 , 7, or 9.
Proof. By Division Algorithm, Algorithm, a = 10 10k k + r + r,, where k is an integer and r 0, 1, . . . , 9 . 2 2 2 2 Therefore a a + 7 = 10 k + 2 10 r + r + r 10 10k k r + 7, which is congruent r congruent r 2 r + 7 (mod 10). Consider the following cases: in case r case r = 0: a2 a + 7 02 0 + 7 7 (mod 10), in case r case r = 1: a2 a + 7 12 1 + 7 7 (mod 10), in case r case r = 2: a2 a + 7 22 2 + 7 9 (mod 10), in case r case r = 3: a2 a + 7 32 3 + 7 13 3 (mod 10), in case r case r = 4: a2 a + 7 42 4 + 7 19 9 (mod 10), in case r case r = 5: a2 a + 7 52 5 + 7 27 7 (mod 10), in case r case r = 6: a2 a + 7 62 6 + 7 37 7 (mod 10), in case r case r = 7: a2 a + 7 72 7 + 7 49 9 (mod 10), in case r case r = 8: a2 a + 7 82 8 + 7 63 3 (mod 10), and finally in case r = 9: a2 a + 7 92 9 + 7 79 9 (mod 10). So we see that in each each case a2 a + 7 is in one of the congruence classes [3], [3], [7], [7], [9] (mod 10), and therefore ends in 3, 7, or 9.
−
−
· ·
− − − − − − − − − −
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
− − − − − − − − − −
−
≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡
≡ ≡ ≡ ≡ ≡ ≡ ≡
13
−
∈ {
−
}
Section 4.4 1. Solve the following linear congruences: (a) 25x 25x 15 (mod 29). (b) 5x 5x 2 (mod 26). (c) 6x 6x 15 (mod 21). (d) 36x 36x 8 (mod 102). (e) 34x 34x 60 (mod 98). (f) 140x 140x 133 (mod 301).
≡ ≡ ≡ ≡ ≡ ≡
Solution Solution.. (a) By Theorem Theorem 4.7, 4.7, 25 25x x 15 (mod 29) has a solution since gcd(25 gcd(25,, 29) = 1 15. The soluti solution on is [18] (mod 29), since 25 18 450 4 50 15 (mod 29). The solutio solution n is unique by Theorem 4.7. (b) Since gcd(5 gcd(5,, 26) = 1, 5x 5x 2 (mod 26) has a unique solution, namely [20] (mod 26). (c) Since gcd(6 gcd(6,, 21) = 3 15, 6x 6x 15 (mod 21) has three solutions. solutions. By trial-and-erro trial-and-errorr we get [6] (mod 21) as a solution, so by Theorem 4.7 the others are [13] (mod 21) and [20] (mod 21). (d) Since gcd(36 gcd(36,, 102) = 6, which does not divide 8, there are no solutions. (e) Since g Since gcd cd(34 (34,, 98) = 2, 34x 34x 60 (mod 98) has two solutions. By trial-and-error we find [45] (mod 98) (because 34 45 1530 60 (mod 60)), so the other one is [94] (mod 98). (f) Since gcd(140 gcd(140,, 301) = 7 by the hint, which divides 133, there are seven congruence classes which elements solve 140x 140x 133 (mod 301). By trial-and-error, we find [16] (mod 301), because 140 16 2240 133 (mod 301). 301). By Theorem Theorem 4.7, the other other soluti solutions ons are [59] (mod 301), [102] (mod 301), [145] (mod 301), [188] (mod 301), [231] (mod 301), and [274] (mod 301).
≡
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≡
≡
≡
≡
≡ · ≡
· ≡
· ≡
≡
≡
2. Using congruences, solve the following Diophantine equations below: (a) 4x 4x + 51y 51y = 9. (b) 12x 12x + 25y 25y = 331. (c) 5x 5x 53 53yy = 17.
