Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Solutions to Math 332 Homework 9 2. Direct computation using (5.10) – (5.13) and (5.18) yields 311−1 14 2 7 311 3 × 7−1 2 2 = = (−1) =− = −(−1) = 1 311 311 311 7 7 313−1 313 165 3 5 11 5 × 11−1 2 2 = = 1 × (−1) × (−1) =− = −1 313 313 313 313 11 11 2 1891 −108 −2 × 33 −3 = = = =1 1999 1999 1999 1999 6. (a) (37/43) ≡ 37 43−1 2 ≡ 3721 ≡ (373 )7 ≡ (−1)7 ≡ −1 (mod 43). So (37/43) = −1. (b) Note (19 − 1)/2 = 9. Now 13 × 1, 13 × 2, . . . , 13 × 9 ≡ 13 , 7, 1, 14 , 8, 2, 15 , 9, 3 (mod 19). Since 13, 14, 15 are the only numbers greater than 9, we have N = 3 and (13/19) = (−1)3 = −1. (c) As in Problem 2, 323 −30 −2 3 5 = = = 1 × (−1) × (−1) = 1. 353 353 353 353 353 25. (a) Note that 175 = 52 × 7. x2 ≡ 6 ≡ −1 (mod 7) has no solution since (−1/7) = −1 by (5.11). So x2 ≡ 6 (mod 175) has no solution. (b) Note that 361 = 192 , so x2 ≡ 361 (mod 32 × 7 × 11) is solvable and thus has 23 = 8 solutions by (5.5). (c) x2 ≡ 41 (mod 26 ) has exactly 4 solutions by (4.14)(iii) since 41 ≡ 1 (mod 8). x2 ≡ 41 (mod 53 ) has exactly 2 solutions by (4.13) since (41/5) = (1/5) = 1. x2 ≡ 41 (mod 372 ) has exactly 2 solutions by (4.13) since (41/37) = (4/37) = 1. x2 ≡ 41 (mod 733 ) has exactly 2 solutions by (4.13) since (41/73) = (73/41) = (−9/41) = (−1/41) = 1. Therefore number of solutions of x2 ≡ 41 (mod 26 × 53 × 372 × 733 ) is 4 × 2 × 2 × 2 = 32. 29. (a) x2 ≡ 17 (mod 25 ) has exactly 4 solutions by (4.14)(iii) since 17 ≡ 1 (mod 8). x2 ≡ 17 (mod 132 ) has exactly 2 solutions by (4.13) since (17/13) = (4/13) = 1. x2 ≡ 17 (mod 19) has exactly 2 solutions by (4.13) since (17/19) = (−2/19) = 1. Therefore number of solutions of x2 ≡ 17 (mod 25 × 132 × 19) is 4 × 2 × 2 = 16. (b) Note that 9 = 32 , so x2 ≡ 9 (mod 24 × 53 × 72 ) is solvable and thus has 23+2 = 16 solutions by (5.5). (c) x2 ≡ 57 (mod 27 ) has exactly 4 solutions by (4.14)(iii) since 57 ≡ 1 (mod 8). x2 ≡ 57 (mod 75 ) has exactly 2 solutions by (4.13) since (57/7) = (1/7) = 1. x2 ≡ 57 (mod 592 ) has exactly 2 solutions by (4.13) since (57/59) = (−2/59) = 1. Therefore number of solutions of x2 ≡ 57 (mod 27 × 75 × 592 ) is 4 × 2 × 2 = 16. 35. Check that 79 is prime. (−3/79) = 1 by (5.13)(iii), so x2 ≡ −3 (mod 79) is solvable. Since 79 = 4 × 19 + 3, the solutions are x ≡ ±(−3)19+1 ≡ ±(34 )5 ≡ ±25 ≡ ±32 (mod 79) by 5-3. 38. (a) Check that 263 is prime. (2/263) = 1 by (5.12), so x2 ≡ 2 (mod 263) is solvable. Since 263 = 4 × 65 + 3, the solutions are x ≡ ±265+1 ≡ ±(28 )8 × 22 ≡ ±(−7)8 × 22 ≡ ±110 (mod 263) by 5-3. (b) Check that 83 is prime. (−53/83) = (30/83) = (2/83)(3/83)(5/83) = −1 × 1 × (−1) = 1 by (5.12) and (5.13), so x2 ≡ −53 (mod 83) is solvable. Since 83 = 4 × 20 + 3, the solutions are x ≡ ±(−53)20+1 ≡ ±3021 ≡ ±(303 )7 ≡ ±257 ≡ ±514 ≡ ±(53 )4 ×52 ≡ ±424 ×52 ≡ ±(422 )2 ×52 ≡ ±212 × 52 ≡ ±26 × 25 ≡ ±14 (mod 83) by 5-3. (c) Check that 79 is prime. (20/79) = (22 × 5/79) = (5/79) = 1 by (5.13)(iv), so x2 ≡ 20 (mod 79) is solvable. Since 79 = 4 × 19 + 3, the solutions are x ≡ ±2019+1 ≡ ±2020 ≡ ±(202 )10 ≡ ±510 ≡ ±(53 )3 × 5 ≡ ±353 × 5 ≡ ±62 × 33 × 5 ≡ ±40 (mod 79) by 5-3. 