−
Solution. (a) 4x 4x + 51 51yy = 9 4x 9 (mod 51). The solution of this linear congruence is 4 15 [15] (mod 51), and with x with x = = 15, we have y have y = = 9 51 = 1 we have a solution of 4x 4x +51 +51yy = 9. The general solution therefore is x is x = = 15 + 51t, 51t, y = 1 4t, where t where t is any integer. (b) 12x 12x + 25y 25 y = 331 12 12x x 331 6 (mod 25). The solution solution of this linear linear congruen congruence ce 331 12 13 is [13] [13] (mod (mod 25). 25). For x = 13, we have y = = 7, so the gen gener eral al solu solutio tion n is 25 x = 13 + 25t, 25t, y = 7 12 12tt for any integer t integer t.. (c) 5x 5x 53 53yy = 17 5x 17 (mod 53). The solution solution of this linear linear congru congruenc encee is (by 17 5 14 trial-and-e trial-and-error) rror) [14] (mod 53). With x = 14, we get y = = 1, whic which h leads leads to the the 53 general solution x solution x = = 14 + 53t, 53t, y = 1 + 5t 5t, where t where t is an arbitrary integer.
⇔ ≡
⇔
−
≡
− ·
≡
− − −
−
− ⇔ ≡
·
− ·
−
14
4. Solve each of the following sets of simultaneous congruences: (a) x (a) x 1 (mod 3), x 2 (mod 5), x 3 (mod 7). (b) x (b) x 5 (mod 11), x 11), x 14 (mod 29), x 15 (mod 31). (c) x (c) x 5 (mod 6), x 4 (mod 11), x 11), x 3 (mod 17). (d) 2x 2x 1 (mod 5), 3x 3 x 9 (mod 6), 4x 4 x 1 (mod 7), 5x 5 x 9 (mod 11).
≡ ≡ ≡ ≡
≡ ≡ ≡ ≡
≡
≡
≡ ≡
≡
Solution Solution.. (a) By Chines Chinesee Remain Remainder der Theorem Theorem,, we have have to solve solve 35 35x x1 1 (mod 3), 21 21x x2 1 (mod 5), 15x 15x3 1 (mod 7). Trial-a rial-andnd-err error or gives gives us x1 = 2, x2 = 1, x3 = 1, therefore x ¯ = 1 35 2 + 2 21 1 + 3 15 1 = 157 is a solution, so the solution is [52] (mod 105). (b) Solving 899x 899x1 1 (mod 11), 341x 341x2 1 (mod 29), 319x 319x3 1 (mod 31), we get x1 = 7, x2 = 4, x3 = 7, so we have x¯ = 5 899 7+14 341 4+15 319 7 = 31465+19096+ 33495 = 84056, so the solution is [4944] (mod 9889). (c) Solving 187x 187x1 1 (mod 6), 102x 102x2 1 (mod 11), 66x 66x3 1 (mod 17), we get x1 = 1, x2 = 4, x3 = 8, so we have x ¯ = 5 187 1 + 4 102 4 + 3 66 8 = 935+ 163 16322 + 1584 1584 = 4151, 4151, so the general solution is [785] (mod 1122) (d) Solving 462x 462x1 1 (mod 5), 385x 385x2 1 (mod 6), 330x 330x3 1 (mod 7), 210x 210x4 1 (mod 11), we get x get x 1 = 3, x2 = 1, x3 = 1, x4 = 1, considering that x1 must be a multiple of 2, we have x have x 1 = 8, considering that x that x 3 must be a multiple of 4, we have x have x 3 = 8, considering that x4 must be a multiple of 5, we have x have x 4 = 45, so we have x ¯ = 12 1 462 8 + 13 9 385 1 + 14 1 330 8 + 15 9 210 45 = 1848 + 1155 + 660 + 17010 = 20673, so the solution is [2193] (mod 2310).
≡
≡ · ·
· ·
·
·
· ·
≡
≡ · ·
≡
≡ · ·
≡
·
≡
·
·
·
·
· ·
≡
≡
·
≡
≡
≡
·
·
·
·
5. Solve the linear congruence 17x 17 x 3 (mod 2 3 5 7) by solving the system 17x 17 x 17 17x x 3 (mod 3), 17x 17 x 3 (mod 5), 17x 17 x 3 (mod 7).
≡
≡
≡
· · ·
≡
·
·
≡ 3 (mod 2),
Solution. By Chinese Remainder Theorem, we have to solve 105x 105x1 1 (mod 2), 70x 70x2 1 (mod 3), 42x 42x3 1 (mod 5), and 30x 30x1 1 (mod 7). By trial-an trial-and-e d-erro rror, r, one solution solution is x1 = 1, x2 = 1, x3 = 3, x4 = 4. Consid Consideri ering ng that that xi has to be divisible by 17, we get x1 = 17 17,, x2 = 34 34,, x3 = 68 68,, x4 = 102, from which we get x¯ = 3 105 17 + 3 70 34 + 3 42 1 6 8 + 3 30 102 = 17 3(178 3(17855 + 2380 2380 + 2856 2856 + 306 3060) 0) = 1779 1779.. He Henc ncee the solut solution ion is [99] [99] (mod 210).
≡
≡
≡
·
· ·
·
≡
· ·
· ·
6. Find the smallest integer a > 2 such that 2 a, 3 a + 1, 1, 4 a + 2, 2, 5 a + 3, 3, 6 a + 4.
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Solution. Since 4 a +2 implies 2 a and 6 a +4 implies 3 a +1, we do not need these conditions. So we only have the conditions a conditions a 2 (mod 4), a 2 (mod 5), a 5), a 2 (mod 6). Since 4 and 6 are not relatively prime, we cannot use the Chinese Remainder Theorem. But by trial-and-err trial-and-error, or, we get 62 as a solution. This is the smallest smallest solution, because a because a 2 (mod 6) implies that if 2 < 2 < a < 62, a 62, a 8, 14 14,, 20 20,, 26 26,, 32 32,, 38 38,, 44 44,, 50 50,, 56 , but 8 2 (mod 5), 14 2 (mod 5), 20 2 (mod 5), 26 2 (mod 5), 32 2 (mod 4), 38 2 (mod 5), 44 2 (mod 5), 50 2 (mod 5), and 56 2 (mod 5).
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7. (a) Obtain three consecutive integers, each having a square factor. (b) Obtain three consecutive integers, the first of which is divisible by a square, the second by a cube, and the third by a fourth power.
Solution Solution.. (a) The conditi condition on holds holds if a 0 (mod 22 ), a 8 (mod 32 ), and a 23 2 (mod 5 ). Using Using the Chinese Chinese Remainder Remainder Theorem Theorem,, we have have x1 = 1, x2 = 1, x3 = 16 as a solution of 225x 225x1 1 (mod 4), 100x 100x2 1 (mod 9), 36x 36x3 1 (mod 25). 25). Ther Theref efor oree ¯ = 8 100 1 + 23 36 16 = 800 + 13248 = 14048 is a solution, i.e. [548] (mod 900) is the x general solution, therefore 548 = 22 137 137,, 549 = 32 61 61,, 550 = 52 22 are three such integers. (b) Similarly to part (a), we get x get x 1 = 18 18,, x2 = 16 16,, x3 = 11 as a solution of 432x 432x1 1 (mod 25), 400x 400x2 1 (mod 27), 675x 675x3 1 (mod 16). Therefore x ¯ = 26 400 16 + 14 675 11 = 166400+103950 = 270350, and the general the solution is [350] (mod 10800), and the three integers could e.g. be 350 = 52 14 14,, 351 = 33 39 39,, 352 = 24 22.
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8. (Brahmagup (Brahmagupta, ta, 7th century A.D.) When eggs in a basket basket are removed removed 2 , 3, 4, 5, 6 at a time there remain, respectively, 1, 1 , 2, 3, 4, 5 eggs. When they are taken out 7 at a time, none are left over. Find the smallest number of eggs that could have been contained in the basket.
Solution. Solution. Let x be the number off eggs. We know that x 1 (mod 2), x 2 (mod 3), x 3 (mod 4), x 4), x 4 (mod 5), x 5), x 5 (mod 6), x 6), x 0 (mod 7), which is equal to x to x 4 (mod 5), x 5), x 11 (mod 12), x 12), x 0 (mod 7). To solve this, we use the Chinese Remainder Theorem: 84 84x x1 1 (mod 5), 35x 35x2 1 (mod 12), 60x 60x3 1 (mod 7), so x1 = 4, x2 = 11 11,, x3 = 2, thus x ¯ = 4 84 4 + 11 35 11 = 1344 + 4235 = 5579, and the general solution is [119] (mod 420). The smallest number of eggs is 119.
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