42. (a) 4x2 − 12x + 5 ≡ 0 (mod 77) is equivalent to ( y 2 ≡ 64 (mod 77) 8x ≡ y + 12 (mod 77) by (5.3). y 2 ≡ 64 (mod 77) is in turn equivalent to ( y 2 ≡ 64 (mod 7) y 2 ≡ 64 (mod 11). Since 64 = (±8)2 , y ≡ ±8 ≡ ±1 (mod 7) and y ≡ ±8 ≡ ±3 (mod 11) are easily seen to be solutions of the respective congruences. Solving 11b1 ≡ 1 (mod 7) and 7b2 ≡ 1 (mod 11), we get b1 ≡ 2 (mod 7) and b2 ≡ −3 (mod 11). By the Chinese Remainder Theorem, the solutions for y are y ≡ 11 × 2 × 1 + 7 × (−3) × 3, 11 × 2 × 1 + 7 × (−3) × (−3), 11 × 2 × (−1) + 7 × (−3) × 3, 11 × 2 × (−1) + 7 × (−3) × (−3) ≡ ±8, ±36 (mod 77). (mod 77) Multiply 8x ≡ y + 12 (mod 77) by suitable numbers to reduce coefficient of x to 1: 10 × 8x ≡ 10 × (y + 12) (mod 77) 3x ≡ 10y − 34 (mod 77) 26 × 3x ≡ 26 × (10y − 34) (mod 77) x ≡ 29y + 40 (mod 77). Hence we get x ≡ −36, −38, 6, −3 (mod 77) upon substituting the values of y found earlier. (b) 2x2 − x + 7 ≡ 0 (mod 91) is equivalent to ( y 2 ≡ 36 (mod 91) 4x ≡ y + 1 (mod 91) by (5.3). y 2 ≡ 36 (mod 91) is in turn equivalent to ( y 2 ≡ 36 (mod 7) y 2 ≡ 36 (mod 13). Since 36 = (±6)2 , y ≡ ±6 ≡ ±1 (mod 7) and y ≡ ±6 (mod 13) are easily seen to be solutions of the respective congruences. Solving 13b1 ≡ 1 (mod 7) and 7b2 ≡ 1 (mod 13), we get b1 ≡ −1 (mod 7) and b2 ≡ 2 (mod 13). By the Chinese Remainder Theorem, the solutions for y are y ≡ 13 × (−1) × 1 + 7 × 2 × 6, 13 × (−1) × 1 + 7 × 2 × (−6), 13 × (−1) × (−1) + 7 × 2 × 6, 13 × (−1) × (−1) + 7 × 2 × (−6) ≡ ±6, ±20 (mod 91). (mod 91) Multiply 4x ≡ y + 1 (mod 91) by suitable numbers to reduce coefficient of x to 1: 23 × 4x ≡ 23 × (y + 1) (mod 91) x ≡ 23 × (y + 1) (mod 91) Hence we get x ≡ −21, −24, 28, 18 (mod 91) upon substituting the values of y found earlier. 43. 7x2 − x + 24 ≡ 0 (mod 36) is equivalent to ( 7x2 − x + 24 ≡ 0 7x2 − x + 24 ≡ 0 (mod 4) (mod 9). Upon reducing the coefficients, we get ( x2 + x ≡ 0 (mod 4) 2x2 + x + 3 ≡ 0 (mod 9). Observe that =⇒ x ≡ 0, 1, 2, 3 (mod 4) x + x ≡ 0, 2, 2, 0 (mod 4). 2 Hence the solutions to x2 + x ≡ 0 (mod 4) are x ≡ 0, 3 (mod 4). Observe that =⇒ x ≡ 0, 1, 2, 3, 4, 5, 6, 7, 8 (mod 9) 2x + x + 3 ≡ 3, 6, 4, 6, 3, 4, 0, 0, 4 (mod 9). 2 Hence the solutions to 2x2 + x + 3 ≡ 0 (mod 9) are x ≡ 6, 7 (mod 9). Solving 9b1 ≡ 1 (mod 4) and 4b2 ≡ 1 (mod 9), we get b1 ≡ 1 (mod 4) and b2 ≡ −2 (mod 9). By the Chinese Remainder Theorem, the solutions to 7x2 − x + 24 ≡ 0 (mod 36) are y ≡ 9 × 1 × 0 + 4 × (−2) × 6, 9 × 1 × 0 + 4 × (−2) × 7, 9 × 1 × 3 + 4 × (−2) × 6, ≡ −12, 16, 15, 7 (mod 36). 9 × 1 × 3 + 4 × (−2) × 7 (mod 36)